Problem & Solutions on Probability & Statistics · Problem & Solutions on Probability & Statistics Problem Set-1 [1] A coin is tossed until for the first time the same result appear

Problem & Solutions on Probability & Statistics

Problem Set-1

[1] A coin is tossed until for the first time the same result appear twice in succession.

To an outcome requiring n tosses assign a probability2−𝑛 . Describe the sample space. Evaluate the

probability of the following events:

(a) A= The experiment ends before the 6th toss.

(b) B= An even number of tosses are required.

(c) A∩ B, 𝐴𝑐 ∩ 𝐵

[2] Three tickets are drawn randomly without replacement from a set of tickets numbered 1 to 100. Show

that the probability that the number of selected tickets are in (i) arithmetic progression is 1/ 66 and (ii)

geometric progression is 105

1003

.

[3] Three players A, B and C play a series of games; none of which can be drawn and their probability of

wining any game are equal. The winner of each game scores 1 point and the series is won by the player

who first scores 4 points. Out of the first three games A won 2 games and B won 1 game. Find the

probability that C will win the series.

[4] A point P is randomly placed in a square with side of 1 cm. Find the probability that the distance from

P to the nearest side does not exceed x cm.

[5] Let there be n people in a room and p denote the probability that there are no common birth days. Find

an approximate value of p for n= 10.

[6] Suppose a lift has 3 occupants A, B and C and there are three possible floors (1, 2 and 3) on which

they can get out. Assuming that each person acts independently of the others and that each person has an

equally likely chance of getting off at each floor, calculate the probability that exactly one person will get

out on each floor.

[7] If n men, among whom are A and B, stand in a row, what is the probability that there will be exactly r

men between A and B ?

[8] In a town of n+ 1 inhabitants, a person tells a rumor to a second persons, who in trun tells it to a third

persons, and so on. At each step the recipient of the rumor is chosen at random from the n people

available. Find the probability that the rumor will be told r times without

(a) returning to the originator,

(b) being repeated to any persons.

Do the same problem when at each step the rumor is told to a gathering of N randomly chosen people.

[9] 2 points are taken at random and independently of each other on a line segment of length m. Find the

probability that the distance between 2 points is less than m/3.

[10] n points are taken at random and independently of one another inside a sphere of radius R. What is

the probability that the distance from the centre of the sphere to the nearest point is not less than r ?

[11] A car is parked among N cars in a row, not at either end. On his return, the owner finds that exactly r

of the N places are still occupied. What is the probability that both neighboring places are empty?

[12] 3 points X, Y, Z are taken at random and independently of each other on a line segment AB. What is

the probability that Y will be between X and Z?

[13] The coefficients of the equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 are determined by throwing an ordinary die. Find

the probability that the framed equation will have real roots.

[14] Let 𝛺= {1, 2, 3, 4}. Check whether any of the following is a 𝜍-field of subsets of 𝛺

ℱ1 = 𝜙, 1, 2 , 3, 4

ℱ2 = {𝜙, 𝛺, 1 , 2, 3, 4 , {3, 4}}

ℱ3 = {𝜙, 𝛺, 1 , 2 , 1, 2 , 3, 4 , 2, 3, 4 , {1, 3, 4} }

[15] Prove that if 𝐹1 𝑎𝑛𝑑 𝐹2 are 𝜍-fields of subsets of 𝛺, then 𝐹1 ∩ 𝐹2 is also a 𝜍-field. Give a counter

example to show that similar result for union 𝜍- fields does not hold.

[16] Let F be a 𝜍- field of subsets of the sample space 𝛺 and let 𝐴 ∊ 𝐹 be a fixed. Show that 𝐹𝐴 = {𝐶: 𝐶 =

𝐴 ∩ 𝐵, 𝐵 ∊ 𝐹} is a 𝜍-field of subsets of A.

Solution Set-1

1) 𝛺= {HH, TT, HTT, THH, HTHH, THTT, ….}

𝑃 𝐻𝐻 =1

4= 𝑃 𝑇𝑇

𝑃 𝐻𝑇𝑇 = 𝑃 𝑇𝐻𝐻 =1

23.

a) 𝑃 𝐴 = 𝑃 exp 𝑒𝑛𝑑𝑠 𝑖𝑛 𝑖 𝑡𝑜𝑠𝑠𝑒𝑠 5𝑖=2

= 𝑃 exp 𝑒𝑛𝑑𝑠 𝑖𝑛 2 𝑡𝑜𝑠𝑠𝑒𝑠 + 𝑃 𝑒𝑛𝑑𝑠 𝑖𝑛 3 + 𝑃 𝑒𝑛𝑑𝑠 𝑖𝑛 4 + 𝑃 … 5

= 2 ×1

22+ 2 ×

1

23+ ⋯

b) P(B)= 2 1

22𝑖∞𝑖=1 = ⋯

c) P (A∩ B)= P(exp ends in 2 tosses) + P(exp ends in 4 tosses)

= ….

𝑃 𝐴𝑐 ∩ 𝐵 = 2 1

22𝑖∞𝑖=3 =…..

2) Total # of cases : 1003

(i) No. in AP Common diff 1, 2, .... , 49

# of cases 98, 96, …, 2

⟹Total # of favorable cases 98+ 96 …+ 2

= 2 49×50

2 = 49 × 50

Reqd prob= 49×50

1003

=1

66

1) No in GP. Common ratio can be integer or fraction

Case 1: C. V. integer

c. r. # f tav cases total #

2 ⟶ (1, 2, 4), ……….. (25, 50, 100) ⟶ 25

3 ⟶ (1, 3, 9), …. (11, 33, 99) ⟶ 11

4 ⟶(1, 4, 6), …. (6, 24, 96) ⟶ 6

5 ⟶(1, 5, 25), …. (4, 20, 100) ⟶ 4

6 ⟶ (1, 6, 36), (2, 12, 72) ⟶ 2

7 ⟶ (1, 7, 49), (2, 14, 98) ⟶ 2

8 ⟶ (1, 8, 64), ⟶ 1

9 ⟶ (1, 9, 81) ⟶ 1

10 ⟶ (1, 10, 100) ⟶ 1

_______________

Total 53

Case 2: c. r. fractional

1st # c. r. tav cases total #

1 3

2⟶ 4, 6, 9 , 8, 12, 18 , … . 44, 66, 99 ⟶ 11

5

2⟶ 4, 10, 25 , 8, 20, 50 , (12, 30, 75)(16, 40, 100) ⟶ 4

7

2⟶ 4, 14, 49 , 8, 28, 98 ⟶ 2

9

2⟶ (4, 18, 81) ⟶ 1

F

9 ⟶ 4

3,

5

3,

7

3,

8

3,

10

3 ⟶ 6 + 4 + 2 + 1 + 1

6 ⟶ 5

4,

7

4,

9

4 ⟶ 4 + 2 + 1

5 ⟶ 6

5,

7

5,

8

5,

9

5 ⟶ 2 + 2 + 1 + 1

6 ⟶ 7

6 ⟶ 2

19 ⟶ 8

7,

9

7,

10

7 ⟶ 1 + 1 + 1

4 ⟶ 9

8 ⟶ 1

:1 ⟶ 10

9 ⟶ 1

__________________________________

Total 52

⟹ reqd prob. 53+52

1003

.

3)C can win in exactly 4 or 5 or 6 or 7 additional games

Case 1: 4 additional games

C wins all ⟶ prob 1

3

4 . ___________(i)

Case 2: 5 additional games.

C wins 3 out of 1st 4 & the 5

th game prob ⟶ 4

3

1

3

3

2

3 ×

1

3 _________(ii)

Case 3: 6 additional games

C wins 3 out of 1st 5 & 6

th game and either (i) B wins 2 and A wins on one or (ii) B wins 1, A wins 1

Prob ⟶ 53

1

3

3

1

3

2×

1

3+ 5

3 2

1

1

3

3 1

3.

1

3×

1

3 ___________(iii)

Case 4 : 7 additional games

Out of 1st 6 games

𝐴 𝑤𝑖𝑛𝑠 1𝐵 …… . .2𝐶 ……… .3

𝑝𝑟𝑜𝑏 63 3

1

1

3

3 1

3

1

3

2

1

3 __________ (iv)

Reqd. prob = (i) + (ii) + (iii) + (iv) ← m. e. ways.

(4) The pt P must lie in the shaded region so that the distance from P to the nearest side does not exceed x

cm.

If 𝑥 ≥1

2, 𝑡𝑕𝑒𝑛 𝑝𝑟𝑜𝑏 = 1

𝐼𝑓 0 < 𝑥 <1

2, 𝑡𝑕𝑒𝑛 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑕𝑒 𝑠𝑕𝑎𝑑𝑒𝑑 𝑟𝑒𝑔𝑖𝑜𝑛 = 1 − (1 − 2𝑥)2

⟹ reqd. prob.= 1-(1 − 2𝑥)2.

5) Reqd prob.= 365×364×…×(365−(𝑛−1))

365𝑛 = 1 −1

365 1 −

2

365 … 1 −

𝑛−1

365 = 𝑝 𝑠𝑎𝑦

log𝑒 𝑝 = log𝑒 1 −𝑘

365

𝑛−1

𝑘=1

≈ −𝑘

365

𝑛−1

𝑘=1

= −1

365.𝑛(𝑛 − 1)

2

For n= 10 log𝑒 𝑝 ≈ −1

365

10×9

2= ⋯

⟹ 𝑝 ≈ ⋯

6) Total # of possible outcomes: 33

Tavarable # of outcomes : 3! = 6

Reqd. prob.= 6

27.

7) Total # of ways in which n men can stand in a row ⟶ n!

# of possible positions for A & B ∋ there are exactly r positions available between them

= 2! × (n – r - 1)

↗ ↖

permutation possible positions

among A&B ({1, r+ 2}, {2, r+ 3}, …., {(n- r-1), n} for A&B)

= Further # of ways that r persons can be chosen to stand between A& B= 𝑛−2𝑟

Favarable # of cases

2! × 𝑛 − 𝑟 − 1 × 𝑛−2𝑟

× 𝑟! × 𝑛 − 𝑟 − 2 !

↙ ↘

Perm of r perm of (n- r- 2) men excluding A< B and r men in

Men betn A & B between

⟹ reqd. prob.

2! × 𝑛 − 𝑟 − 1 ! × 𝑟! × 𝑛−2𝑟

𝑛!

8)

(a) Total # of of ways 𝑛𝑟

𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑡𝑜𝑟 ⟶ 𝑛 𝑤𝑎𝑦𝑠

2𝑛𝑑 𝑝𝑒𝑟𝑠𝑜𝑛 ⟶ 𝑛 − 1 𝑤𝑎𝑦𝑠⋮

𝑟𝑡𝑕 𝑝𝑒𝑟𝑠𝑜𝑛 ⟶ 𝑛 − 1 𝑤𝑎𝑦𝑠

} ⟶ 𝑛(𝑛 − 1)𝑟−1

Reqd. prob.= 𝑛 𝑛−1 𝑟−1

𝑛𝑟

(b) Originator ⟶ n options

2nd

person ⟶ n- 1

3rd

person ⟶ n- 2

⋮

𝑟𝑡𝑕 𝑝𝑒𝑟𝑠𝑜𝑛 ⟶ (n- r+ 1)

Reqd. prob. = 𝑛 𝑛−1 …(𝑛−𝑟+1)

𝑛𝑟

Second part:- Total # of cases 𝑛𝑁 𝑟

.

Case tavarable to 1st event 𝑛

𝑁 𝑛−1

𝑁 𝑟−1

𝑅𝑒𝑞𝑑 𝑝𝑟𝑜𝑏 = 𝑛𝑁 𝑛−1

𝑁 𝑟−1

𝑛𝑁 𝑟

Apply for 2nd

event

𝑅𝑒𝑞𝑑 𝑝𝑟𝑜𝑏 = 𝑛𝑁 𝑛−𝑁

𝑁 𝑛−2𝑁𝑁 … 𝑛−(𝑟−1)𝑁

𝑁

𝑛𝑁

𝑟 ↗

𝑤𝑖𝑡 𝑕 𝑜𝑏𝑣𝑖𝑜𝑛𝑠 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛

9) Let the distance of 2 randomly chosen pts from a fixed pt A on the line segment be denoted as x& y

Reqd. condition is |x- y|< 𝑚

3.

𝑖. 𝑒. −𝑚

3< 𝑥 − 𝑦 <

𝑚

3

Note that x, y ∊ [0, m]

Inside the rect bounded by x axis, y axis, x= m and y= m, the area favorable to |x- y|< 𝑚

3 is clearly the

region OABCDE

Area of OABCDE = 𝑚2 − 2

3

2𝑚2

= 𝑚2 − 2 1

2

2𝑚

3×

2𝑚

3

⟹ 𝑟𝑒𝑞𝑑 𝑝𝑟𝑜𝑏 = 𝑚2−

4

9𝑚2

𝑚2 =5

9.

10) n pt must lie on or outside a sphere of raidus r, having same centre as the original sphere of raidus R.

For any of the n pts,

P(lie inside the smaller sphere)= 𝑣𝑜𝑙𝑚 𝑜𝑓 𝑠𝑝𝑕 𝑤𝑖𝑡 𝑕𝑟𝑎

𝑣𝑜𝑙𝑚 𝑜𝑓 𝑠𝑝𝑕 𝑤𝑖𝑡 𝑕𝑟𝑎

𝑟

𝑅=

𝑟3

𝑅3 .

⟹ P(lie on or outside the smaller sphere)= 1 −𝑟3

𝑅3

As the pts are taken independently, the reqd. prob.= 1 −𝑟3

𝑅3 𝑛

.

11) Owner’s car can be in any of the (N- 2) places (leaving 2 ends)

Remaining (r- 1) cars in (N- 1) remaining place

⟹ Total # of cases= (N- 2) 𝑁−1𝑟−1

Favorable # of cases:

Owner’s car in any of the (N- 2) places and 2 neighboring places are empty

⟹remaining (r-1) cars can be in (N- 3) remaining places.

⟹favorable # of cases: (N- 2) 𝑁−3𝑟−1

.

Reqd. prob. = 𝑁−3

𝑟−1

𝑁−1𝑟−1

.

12) Let x, y, 3 be the distances of X, Y, Z from a fixed pt P on the line segment the six possibilities are

x < y < 3 ; x < 3 < y ; 3 < x < y;

y < x < 3 ; y < 3 < x ; 3 < y < x;

The above 6 possibilities are equally likely due to random draws

Y lie between X & Z in 2 cases (x < y < 3 & 3 < y < x)

Reqd. prob. =2

6

13) Coefficient a, b or c can take any of the values 1, 2, …

Total # of (a, b, c) combinations 6× 6 × 6 = 216

Real roots ⟶ requirement 𝑏2 ≥ 4𝑎𝑐

Listing of favorable # of cases

ac (a , c) 4ac 𝑏⃝ ∋ 𝑏2 ≥ 4𝑎𝑐 # of cases

1 (1 , 1) 4 2, 3, 4, 5, 6 5

2 (1, 2)(2, 1)

⟶ 8 3, 4, 5, 6 2× 4 = 8

3 (1 , 3)(3, 1)

⟶ 12 4, 5, 6 2× 3 = 6

4

(1, 4) 4, 1 (2, 2)

⟶ 16 4, 5, 6 3× 3 = 9

5 (1, 5)(5, 1)

⟶ 20 5, 6 2× 2 = 4

6

(1, 6)(6, 1)(2, 3)(3, 2)

⟶ 24 5 ; 6 4× 2 = 8

____ not possible to obtain ac= 7

(2, 4)(4, 2)

⟶ 32 6 2× 1 = 2

ac values higher than 9 will not have any b∋ 𝑏2 ≥ 4𝑎𝑐

⟹ # of favorable case for 𝑏2 ≥ 4𝑎𝑐 = 5 + 8 + 6 + 9 + 4 + 8 + 2 + 1 = 43

𝑟𝑒𝑞𝑑 𝑝𝑟𝑜𝑏 =43

216.

14) (i) 𝜙𝑐 = 𝛺 ∈ ℱ1 ⟹ ℱ1 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝜍 − 𝑓𝑖𝑒𝑙𝑑

(ii) {1} ∪ {3, 4} = {1, 3, 4} ∉ℱ2

ℱ2 is not closed under union ⟹ℱ2 is not a 𝜍-field

Or {1, 2}∩{2, 3, 4}= {2} ∈ℱ2

⟹ℱ2 is not 𝜍- field.

(𝑖𝑖𝑖)ℱ3 contain 𝛺 and is closed under complementation and union ⟹ℱ3 is a 𝜍-field.

15) 𝛺∈ ℱ1, ℱ2 ⟹ 𝛺 ∈ ℱ1 ∩ ℱ2______(i)

Let A ∈ ℱ1 ∩ ℱ2 , 𝑡𝑕𝑒𝑛 𝐴 ∈ ℱ1 & 𝐴 ∈ ℱ2

⟹ 𝐴𝑐 ∈ ℱ1& 𝐴𝑐 ∈ ℱ2

⟹ 𝐴𝑐 ∈ ℱ1 ∩ ℱ2______(ii)

If 𝐴1 , 𝐴2 , … ∈ ℱ1 ∩ ℱ2 , 𝑡𝑕𝑒𝑛

𝐴1 , 𝐴2 , … ∈ ℱ1 & ⟹∪ 𝐴𝑖 ∈ ℱ1

𝐴1 , 𝐴2 , … ∈ ℱ2 & ⟹∪ 𝐴𝑖 ∈ ℱ2

⟹ ∪ 𝐴𝑖 ∈ ℱ1 ∩ ℱ2_________(iii)

(i), (ii) & (iii) ⟹ ℱ1 ∩ ℱ2 is a 𝜍-field.

Counter example

𝛺= {1, 2, 3}

i.e. ℱ1 = 𝜙, 𝛺, 1 , 2, 3 ⟶ 𝜍 − 𝑓𝑖𝑒𝑙𝑑

ℱ2 = 𝜙, 𝛺, 2 , 1, 3 ⟶ 𝜍 − 𝑓𝑖𝑒𝑙𝑑

ℱ1 ∪ ℱ2 = {𝜙, 𝛺, 1 , 2 , 1, 3 , {2, 3}}

ℱ1 ∪ ℱ2 is not a 𝜍-field ({1}∪{2}∉ ℱ1 ∪ ℱ2)

16)(i) A∈ ℱ & A ∩ A= A ⟹A∈ℱ𝐴.

(ii) Let C ∈ℱ𝐴 𝑡𝑕𝑒𝑛 𝐶 = 𝐴 ∩ 𝐵 𝑓𝑜𝑟 𝐵 ∈ ℱ

𝐶𝐶𝐴 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 𝑤. 𝑟. 𝑡. 𝐴 = 𝐴 − 𝐶

= 𝐴 − 𝐴 ∩ 𝐵

= 𝐴 ∩ 𝐵𝐶 ∈ ℱ𝐴 (𝑎𝑠 𝐵𝐶 ∈ ℱ𝐴 )

(iii)Let 𝐶1, 𝐶2 , … ∈ ℱ𝐴 , 𝑡𝑕𝑒𝑛

𝐶𝑖 = 𝐴 ∩ 𝐵𝑖 , 𝑖 = 1, 2, … . 𝑓𝑜𝑟 𝐵𝑖 ∈ ℱ

∪ 𝐶𝑖 = 𝑈

𝑖 𝐴 ∩ 𝐵𝑖 = 𝐴 ∩

𝑈

𝑖𝐵𝑖 ∈ ℱ𝐴 𝑎𝑠 𝑈𝐵𝑖 ∈ ℱ

⟹ ℱ𝐴 𝑖𝑠 𝑎 𝜍 − 𝑓𝑖𝑒𝑙𝑑 𝑜𝑓 𝑠𝑢𝑏𝑠𝑒𝑡𝑠 𝑜𝑓 𝐴.

Problem Set -2

[1] Let 𝛺= {0, 1, 2, …}. If for an event A,

(a) P(A)= 𝑒−𝜆𝜆𝑥

𝑥 !, 𝜆 > 0.𝑥∈𝐴

(b) P(A) = 𝑝(1 − 𝑝)𝑥𝑥∈𝐴 , 0 < 𝑝 < 1.

(c) P(A)= 1 𝑖𝑓 𝑡𝑕𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝐴 𝑖𝑠 𝑓𝑖𝑛𝑖𝑡𝑒

0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒.

Determine in each of the above cases whether P is a probability measure.

In case where your answer is in the affirmative, determine P(E), P(F), P(G), P(E∩F), P(E ∪F), P(F∪G),

P(E∩G) and P(F∩G) , where E= {x ∈𝛺 : x > 2},

F= {x ∈ 𝛺 : 0< x < 3} and G= {x ∈ 𝛺: 3 < x < 6}.

[2] Let 𝛺= ℜ. In each of the following cases determine whether P(.) is a probability measure. For an

interval I,

(a) P(I)= ∫1

2𝑒|𝑥|

𝐼𝑑𝑥

(b) P(I)= 0, 𝑖𝑓 𝐼 ⊂ −∞, 0 ,

∫ 2𝑥𝑒−𝑥2𝑑𝑥

𝐼, 𝑖𝑓 𝐼 ⊂ 0,∞ .

(c) P(I)= 1 𝑖𝑓 𝑙𝑒𝑛𝑔𝑡𝑕 𝑜𝑓 𝐼 𝑖𝑠 𝑓𝑖𝑛𝑖𝑡𝑒

0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒

[3] Show that the probability of exactly one of the events A or B occurring is P(A)+ P(A)- 2P(A ∩B).

[4] Prove that

P(A ∩B)- P(A) P(B)= P(A)P(𝐵𝑐)- P(A∩ 𝐵𝑐)

= 𝑃 𝐴𝑐 𝑃 𝐵 − 𝑃 𝐴𝑐 ∩ 𝐵

= 𝑃 𝐴 ∪ 𝐵𝑐 − 𝑃(𝐴𝑐)𝑃(𝐵𝑐)

[5] For events 𝐴1 , 𝐴2 , … . . , 𝐴𝑛 𝑠𝑕𝑜𝑤 𝑡𝑕𝑎𝑡 𝑃 𝐴𝑖𝑛𝑖=1 = 𝑃 𝐴𝑖

𝑛𝑖=1 − 𝑃(𝐴𝑖1

∩ 𝐴𝑖2∩ 𝐴𝑖3

)1≤𝑖1<𝑖2≤𝑛 −

⋯ + −1 𝑛−1𝑃( 𝐴𝑖𝑛𝑖=1 )

[6] Consider the sample space 𝛺= {, 1, 2, ….}and ℱ the 𝜍-field of subsets of 𝛺. To the elementary event

{j} assign the probability

P({j})= 𝐶2𝑗

𝑗 !, 𝑗 = 0, 1, 2, … ..

(a) Determine the constant c.

(b) Define the events A, B and C by A= {j :2≤j ≤ 4}, B= {j : j ≥ 3} and C= {j : j is an odd integer}.

Evaluate P(A), P(B), P(C), P(A∩B), P(A∩C), P(C∩B), P(A∩B∩C) and verify the formula for P(A∪

B∪C).

[7] Each packet of a certain cereal contains a small plastic model of one of the five different dinosaurs; a

given packet is equally likely to contain any one of the five dinosaurs. Find the probability that someone

buying six packets of the cereal will acquire models of three favorite dinosaurs.

[8] Suppose n cards numbered 1, 2, …, n are laid out at random in a row. Let 𝐴𝑖 denote the event that

‘card I appears in the ith position of the row’, which is termed as a match. What is the probability of

obtaining at least one match?

[9] A man addresses n envelopes and writes n cheques for payment of n bills.

(a) If the n bills are placed at random in the n envelopes, what would be the probability that eaxch bill

would be placed in the wrong envelope?

(b) If the bills and n cheques are placed at random in the n envelopes, one bill and one cheque in each

envelope, what would be the probability that in no instance would the enclosures be completely correct?

[10] For events A, B and C such that P(C) > 0, prove that

(a) P(A∪B| c)= P(A|C)+ P(B|C)- P(AB|C)

(b) P(𝐴𝐶|𝐶)= 1- P(A|C).

[11] Let A and B be two events such that 0 < P(A) < 1. Which of the following statements are true?

(a) P(A|B) + P(𝐴𝐶|𝐵)= 1; (b) P(A|B) + P(A|𝐵𝐶)= 1; (c) P(A|B)+ P(𝐴𝐶| 𝐵𝐶)= 1

[12] Consider the two events A and B such that P(A)= 1

4, P(B|A)= ½ and P(A|B)=

1

4.

Which of the following statements are true?

(a) A and B are mutually exclusive events,

(b) A ⊂ B,

(c) P(𝐴𝐶 𝐵𝐶 =3

4,

(d) P(A|B) + P(A| 𝐵𝐶)= 1

[13] Consider an urn in which 4 balls have been placed by the following scheme. A fair coin is tossed, if

the coin comes up heads, a white ball is placed in the urn otherwise a red ball is placed in the urn.

(a) What is the probability that the urn will contain exactly 3 white balls?

(b) What is the probability that the urn will contain exactly 3 white balls, given that the first ball placed in

the urn was white?

[14] A random experiment has three possible outcomes, A, B and C, with probabilities𝑝𝐴 , 𝑝𝐵 , 𝑎𝑛𝑑 𝑝𝐶 .

What is the probability that, in independent performances of the experiment, A will occur before B?

[15] a system composed of n separate components is said to be a parallel system if it functions when at

least one of the components functions. For such a system, if component I, independent of other

components, functions with probability𝑝𝑖 , 𝑖 = 1(1)𝑛, what is the probability that the system functions?

[16] A student has to sit for an examination consisting of 3 questions selected randomly from a list of 100

questions. To pass, the student needs to answer correctly all the three questions. What is the probability

that the student will pass the examination if he remembers correctly answer to 90 questions o the list?

[17] A person has three coins in his pocket, two fair coins (heads and tails are equally likely) but the third

one is baised with probability of heads 2/3. One coin selected at random drops on the floor, landing heads

up. How likely is it that it is one of the fair coins?

[18] A slip of paper is given to A, who marks it with either a+ or a- sign, with a probability 1/3 of writing

a+ sign. A passes the slip to B, who may either leave it unchanged or change the sign before passing it to

C. C in turn passes the slip to D after perhaps changing the sign; finally D passes it to a referee after

perhaps changing the sign. It is further known that B, C and D each change the sign with probability 2/3.

Find the probability that A originally wrote a+ given that the referee sees a+ sign o the slip.

[19] Each of the three boxes A, B, and C, identical in appearance, has two drawers. Box A contains a gold

coin in each drawer, Box B contains a silver coin in each drawer and box C contains a gold coi in one

drawer and silver coin in the other. A box is chosen at random and one of its drawers is then chosen at

random and opened, and a gold coin is found. What is the probability that the other drawer of this box

contains a silver coin?

[20] Each of four persons fires one shot at a target. Let 𝐶𝑘 denote the event that the target is hit by person

k, k= 1, 2, 3, 4. If the events 𝐶1, 𝐶2 , 𝐶3 , 𝐶4 are independent and if P(𝐶1)= P(𝐶2)= 0.7,

P(𝐶3)= 0.9 and P(𝐶4)= 0.4, compute the probability that : (a) all of them hit the target; (b) no one hits the

target; (c) exactly one hits the target; (d) at least one hits the target.

[21] Let 𝐴1 , 𝐴2 , … . . , 𝐴𝑛 be n independent events. Show that 𝑃 𝐴𝑖𝐶𝑛

𝑖=1 ≤ exp − 𝑃 𝐴𝑖 𝑛𝑖=1 .

[22] Give a counter example to show that pair wise independence of a set of events 𝐴1 , 𝐴2 , … . . , 𝐴𝑛 does

not imply mutual independence.

[23] We say that B carries negative information about event A if P(A|B) < P(A). Let A, B and C be three

events such that B carries negative information about A and C carries negative information about B. Is it

true C carries negative information about A ? Prove your assertion.

[24] Suppose in a class there are 5 boys and 3 girl students. A list of 4 students, to be interviewed, is made

by choosing 4 students at random from this class. If the first student selected at random from the list, for

interview, is a girl, then find the conditional probability of selecting a boy next from among the remaining

3 students in the list.

[25] During the course of an experiment with a particular brand of a disinfectant on files, it is found that

80% are killed in the first application. Those which survive develop a resistance, so that the percentage of

survivors killed in any later application is half of that in the preceding application. Find the probability

that (a) a fly will survive 4 applications; (b) it will survive 4 applications, given that it has survived the 1st

one.

[26] An art dealer receives a group of 5 old paintings and on the basis of past experience, the thinks that

the probabilities are, 0.76, 0.09, 0.02, 0.01, 0.02 and 0.10 that 0, 1, 2, 3, 4 or all 5 of them, respectively,

are forgeries. The art dealer sends one randomly chosen (out of 5) paintings for authentication. If this

painting turns out to be a forgery, then what probability should he now assign to the possibility that the

other 4 are also forgeries?

Solution Key

(1) (a) 𝛺= {0, 1, 2, ….}

P(A)= 𝑒−𝜆𝜆𝑥

𝑥!, 𝜆 > 0.𝑥∈𝐴

P(A)≥ 0 obv ____(i)

Define 𝐴𝑖 = 𝑥 ∈ 𝛺: 𝑖 − 1 < 𝑥 < 𝑖 + 1 = 𝑖

𝑃 𝐴𝑖∞1 = 𝑃 1, 2, … = 1 − 𝑃 𝑋 = 0

𝑃 𝐴𝑖 ∞1 = 𝑃 𝐴1 + ⋯ = 1 − 𝑃 𝑋 = 0

𝑃 𝐴𝑖∞1 = 𝑃 𝐴𝑖

∞1 _________(iii)

⟹ P is prob. Measure.

(b)Similar to (a).

(c) P(A)≥ 0

P(𝛺)= 0 (∵𝛺 is infinite).

𝜕𝐴𝑖 ∩ 𝐴𝑗 = 𝜙 𝑖 ∀𝑖 = 𝑗

𝑃 𝐴𝑖

𝑖

= 𝑃 𝑥

𝑥∈𝑈𝐴𝑖

= 𝑃({𝑥})

𝑥∈𝐴𝑖𝑖

= 𝑃(𝐴𝑖)

𝑖

P(𝛺)= 𝑒−𝜆𝜆𝑥

𝑥!𝑥∈𝛺 = 𝑒−𝜆 𝜆𝑥

𝑥!∞0 = 1 _______(i)

In general, 𝐴1 , 𝐴2, ….

P is not prob. Measure.

Note:- Also take, 𝐴𝑖 = {𝑖} i= 1, 2, …

Then 𝑈𝐴𝑖 = {1, 2, … } -infinite # of elements

𝑃 𝐶𝑖

∞

1

= 0 ≠ 𝑃 𝐶𝑖

∞

1

= 1

∞

1

2nd

part

(a) P(E)= 𝑒−𝜆𝜆𝑥

𝑥 !∞3 = 1 − 𝑒−𝜆 1 + 𝜆 +

𝜆2

2!

𝑃 𝐹 = 𝑒−𝜆𝜆𝑥

𝑥!

2

1

= 𝑒−𝜆 𝜆 +𝜆2

2!

𝑃 𝐸 ∪ 𝐹 = 𝑒−𝜆𝜆𝑥

𝑥!

∞

1

= 1 −𝑒−𝜆𝜆0

0!= 1 − 𝑒−𝜆

𝑜𝑡𝑕𝑒𝑟𝑠.

( 2) 𝛺= R

𝑎 𝑃 𝐼 = ∫1

2𝑒|𝑥|

𝐼𝑑𝑥 ≥ 0 ∀ 𝐼

P(𝛺)= ½ ∫ 𝑒|𝑥|∞

−∞𝑑𝑥 =

1

2 ∫ 𝑒𝑥0

−∞𝑑𝑥 + ∫ 𝑒−𝑥∞

0𝑑𝑥 = 1

𝐼1 ∩ 𝐼2 = 𝜙 ∶ 𝑃 𝐼1 ∪ 𝐼2 =1

2 ∫ + ∫𝐼2𝐼1

= 𝑃 𝐼1 + 𝑃 𝐼2 𝑒𝑥𝑡𝑒𝑛𝑑𝑒𝑑 …. P is prob measure

𝑏 𝑆𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑜 (𝑎)

𝑐 P(𝛺)= P(ℜ)= 0 ≠ 1 P is not prob measure.

(3) P(exactly me of A or B)

= 𝑃( 𝐴 ∩ 𝐵𝑐 ∪ 𝐴𝑐 ∩ 𝐵

= 𝑃 𝐴 + 𝑃 𝐵 − 2𝑃(𝐴 ∩ 𝐵) ___ on simplification.

𝑃 𝐴𝐵 − 𝑃 𝐴 𝑃 𝐵 = 𝑃 𝐴 𝑃 𝐵𝑐 − 𝑃(𝐴𝐵𝑐). _____ 1st equation.

[𝑈𝑠𝑖𝑛𝑔 𝑃 𝐴 = 𝑃 𝐴𝐵 + 𝑃 𝐴𝐵𝑐 . ]

2nd

& 3rd

equation is can be proved in a similar way.

(5) For n = 2

𝑃 𝐴1 ∪ 𝐴2 = 𝑃 𝐴1 ∪ 𝐴1𝑐𝐴2

= 𝑃 𝐴1 + 𝑃 𝐴1𝑐𝐴2

= 𝑃 𝐴1 + [𝑃 𝐴2 − 𝑃 𝐴1𝐴2 ] ___true for n= 2

𝑃𝑟𝑜𝑜𝑓 by indention

𝐴𝑟𝑢𝑚 that it is true for n= m, then

𝑃 𝐴𝑘

𝑚+1

1

= 𝑃 𝐴𝑘

𝑚

1

∪ 𝐴𝑚+1

= 𝑃 𝐴𝑘

𝑚

1

+ 𝑃 𝐴𝑚+1 − 𝑃 𝐴𝑘

𝑚

1

∩ 𝐴𝑚+1

𝑟. 𝑕. 𝑠 = 𝑃 𝐴𝑘

𝑛

1

− 𝑃 𝐴𝑘1∩ 𝐴𝑘2

𝑘1<𝑘2

+ 𝑃 𝐴𝑘1𝐴𝑘2

𝐴𝑘3

𝑘1<𝑘2<𝑘3

− ⋯ + −1 𝑚−1𝑃 𝐴𝑘

𝑚

1

+ 𝑃 𝐴𝑚+1 − 𝑃 𝐴𝑘 𝐴𝑚+1

𝑚

1

_______(1)

𝑃 𝐴𝑘 𝐴𝑚+1

𝑚

1

= 𝑃 𝐴𝑘𝐴𝑚+1

𝑛

1

− 𝑃 𝐴𝑘1𝐴𝑚+1

𝑘1<𝑘2

∩ 𝐴𝑘2𝐴𝑚+1

+ 𝑃 𝐴𝑘1𝐴𝑚+1 ∩ 𝐴𝑘2

𝐴𝑚+1 ∩ 𝐴𝑘3𝐴𝑚+1

𝑘1<𝑘2<𝑘3

− ⋯

+ −1 𝑚−1𝑃 𝐴𝑘𝐴𝑚+1

𝑚

1

____(2)

𝑈𝑠𝑖𝑛𝑔 2 𝑖𝑛 (1) gives

𝑃 𝐴𝑘

𝑚+1

1

= 𝑃 𝐴𝑘

𝑚+1

1

− 𝑃 𝐴𝑘1𝐴𝑘2

𝑚+1

𝑘1<𝑘2

+ 𝑃 𝐴𝑘1𝐴𝑘2

𝐴𝑘3

𝑘1<𝑘2<𝑘3

− ⋯

+ −1 𝑚−1𝑃 𝐴𝑘

𝑚

1

.

6) 𝛺= {0, 1, 2, ….}. P({j})= 𝑐2𝑗

𝑗 != 𝐶𝑒2 ⟹ 𝐶 = 𝑒−2

𝑏 𝑃 𝐴 = 𝑒−2

4

2

2𝑗

𝑗!; 𝑃 𝐵 = 𝑒−2

∝

3

2𝑗

𝑗! ; 𝑃 𝐶 = 𝑒−2

∝

0

22𝑗+1

2𝑗 + 1 !

𝑃 𝐵 ∩ 𝐶 = 𝑃 3 + 𝑃 5 + ⋯ = ⋯

Other probs can be completed in a similar manner

(7) Favorite models of dinosaurs numbered 1, 2, 3 (say) define events

𝐴𝑖= model # I not found in 6 pockets.

𝑖 = 1, 2, 3

Reqd. prob. = 𝑃 𝐴1𝑐 ∩ 𝐴2

𝑐 ∩ 𝐴3𝑐 = 1 − 𝑃 𝐴1

𝑐 ∩ 𝐴2𝑐 ∩ 𝐴3

𝑐 𝑐 = 1 − 𝑃 𝐴1 ∪ 𝐴2 ∪ 𝐴3

= 1 − 𝑃 𝐴1 + 𝑃 𝐴2 + 𝑃 𝐴3 − 𝑃 𝐴1𝐴2 − 𝑃 𝐴1𝐴3 − 𝑃 𝐴1𝐴3 + 𝑃 𝐴1𝐴2𝐴3 ______(1)

𝑁ote that

𝑃 𝐴𝑖 = 4

5

6 ∀ 𝑖

𝑃 𝐴𝑖𝐴𝑗 = 3

5

6 ∀ 𝑖 ≠ 𝑗

𝑃 𝐴1 𝐴2 𝐴3 = 2

5

6

_____(2)

Use (2) in (1) get the desired prob.

(8) 𝐴𝑖 : match at position i

P(at least one match)

= P(𝐴1 ∪ 𝐴2 ∪ …∪ 𝐴𝑛) = 𝑃(𝐴𝑖)𝑛1 − 𝑃(𝐴𝑖 − 𝐴𝑗 )𝑖<𝑗 + ⋯ + −1 𝑛−1𝑃( 𝐴𝑖

𝑛1 )

𝑃(𝐴𝑖1∩ 𝐴𝑖2

∩ …∩ 𝐴𝑖𝑟 = 𝑛 − 𝑟 !

𝑛!; 1 ≤ 𝑖1 < 𝑖2 < ⋯ < 𝑖𝑟 ≤ 𝑛 𝑟 = 1(1)𝑛

𝑟𝑒𝑞𝑑 𝑝𝑟𝑜𝑏 = 1 −1

2!+

1

3!+ ⋯ + −1 𝑛−1

1

𝑛!

(9)

(a) 𝐴𝑖 : event that ith bill goes to ith envelope (i= 1(1) n)

𝑟𝑒𝑞𝑑. Prob.=𝑃 𝐴𝑖𝑐𝑛

𝑖=1 = 1 − 𝑃 𝐴𝑖

= 1 − 𝑃 𝐴𝑖 − 𝑃 𝐴𝑖𝐴𝑗 𝑖<𝑗 + ⋯ + −1 𝑛−1𝑃 𝐴1 , … , 𝐴𝑟 =

1 − 𝑃 𝐴𝑖 ˽𝑄1 𝑃 𝐴𝑖𝐴𝑗 𝑖<𝑗 ˽𝑄2 + ⋯ + (−1)𝑛𝑃 𝐴1 , … , 𝐴𝑟 ˽𝑄𝑛

𝐼𝑛 𝑄𝑖 ⟶ 𝑛

𝑖 𝑡𝑒𝑟𝑚𝑠 𝑒𝑎𝑐𝑕 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜

𝑛 − 𝑖 !

𝑛! =

1

(𝑛)𝑖

⟹ 𝑃 𝐴𝑖𝑐 = 1 −

𝑛

1

1

𝑛 1+

𝑛

2

1

𝑛 2− ⋯ + −1 𝑛

𝑛

𝑛

1

𝑛 𝑛

= 1 −𝑛!

1! 𝑛 − 1 !. 𝑛 − 1 !

𝑛!+

𝑛!

2! 𝑛 − 2 !. 𝑛 − 2 !

𝑛!… + −1 𝑛 .

1

𝑛!

= 1 −1

1!+

1

2!− ⋯ + −1 𝑛

1

𝑛!= (−1)𝑖

1

𝑛!

𝑛

𝑖=0

(b) 𝐵𝑖 : ith envelope get ith bill & ith cheque

𝑟𝑒𝑞𝑑. Prob. 𝑃 𝐵𝑖𝑐 = 1 − 𝑃 𝐵𝑖 = 1 − 𝑃 𝐵𝑖 + 𝑃 𝐵𝑖𝐵𝑗 𝑖<𝑗 − ⋯ + −1 𝑛𝑃(𝐵1 …𝐵𝑛)

⟷ ⎵ ↓

𝑅1 𝑅2 𝑅𝑛

In 𝑅𝑖 ⟶ 𝑛𝑖 𝑡𝑒𝑟𝑚 𝑒𝑎𝑐𝑕 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜

𝑛−𝑖 !

𝑛!. 𝑛−𝑖 !

𝑛! .

⟹𝑃 𝐵𝑖𝑐

𝑖 = 1 − 𝑃 𝐵𝑖𝑖 = 1 − 𝑃 𝐵𝑖 𝑖 + 𝑃 𝐵𝑖𝐵𝑗 𝑖<𝑗 − ⋯ + −1 𝑛𝑃 𝐵1 …𝐵𝑛

= 1 − 𝑅1 + 𝑅2 − 𝑅3 + ⋯ + −1 𝑛𝑅𝑛

𝑅𝑖 = 𝑛

𝑖 𝑛 − 𝑖 !

𝑛! ×

𝑛 − 𝑖 !

𝑛!

=𝑛!

𝑖! 𝑛 − 𝑖 !. 𝑛 − 𝑖 !

𝑛!. 𝑛 − 𝑖 !

𝑛!=

1

𝑖!.

1

𝑛 𝑖 | 𝑛 𝑖 =

𝑛!

𝑛 − 𝑖 !

⟹ 𝑃 𝐵𝑖𝑐

𝑖

= −1 𝑖

𝑛

𝑖=0

1

𝑖! 𝑛 𝑖 .

(10) (i) P(A∪ B | C)= 𝑃 𝐴∪𝐵 ∩𝐶

𝑃 𝐶 =

𝑃 𝐴𝐶∩𝐵𝐶

𝑃 𝐶

= 𝑃 𝐴 𝐶) + 𝑃 𝐵 𝐶 − 𝑃 𝐴𝐵 𝐶)

(ii) 𝑃 𝐴𝑐 𝐶) =𝑃(𝐴𝑐𝑐)

𝑃(𝐶)=

𝑃 𝐶 −𝑃(𝐴𝐶)

𝑃(𝐶)= 1 − 𝑃(𝐴|𝐶)

(11) (a) true – 𝑃 𝐴 𝐵 + 𝑃 𝐴𝑐 𝐵 =𝑃(𝐴𝐵)

𝑃(𝐵)+

𝑃(𝐴𝑐𝐵)

𝑃(𝐵)=

𝑃(𝐵)

𝑃 𝐵 = 1

(b) 𝑃 𝐴 𝐵 =𝑃 𝐴𝐵

𝑃 𝐵 ; 𝑃 𝐴 𝐵𝑐 =

𝑃 𝐴𝐵𝑐

𝑃 𝐵𝑐 =

𝑃 𝐴 −𝑃 𝐴𝐵

1−𝑃 𝐵

𝑇𝑎𝑘𝑒 𝐴 ⊂ 𝐵, 𝑃 𝐴 > 0, 𝑃 𝐵 − 𝐴 > 0

𝑃 𝐴 𝐵 + 𝑃 𝐴 𝐵𝑐 =𝑃(𝐴𝐵)

𝑃(𝐵)+

𝑃(𝐴𝐵𝑐)

𝑃(𝐵𝑐)=

𝑃 𝐴

𝑃(𝐵)< 1 𝑓𝑎𝑙𝑠𝑒.

(c) Take A⊂B, i.e. 𝐵𝑐 ⊂ 𝐴𝑐

𝑃 𝐴 𝐵 + 𝑃 𝐴𝑐 𝐵𝑐 .

=𝑃(𝐴𝐵)

𝑃(𝐵)+

𝑃 𝐴𝑐 𝐵𝑐

𝑃(𝐵𝑐)> 1 𝑓𝑎𝑙𝑠𝑒.

(12) (a) 𝑃 𝐴 𝐵 =1

4⟹ 𝑃 𝐴𝐵 ≠ 0

↖(i.e. AB ≠ 0)

(b) 𝐴𝐶𝐵 ⟹ 𝑃 𝐴𝐵 = 𝑃 𝐴

𝐺𝑖𝑣𝑒𝑛 𝑃 𝐵 𝐴 =𝑃(𝐴𝐵)

𝑃(𝐴)=

1

2 ⟹ 𝑃 𝐴𝐵 ≠ 𝑃 𝐴 𝑓𝑎𝑙𝑠𝑒.

(c) 𝑃 𝐴𝑐 𝐵𝑐 =𝑃 𝐴𝑐𝐵𝑐

𝑃 𝐵𝑐 =

1−𝑃 𝐴 − 𝑃 𝐵 +𝑃 𝐴𝐵

1−𝑃 𝐵 _____(∗)

𝑃 𝐵 𝐴 =1

2

& 𝑃 𝐴 =1

4, ⟹ 𝑃 𝐴𝐵 =

1

8

& 𝑃 𝐴 𝐵 =𝑃(𝐴𝐵)

𝑃(𝐵)=

18

𝑃(𝐵)=

1

4⟹ 𝑃 𝐵 =

1

2

(*) ⟹ 𝑃 𝐴𝑐 𝐵𝑐 =1−

1

4−

1

2+

1

81

2

=3

4.

(13) (a) P(exactly 3 white balls, out of 4)

= 43

1

2

3.

1

2= ⋯

(b)A : first ball placed is white

𝑃 𝐴 =1

2

𝐵: 𝑢𝑟𝑛 𝑐𝑜𝑛𝑡𝑎𝑖𝑛 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 3 𝑤𝑕𝑖𝑡𝑒 𝑏𝑎𝑙𝑙𝑠 𝑃 𝐵 𝐴 =𝑃(𝐴𝐵)

𝑃(𝐴)=

12 . 3

2

12

3

12

=3

8.

(14) D: A occurs before B

𝑃 𝐷 = 𝑝𝐴 + 𝑝𝐴𝑝𝐶 + 𝑝𝐴𝑝𝐶2 + ⋯ =

𝑝𝐴

1 − 𝑝𝐶=

𝑝𝐴

𝑝𝐴 + 𝑝𝐵

𝐴𝑖 : 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑖 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛

P(system function)= 1 − 𝑃 𝐴𝑖𝑐 = 1 − (1 − 𝑝)𝑛

(16) 𝐴𝑖 : Question I is among the 90 questions that the student can answer correctly.

Reqd. prob. 𝑃 𝐴1𝐴2𝐴3 = 𝑃 𝐴1 𝑃 𝐴2 𝐴1 𝑃 𝐴3 𝐴1𝐴2)

=90

100.89

99.88

98….

(17) Apply bayes theorem

Reqd. prob. =

1

2×

2

31

2×

2

3+

2

3×

1

3

= ⋯

(18) 𝐴+, 𝐵+, 𝐶+, 𝐷+-events that A, B, C, D passes the paper with (t) sign

bayes theorem

Reqd. prob. =P 𝐴+ 𝐷+ =𝑃 𝐴+ 𝑃 𝐷+ 𝐴+

𝑃 𝐷+

𝑃 𝐷+ 𝐴+) = 1

3

3

+ 3

2

2

3

2

.1

3=

13

27

𝑎𝑙𝑠𝑜 𝑃 𝐷+ = 𝑃 𝐷+ 𝐴+ 𝑃 𝐴+ + 𝑃 𝐷+ 𝐴+𝑐 𝑃 𝐴+

𝑐

𝑃 𝐷+ 𝐴+𝑐 = 𝑃 𝐷 𝑝𝑎𝑠𝑠𝑒𝑠 𝑤𝑖𝑡𝑕 + 𝐴 𝑝𝑎𝑠𝑠𝑒𝑠−).

= 3

1

2

3

1

3

2

+ 3

3

2

3

3

1

3

0

=14

27

𝑃 𝐷+ =13

27.1

3+

14

27.2

3=

41

81

⟹ 𝑃 𝐴+ 𝐷+ =13

27.1

341

81

=13

41.

(19) silver coin in other

P(h| Gold coin in one drawer).

Bayes thm

=1

3×

1

21

3×1+

1

3×

1

2+

1

3×0

=1

3 .

(20) (a) 𝑃 𝐶1𝐶2𝐶3𝐶4 = 𝑃(𝐶𝑖)41 = ⋯

(b) 𝑃 𝐶1𝑐 ∩ 𝐶2

𝑐 ∩ 𝐶3𝑐 ∩ 𝐶4

𝑐 = 𝑃(𝐶𝑖𝑐)4

1 [𝑒𝑥𝑝𝑙𝑎𝑖𝑛 𝑤𝑕𝑦 𝑖𝑠 𝑠𝑜 𝐶1 , 𝐶2, 𝐶3 , 𝐶4 𝑖𝑛𝑑𝑒𝑝𝑒𝑛 ⟹

𝐶1𝑐 , 𝐶2

𝑐 , 𝐶3𝑐 , 𝐶4

𝑐 𝑎𝑟𝑒 𝑎𝑙𝑠𝑜 𝑖𝑛𝑑𝑒𝑝𝑒𝑛. ]

(c) 𝑃 𝐶1𝐶2𝑐𝐶3

𝑐𝐶4𝑐 + 𝑃 𝐶1

𝑐𝐶2𝐶3𝑐𝐶4

𝑐 + 𝑃 𝐶1𝑐𝐶2

𝑐𝐶3𝐶4𝑐 + 𝑃 𝐶1

𝑐𝐶2𝑐𝐶3

𝑐𝐶4𝑐

= 𝑃 𝐶1 𝑃 𝐶𝑖3

4

2

+ ⋯

(d) P(at least one hits)= 1 − 𝑃 𝑛𝑜 𝑜𝑛𝑒 𝑕𝑖𝑡𝑠

= 1 − 𝑃 𝐶1𝑐𝐶2

𝑐𝐶3𝑐𝐶4

𝑐 = ⋯

(21) 𝑃 𝐴𝑖𝑐𝑛

1 = 𝑃 𝐴𝑖𝑐 𝑛

1 = 1 − 𝑃 𝐴𝑖 𝑛1 ≤ exp −𝑃 𝐴𝑖

𝑛1 0 < 𝑥 < 1, 1 − 𝑥 < 𝑒−𝑥

𝑖. 𝑒. 𝑃 𝐴𝑖𝑐𝑛

1 ≤ exp(− 𝑃(𝐴𝑖)𝑛1 ).

(22) 𝛺= {1, 2, 3, 4} 7: Proper set

𝑃 𝑖 =1

4 𝑖 = 1, 2, 3, 4

A= {1, 4} , B= {2, 4}, C= {3, 4}.

P(A)= P(B)= P(C)= ½

𝑃 𝐴𝐵 = 𝑃 𝐴𝐶 = 𝑃 𝐵𝐶 =1

4; 𝑃 𝐴𝐵𝐶 =

1

4

⟹ 𝑃 𝐴𝐵 = 𝑃 𝐴 𝑃 𝐵 , 𝑃 𝐴𝐶 = 𝑃 𝐴 𝑃 𝐶 & 𝑃 𝐵𝐶 = 𝑃 𝐵 𝑃(𝐶)

i.e. A, B, C are pair wise independent

but P(ABC)= 1

4 ≠ 𝑃 𝐴 . 𝑃 𝐵 . 𝑃 𝐶 =

1

8

⟹ 𝐴, 𝐵, 𝐶 𝑛𝑜𝑡 𝑚𝑢𝑡𝑢𝑎𝑙𝑠 𝑖𝑛𝑑𝑒𝑝.

(23) Counter example

In prv prob set up take

A= {1, 2}, B= {3, 4}, C= {1}

P(A| B) < P(A)

P(B| C) < P(B) but P(A|C) > P(A)

⟹ C does not carry negative information about A.

(24) 𝐴𝑖 : I gives are in the list i= 0, 1, 2, 3

B : 1st student is girl.

C : 2nd

student is boy:- to obtain P(C|B)

P(B)= 𝑃 𝐴1 𝑃 𝐵 𝐴1 + 𝑃 𝐴2 𝑃 𝐵 𝐴2 + 𝑃 𝐴3 𝑃 𝐵 𝐴3 = 3

1 53

84

×1

4+

32 5

2

84

×2

4+

33 5

1

84

×3

4

𝑃 𝐵 =105

4 × 84

.

𝑃 𝐶 𝐵 =𝑃 𝐶𝐵

𝑃 𝐵 =

𝑃 𝐶 ∩ 𝐴𝑖𝐵31

𝑃 𝐵 =

𝑃 𝐶𝐴𝑖𝐵31

𝑃 𝐵

= 𝑃 𝐶 𝐴𝑖𝐵

3

1

𝑃 𝐴𝑖 𝐵

= 1 × 𝑃 𝐴1 𝐵 +2

3× 𝑃 𝐴2 𝐵 +

1

3× 𝑃 𝐴3 𝐵

𝑃 𝐴1 𝐵 =

𝑃 𝐴1 𝑃 𝐵 𝐴1

𝑃 𝐵 =

2

7

𝑃 𝐴2 𝐵 =𝑃 𝐴2 𝑃 𝐵 𝐴1

𝑃 𝐵 =

4

7 𝑃 𝐴3 𝐵) =

1

7

⟹ 𝑃 𝐶 𝐵 = 1 ×2

7+

2

3×

4

7+

1

3×

1

7= ⋯

(25)

𝐴𝑖 : event that a fly survives ith application i= 1, 2, 3, 4

Note that 𝐴4 ⊂ 𝐴3 ⊂ 𝐴2 ⊂ 𝐴1

⟹ 𝐴4 = 𝐴1 ∩ 𝐴2 ∩ 𝐴3 ∩ 𝐴4

(a) Reqd. prob. = P(a fly survives 4 applications)

= P 𝐴1 𝐴2 𝐴3 𝐴4 = 𝑃 𝐴4

= 𝑃 𝐴1 𝑃 𝐴2 𝐴1 𝑃 𝐴3 𝐴1𝐴2 𝑃(𝐴4|𝐴1 𝐴2 𝐴3)

↓ ↓ ↓ ↓

= (1- 0.8) (1- 0.4) (1- 0.2) (1- 0.1)

(for the given conditions)

= 0.2 × 0.6 × 0.8 × 0.9

(b) 𝑃 𝐴4 𝐴1 =𝑃 𝐴4∩𝐴1

𝑃 𝐴1 =

𝑃 𝐴4

𝑃 𝐴1

= 0.6 × 0.8 × 0.9

(26) 𝐵𝑖 : event that I of the paintings are forgeries i= 0(1) 5

P(𝐵0) = 0.76, 𝑃 𝐵1 = 0.09, 𝑃 𝐵2 = 0.02, 𝑃 𝐵3 = 0.01, 𝑃 𝐵4 = 0.02 & 𝑃 𝐵5 =

0.1 (𝑔𝑖𝑣𝑒𝑛 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑠)

A : event that the painting sent for authentication turns out to be a forgery.

Reqd. prob. = 𝑃 𝐵5 𝐴 =𝑃 𝐵5 𝑃 𝐴 𝐵5)

𝑃 𝐵𝑖 𝑃(𝐴| 𝐵𝑖)5𝑖=0

↑

Bayes them

𝑃 𝐴 = 𝑃 𝐵𝑖 𝑃(𝐴| 𝐵𝑖)

5

𝑖=0

= 0.76 × 0 + 0.09 ×1

5+ 0.02 ×

2

5+ 0.01 ×

3

5+ 0.02 ×

4

5+ 0.10 × 1

= ⋯

⤥

𝑃 𝐵5 𝐴 =0.10 × 1

𝑃(𝐴)= ⋯

Problem Set-3

[1] Let X be a random variable defined on (𝛺, ℑ, 𝑃).Show that the following are also random variables;

(a) |X|, (b) 𝑋2 𝑎𝑛𝑑 𝑐 𝑋, given that {x< 0}=𝜙.

[2] Let 𝛺= [0, 1] and ℑ be the Boral 𝜍-field of subsets of 𝛺. Define X on 𝛺 as follows:

𝑋 𝜔 = 𝜔 𝑖𝑓 0 ≤ 𝜔 ≤

1

2

𝜔 −1

2 𝑖𝑓

1

2< 𝜔 ≤ 1

Show that X defined above is a random variable.

[3] Let 𝛺= {1, 2, 3, 4} and ℑ= {𝜙, 𝛺, {1}, {2, 3, 4}} be a 𝜍-field of subsets of 𝛺. Verify whether

𝑋 𝜔 = 𝜔 + 1; ∀ 𝜔 ∈ 𝛺, is a random variable with respect to ℑ.

[4] Let a card be selected from an ordinary pack of playing cards. The outcome 𝜔 is one of these 52 cards.

Define X on 𝛺 as :

𝑋 𝜔 =

4 𝑖𝑓 𝜔 𝑖𝑠 𝑎𝑛 𝑎𝑐𝑒3 𝑖𝑓 𝜔 𝑖𝑠 𝑎 𝑘𝑖𝑛𝑔

2 𝑖𝑓 𝜔 𝑖𝑠 𝑎 𝑞𝑢𝑒𝑒𝑛1 𝑖𝑓 𝜔 𝑖𝑠 𝑎 𝑗𝑎𝑐𝑘

0 𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒.

Show that X is a random variable. Further, suppose that P(.) assigns a probability of 1/52 to each outcome

𝜔. Derive the distribution function of X.

[5] Let F(x)=

0 𝑖𝑓 𝑥 < −1(𝑥+2)

4 𝑖𝑓 − 1 ≤ 𝑥 < 1

1 𝑖𝑓 𝑥 ≥ 1

Show that F(.) is a distribution function. Sketch the graph of F(x) and compute the probabilities P(- ½ < X

≤1

2) , 𝑃 𝑋 = 0 , 𝑃 𝑋 = 1 𝑎𝑛𝑑𝑃(−1 ≤ 𝑥 < 1). Further, obtain the decomposition F(x)= 𝛼 𝐹𝑑 𝑥 +

1 − 𝛼 𝐹𝑐(𝑥); where 𝐹𝑑 𝑥 𝑎𝑛𝑑𝐹𝑐(𝑥) are purely discrete and purely continuous distribution functions,

respectively.

[6] Which of the following functions are distribution function?

(a) F(x)=

0, 𝑥 < 0

𝑥, 0 ≤ 𝑥 ≤1

2

1, 𝑥 >1

2

; (b) 𝐹 𝑥 = 0, 𝑥 < 0

1 − 𝑒𝑥 , 𝑥 ≥ ; (c)𝐹 𝑥 =

0 𝑥 ≤ 1

1 −1

𝑥 𝑥 > 1

[7]𝐿𝑒𝑡 𝐹 𝑥 = 0 𝑖𝑓 𝑥 ≤ 0

1 −2

3𝑒−𝑥/3 −

1

3𝑒−[𝑥/3] 𝑖𝑓 𝑥 > 0

Where, [x] is the largest integer≤ x. show that F(.) is a distribution function and compute P(X > 6), P(X =

5) and P(5 ≤ 𝑋 ≤ 8).

[8] The distribution function of a random variable X is given by

F(x)=

0, 𝑥 < −2,1

3, −2 ≤ 𝑥 < 0,

1

2, 0 ≤ 𝑥 < 5,

1

2+

𝑥−5 2

2, 5 ≤ 𝑥 < 6

1, 𝑥 ≥ 6

𝐹𝑖𝑛𝑑 𝑃 −2 ≤ 𝑋 < 5 , 𝑃 0 < 𝑋 < 5.5 𝑎𝑛𝑑 𝑃 1.5 < 𝑋 ≤ 5.5 𝑋 > 2).

[9] Prove that if 𝐹1 . , … . , 𝐹𝑛 . 𝑎𝑟𝑒 𝑛 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠, 𝑡𝑕𝑒𝑛 𝐹 𝑥 = 𝛼𝑖𝐹𝑖(𝑥)𝑛𝑖=1 is also a

distribution function for any (𝛼1 , … , 𝛼𝑛 ), such that 𝛼𝑖 ≥ 0and 𝛼𝑖𝑛𝑖=1 = 1.

[10] Suppose 𝐹1 𝑎𝑛𝑑 𝐹2 are distribution functions. Verify whether G(x) = 𝐹1 𝑥 + 𝐹2(𝑥)is also a

distribution function.

[11] Find the value of 𝛼 and k so that F given by

F(x)= 0 𝑖𝑓 𝑥 ≤ 0

+𝑘𝑒−𝑥2/2 𝑖𝑓 𝑥 > 0

Is distribution function of a continuous random variable.

[12] Let F(x)=

0 𝑖𝑓 𝑥 < 0(𝑥+2)

8 𝑖𝑓 0 ≤ 𝑥 < 1

𝑥2+2

8 𝑖𝑓 1 ≤ 𝑥 < 2

2𝑥+𝑐

8 𝑖𝑓 2 ≤ 𝑥 ≤ 3

1 𝑖𝑓 𝑥 > 3

Find the value of c such that F is a distribution function. Using the obtained value of c, find the

decomposition F(x)= 𝛼 𝐹𝑑 𝑥 + 1 − 𝛼 𝐹𝑐(𝑥); where 𝐹𝑑 𝑥 𝑎𝑛𝑑𝐹𝑐(𝑥) are purely discrete and purely

continuous distribution functions, respectively.

[13] Suppose 𝐹𝑋 is the distribution function of a random variable X. Determine the distribution function

of a 𝑋+𝑎𝑛𝑑 𝑏 |𝑋|. Where

𝑋+ = 𝑋 𝑖𝑓 𝑋 ≥ 00 𝑖𝑓 𝑋 < 0

[14]The convolution F of two distribution functions 𝐹1 𝑎𝑛𝑑 𝐹2 is defined as follows;

F(x)= ∫ 𝐹1 𝑥 − 𝑦 ∞

−∞𝑑𝐹2 𝑦 ; 𝑥 ∈ ℝ

And is denoted by F= 𝐹1 ∗ 𝐹2. Show that is F is also a distribution function.

[15] Which of the following functions are probability mass functions?

(a) f(x)= (𝑥−2)

2 𝑖𝑓 𝑥 = 1, 2, 3, 4

0 𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒.

; 𝑏 𝑓 𝑥 = (𝑒−𝜆𝜆𝑥)

𝑥! 𝑖𝑓 𝑥 = 0, 1, 2, 3, 4, …


𝑤𝑕𝑒𝑟𝑒, 𝜆 > 0 ; 𝑐 𝑓 𝑥 =

(𝑒−𝜆𝜆𝑥)

𝑥! 𝑖𝑓 𝑥 = 1, 2, 3, 4, …


𝑤𝑕𝑒𝑟𝑒, 𝜆 > 0

[16] Find the value of the constant c such that f(x)= (1- c)𝑐𝑥 ; x= 0, 1, 2,3, … defines a probability mass

function.

[17] Let X be a discrete random variable taking values in 𝒳 = −3, −2, −1, 0, 1, 2, 3 𝑠𝑢𝑐𝑕 𝑡𝑕𝑎𝑡 𝑃 𝑋 =

−3 = 𝑃 𝑋 = −2 = 𝑃 𝑋 = −1 = 𝑃 𝑋 = 1 = 𝑃 𝑋 = 2 = 𝑃(𝑋 = 3)

And P(X < 0)= P(X= 0)= P(X > 0). Find the distribution function of X.

[18] A battery cell is labeled as good if it works for at least 300 days in a clock, otherwise it is labeled as

bad. Three manufacturers, A, B, and C make cells with probability of marking good cells as 0.95, 0.90

and 0.80 respectively. Three identical clocks are selected and cells made by A, B, and C are used in clock

numbers 1, 2 and 3 respectively. Let X be the total number of clocks working after 300 days. Find the

probability mass function of X and plot the corresponding distribution function.

[19] Prove that the function 𝑓𝜃(𝑥) = 𝜃2𝑥𝑒−𝜃𝑥 𝑖𝑓 𝑥>0


Defines a probability density function for 𝜃 > 0. Find the corresponding distribution function and hence

compute P(2 < X < 3) and P(X > 5).

[20] Find the value of the constant c such that the following function is probability density function.

𝑓𝜆 𝑥 = 𝑐 𝑥 + 1 𝑒−𝜆𝑥 𝑖𝑓 𝑥 ≥ 0

0 𝑖𝑓 𝑥 < 0

Where 𝜆 > 0. Obtain the distribution function of the random variable associated with probability density

function𝑓𝜆 𝑥 .

[21] Show that f(x)= 𝑥2

18 𝑖𝑓 − 3 < 𝑥 < 3


Defines a probability density function. Find the corresponding distribution function and hence find P(|X|<

1) and P(𝑥2 < 9)

Solution Key

(1) (a) y= |X| ∀𝑥 ∈ ℝ 𝑦−1 −∞, 𝑥 = 𝜔: 𝑦 𝜔 ≤ 𝑥 = 𝜔: 𝑋 𝜔 ≤ 𝑥

= {𝜔: −𝑥 ≤ 𝑋(𝜔) ≤ 𝑥}

= 𝑋−1 −𝑥, ∞ ∩ 𝑋−1(−∞, 𝑥]

As X is a.r.v.

𝑋−1 −𝑥, ∞ ∩ 𝑋−1(−∞, 𝑥] ∈ 𝓕

[X is a. r. v. ⟹𝑋−1 𝐵 ∈ℱ∀ 𝐵 ∈ 𝔅

⟹ 𝑋−1 −𝑥, ∞ ∈ ℱ &𝑋−1 −∞, 𝑥 ∈ ℱ]

⟹ 𝑦−1 −∞, 𝑥 ∈ ℱ ∀ 𝑥 ∈ ℝ

⟹ y = X 𝑖𝑠 𝑎. 𝑟. 𝑣.

(b)y= 𝑋2; ∀ 𝑥 ∈ ℝ 𝑦−1 −∞, 𝑥 = 𝜔: 𝑦 𝜔 ≤ 𝑥

= 𝜔: 𝑋2 𝜔 ≤ 𝑥

= 𝜔: − 𝑥 ≤ 𝑋 𝜔 ≤ 𝑥

= 𝑋−1 − 𝑥, ∞ ∩ 𝑋−1 −∞, 𝑥 ∈ ℱ

⟹ 𝑦 = 𝑋2 𝑖𝑠 𝑎. 𝑟. 𝑣.

Sly part (c).

(2) 𝛺= [0, 1] ℱ : Boral 𝜍-field of student of 𝛺

X(𝜔)= 𝜔, 0 ≤ 𝜔 ≤

1

2

𝜔 −1

2,

1

2< 𝜔 ≤ 1

𝑋−1 −∞, 𝑥 =

𝜙 ∈ ℱ, 𝑥 < 0

0, 𝑥 ∪ 1

2,1

2+ 𝑥 0 ≤ 𝑥 <

1

2∈ ℱ

𝛺 ∈ ℱ , 𝑥 ≥1

2

⟹ 𝑋−1 −∞, 𝑥 ∈ ℱ ∀ 𝑥 ∈ ℝ ⟹ X is a. r. v.

(3) 𝛺= {1, 2, 3, 4}

ℑ= {𝜙, 𝛺, {1}, {2, 3, 4}}

X(𝜔)= 𝜔+1 → 2, 3, 4, 5

𝑋−1 −∞, 𝑥 =

𝜙 ∈ ℑ 𝑥 < 2 1 ∈ ℑ 2 ≤ 𝑥 < 3 1, 2 ∉ ℑ 3 ≤ 𝑥 < 4

↓

⟹ 𝑋 𝑖𝑠 𝑛𝑜𝑡 𝑎. 𝑟. 𝑣.

(4) Ace king

↓ ↓

𝛺= {1 H, ……, k H, 1 S, ….., k S, 1D, …., k D, 1C, …. K C }

↔ ↔ ↔ ↔

All hearts Spades Diamond club

X(𝜔)=

4 𝑎𝑐𝑒 ℑ = 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 3 𝑘𝑖𝑛𝑔 2 𝑞𝑢𝑒𝑒𝑛1 𝑗𝑎𝑐𝑘


𝑋−1 −∞, 𝑥 =

𝜙 𝐼𝐹 𝑋 < 0 2𝐻, … 10𝐻, 2𝑆, … 10𝑆, 2𝐷, … 10�͍�, 2𝐶, … 10𝐶 𝑖𝑓 0 ≤ 𝑥 < 1

2𝐻, 𝐽�͍� , 2𝑆, …𝐽�͍� , 2𝐷, …𝐽�͍� , 2𝐶, …𝐽�͍� 𝑖𝑓 1 ≤ 𝑋 < 2

2𝐻, …𝑄�͍� , 2𝑆, …𝑄�͍� , 2𝐷, …𝑄�͍� , 2𝐶, … , 𝑄�͍� , 2 ≤ 𝑋 < 3

2𝐻, …𝐾�͍�, 2𝑆, …𝐾�͍� , 2𝐷, …𝐾�͍� , 2𝐶, …𝐾�͍� 3 ≤ 𝑋 < 4

𝛺 𝑋 ≥ 4

⟹ 𝑋−1 −∞, 𝑥 ∈ ℱ∀ 𝑥 ∈ ℝ

⟹ X is a. r. v.

(5)

(i) F(.) is non decreasing

(ii) F(.) is right contain everywhere

(iii) F(-∞)= 0 & F(∞)= 1

⟹ F(.) is a d. f.

P(- ½ <X ≤ ½ )= 𝐹 1

2 − 𝐹 −

1

2

=5

8−

3

8=

2

8

𝑃 𝑋 = 0 = 𝐹 0 − 𝐹 0 − = 0

𝑃 𝑋 = 1 = 𝐹 1 − 𝐹1—𝐹(−1−)

=3

4− 0 =

3

4

(6) (a) F(x) is not right continuous at x = ½

⟹ F(.) is not a d. f.

(b)& (c) the n. s. c. holds for these 2 and hence they are d. f. s.

(7) F(.) is non decreasing

F(∞)= 1 & F(-∞)= 0

F(x) is right contain ∀ 𝑥 ∈ ℝ

⟹ F(.) is a. d. f.

𝑃 𝑋 > 6 = 1 − 𝑃 𝑋 ≤ 6 = 1 − 𝐹 6 = 1 − 1 −2

3𝑒−

63 −

1

3𝑒

− 63

= 1 − 1 −2

3𝑒−2 −

1

3𝑒−2 = 𝑒−2

𝑃 𝑋 = 5 = 𝐹 5 − 𝐹 5 −

= 1 −2

3𝑒−

53 −

1

3𝑒

− 53 − 1 −

2

3𝑒−

53 −

1

3𝑒

− 53

= 1 −2

3𝑒−

53 −

1

3𝑒−1 − 1 −

2

3𝑒−

53 −

1

3𝑒−1 = 0

𝑃 5 ≤ 𝑋 ≤ 8 = 𝐹 8 − 𝐹 5 −

= ⋯

(8) (6) (a) F(x) is not right continuous at x = ½

⟹ F(.) is not a d. f.

(b)& (c) the n. s. c. holds for these 2 and hence they are d. f. s.

(7) F(.) is non decreasing

F(∞)= 1 & F(-∞)= 0

F(x) is right contain ∀ 𝑥 ∈ ℝ

⟹ F(.) is a. d. f.

𝑃 𝑋 > 6 = 1 − 𝑃 𝑋 ≤ 6 = 1 − 𝐹 6 = 1 − 1 −2

3𝑒−

63 −

1

3𝑒

− 63

= 1 − 1 −2

3𝑒−2 −

1

3𝑒−2 = 𝑒−2

𝑃 𝑋 = 5 = 𝐹 5 − 𝐹 5 −

= 1 −2

3𝑒−

53 −

1

3𝑒

− 53 − 1 −

2

3𝑒−

53 −

1

3𝑒

− 53

= 1 −2

3𝑒−

53 −

1

3𝑒−1 − 1 −

2

3𝑒−

53 −

1

3𝑒−1 = 0

𝑃 5 ≤ 𝑋 ≤ 8 = 𝐹 8 − 𝐹 5 −

= ⋯

(8) P(-2≤ X < 5)= F(5-)- F(-2 -)

= 1

2− 0 =

1

2

P(0< X < 5.5)= F(5.5-) – F(0)

= 1

2+

1

8 −

1

2=

1

8

P(1.5 < X ≤ 5.5| X > 2)= 𝑃 2<𝑋 ≤5.5

𝑃 𝑋>2

=𝐹 5.5 − 𝐹 2

1 − 𝐹 2

=

12

+18 −

12

1 −12

=1

4

(9) for 𝑥1 < 𝑥2

𝐹 𝑥2 − 𝐹 𝑥1 = 𝛼𝑖𝑛1 𝐹𝑖 𝑥2 − 𝐹𝑖 𝑥1 ≥ 0 ≥ 0∀ 𝑖

⟹ F(x) is non-decreases.

𝐹 −∞ = 𝛼𝑖 𝐹𝑖 −∞ = 0 & 𝐹 ∞ = 1

𝐹 𝑥+ = lim𝑧↓𝑥

𝐹 𝑧 = lim𝑧↓𝑥

𝛼𝑖 𝐹𝑖 𝑧

= 𝛼𝑖 𝐹𝑖 𝑥+ = 𝛼𝑖 𝐹𝑖 𝑥 = 𝐹(𝑥)

⟹ F(.) is a d. f.

(10) 𝐺 ∞ = 𝐹1 ∞ + 𝐹2 ∞ = 2 ≠ 1

⟹ 𝐺 . 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑑. 𝑓.

(11) Right cont at x= 0 ⟹ F(0)= F(0+)

⟹ 0= 𝛼+ k ________(i)

𝐹 ∞ = 1 ⟹ 𝛼 = 1 ⟹ 𝑘 = −𝑣

(12) Right cont at x= 3 ⟹ F(3)= F(3+)

⟹6+𝐶

8= 1 ⟹ 𝐶 = 2

𝑓𝑛 𝐹 . 𝑖𝑠 𝑕𝑎𝑣𝑖𝑛𝑔 𝑗𝑢𝑚𝑝 𝑎𝑡 𝑥 = 0 𝑜𝑛𝑙𝑦 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒2

8

Discrete part of d. f.

𝛼𝐹𝑑 𝑥 = 0 𝑥 < 02

8 𝑥 ≥ 0

𝑐𝑜𝑛𝑡 𝑝𝑎𝑟𝑡

1 − 𝛼 𝐹𝑐 𝑥 =

0 𝑖𝑓 𝑥 < 0𝑥

8 0 ≤ 𝑥 < 1

𝑥2

8 1 ≤ 𝑥 < 2

2𝑥

8 2 ≤ 𝑥 ≤ 3

6

8 𝑖𝑓 𝑥 > 3

⟹ 𝛼 =2

8=

1

4 & 1 − 𝛼 =

3

4

𝐹𝑑 𝑥 = 0 𝑥 < 01 𝑥 ≥ 0

& 𝐹𝑐 𝑥 =

0 𝑖𝑓 𝑥 < 0𝑥

6 0 ≤ 𝑥 < 1

𝑥2

6 1 ≤ 𝑥 < 2

𝑥

3 2 ≤ 𝑥 ≤ 3

1 𝑖𝑓 𝑥 > 3

(13) Y= 𝑋+

P(Y ≤ y)= 0 if y < 0

If y= 0 𝑃 𝑌 ≤ 0 = 𝑃 𝑋+ ≤ 0 = 𝑃 𝑋+ = 0 = 𝑃 𝑋 ≤ 0 = 𝐹(0)

If y> 0 𝑃 𝑌 ≤ 𝑦 = 𝑃 𝑋+ ≤ 𝑦 = 𝑃 𝑋 ≤ 𝑦 = 𝐹 𝑦

𝐹𝑌 𝑦 = 0 𝑦 < 0

𝐹 𝑦 𝑦 ≥ 0

𝑍 = 𝑋

𝐹𝑍 𝔷 = 𝑃 1 × 1 ≤ 𝔷

= 𝑃 −𝔷 ≤ 𝑋 ≤ 𝔷

= 𝐹 𝔷 − 𝐹 −𝔷 − 𝑖𝑓 𝔷 ≥ 0

0 𝑖𝑓 𝔷 < 0

(14) for 𝑥1 < 𝑥2

𝐹 𝑥2 − 𝐹 𝑥1 = 𝐹1 𝑥2 − 𝑦 ∞

−∞

𝑑𝐹2 𝑦 − 𝐹1 𝑥1 − 𝑦 ∞

−∞

𝑑𝐹2 𝑦 ∞

−∞

= 𝐹1 𝑥2 − 𝑦 − 𝐹1 𝑥1 − 𝑦 ∞

−∞

𝑑𝐹2 𝑦

≥ 0 ∀𝑥1 < 𝑥2

𝐹 𝛼 = 𝐹1

∞

−∞

∞ 𝑑 𝐹2 𝑦 = 𝐹2 ∞ − 𝐹2 −∞ = 1

𝐹 −∞ = 0

𝐹 𝑋+ = 𝐹1( 𝑋+ − 𝑦)∞

−∞

𝑑𝐹2 𝑦 = 𝐹1(𝑥 − 𝑦)∞

−∞

𝑑𝐹2 𝑦 = 𝐹(𝑥)

⟹ F(.) as defined is a d. f.

(15) (a) f(1) < 0 ⟹ f(.) is not a p. m. f.

(b) f(x) ≥ 0 ∀ x

𝑓 𝑥 =

𝑥

𝑒−𝜆 𝜆𝑥

𝑥!

∞

𝛼

= 1

⟹ f(.) is a p. m. f.

(c) 𝑓 𝑥 𝑥 ≠ 1 ⟹ 𝑓 . 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑝. 𝑚. 𝑓.

(16) f(x)≥ 0 ∀ C∈[0, 1]

𝑓 𝑥 𝑥 = 1 𝑎𝑙𝑠𝑜 ∀ C∈[0, 1]

⟹ ∀ C → 0≤C ≤ 1; f(.) is p. m .f.

(17) Suppose

P(X= -3)= P(X= -2) = P(X= -1)= P(X= 1) = P(X= 2) = P(X= 3)= p

⟹ P(X < 0)= 3p = P(X > 0)= P(X = 0)

P(X < 0) + P(X = 0) + P(X > 0)= 1

⟹ 𝑝 =1

9

p. m. f.

X= x -3 -2 -1 0 1 2 3

P(X=x) 1

9

1

9

1

9

3

9

1

9

1

9

1

9

𝐹𝑋 𝑥 =

0 𝑥 < −31

9 − 3 ≤ 𝑥 < −2

2

9 − 2 ≤ 𝑥 < −1

3

9 − 1 ≤ 𝑥 < 0

6

9 0 ≤ 𝑥 < 1

7

9 1 ≤ 𝑥 < 2

8

9 2 ≤ 𝑥 < 3

1 𝑥 ≥ 3

(18) X: r. v. denoting total # of clocks working after 300days

X can take values in {0, 1, 2, 3}

𝑃 𝑋 = 0 = 𝑃 𝐴𝑐𝐵𝑐𝐶𝑐 = 𝑃 𝐴𝑐 𝑃 𝐵𝑐 𝑃 𝐶𝑐

= 1 − 0.95 1 − 0.9 1 − 0.8 = 𝑝0 , 𝑠𝑎𝑦

𝑃 𝑋 = 1 = 𝑃 𝐴𝐵𝑐𝐶𝑐 ∪ 𝐴𝑐𝐵𝐶𝑐 ∪ 𝐴𝑐𝐵𝑐𝐶

= 𝑃 𝐴 𝑃 𝐵𝑐 𝑃 𝐶𝑐 + 𝑃 𝐴𝑐 𝑃 𝐵 𝑃 𝐶𝑐 + 𝑃 𝐴𝑐 𝑃 𝐵𝑐 𝑃 𝐶

= ⋯

= 𝑝1 𝑠𝑎𝑦

𝑃 𝑋 = 2 = 𝑃 𝐴𝐵𝐶𝑐 ∪ 𝐴 𝐵𝑐𝐶 ∪ 𝐴𝑐𝐵𝐶

= 𝑃 𝐴 𝑃 𝐵 𝑃 𝐶𝑐 + 𝑃 𝐴 𝑃 𝐵𝑐 𝑃 𝐶 + 𝑃 𝐴𝑐 𝑃 𝐵 𝑃 𝐶

= ⋯

= 𝑝2 𝑠𝑎𝑦

𝑃 𝑋 = 3 = 𝑃 𝐴𝐵𝐶 = 𝑃 𝐴 𝑃 𝐵 𝑃 𝐶 = 0.95 × 0.9 × 0.8 = 𝑝3 𝑠𝑎𝑦

p. m. f.

X= x 0 1 2

3

P(X=

x) 𝑝0 𝑝1 𝑝2

𝑝3

d. f.

(19) f(x) ≥0∀ x

𝑓(𝑥)∞

−∞

𝑑𝑥 = 𝜃2 𝑥𝑒−𝜃𝑥∞

0

𝑑𝑥 = 𝜃2 .⎾2

𝜃2= 1

⟹ f(.) is a p. d. f.

P(2 < X < 3)= F(3)- F(2)

& P(X> 5)= 1- F(5)

Here, F(x)= ∫ 𝑓 𝑥 𝑥

−∞𝑑𝑥 = 𝜃2 ∫ 𝑥𝑒−𝜃𝑥𝑥

0𝑑𝑥 𝑖𝑓 𝑥 > 0

= 0 𝑖𝑓 𝑥 ≤ 0

↗

Integration by parts.

(20) If C ≥ 0 then f(x) ≥ 0 ∀ x

𝑓 𝑥 ∞

0

𝑑𝑥 = 1

⟹ 𝐶 (𝑥 + 1)∞

0

𝑒−𝜆𝑥𝑑𝑥 = 1

i.e. 𝐶 ⎾2

𝜆2 +⎾1

𝜆 = 1 ⟹ 𝐶 =

𝜆2

1+𝜆

d. f. F(x)= ∫ 𝑓 𝑥 𝑥

0𝑑𝑥 =

𝜆2

1+𝜆 ∫ 1 + 𝑥

𝑥

0𝑒−𝜆𝑥𝑑𝑥 𝑖𝑓𝑥 ≥ 0

= 0 𝑖𝑓 𝑥 < 0

↑

By parts.

(21) f(x) ≥ 0 ∀ x

𝑓 𝑥 𝑥

−∞

𝑑𝑥 = 𝑥2

18

3

−3

𝑑𝑥 =1

18 𝑥3

3|

3

−3=

1

18. 18 = 1

⟹ f(.) is a p. d. f.

𝐹𝑋 𝑥 =

0 𝑖𝑓 𝑥 < −3

𝑥3 + 27

54 − 3 ≤ 𝑥 ≤ 3

1 𝑥 > 3

𝑃 𝑋 < 1 = 𝐹 1 − 𝐹 −1 =28

7−

26

7

𝑃 𝑋2 < 9 = 1

Problem Set -4

[1] Find the expected number of throws of a fair die required to obtain a 6.

[2] Consider a sequence of independent coin flips, each of which has a probability p of being heads.

Define a random variable X as the length of the urn (of either heads or tails) started by the first trial. Find

E(X).

[3] Verify whether E(X) exists in the following cases:

(a) X has the p. m. f. P(X= x)= 𝑥 𝑥 + 1 −1

, 𝑖𝑓 𝑥 = 1, 2, …


𝑏 𝑋 𝑕𝑎𝑠 𝑡𝑕𝑒 𝑝. 𝑑. 𝑓. 𝑓 𝑥 = (2𝑥2)−1, 𝑖𝑓 𝑋 > 1

0, 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒.

(c)X (Cauchy r. v. ) has the p. d. f. f(x)= 1

𝜋

1

1+𝑥2 ; −∞ < 𝑥 < ∞ .

[4] Find the mean and variance of the distributions having the following p. d. f. / p. m. f.

(a) 𝑓 𝑥 = 𝑎𝑥𝑎−1 , 0 < 𝑥 < 1, 𝑎 > 0

𝑏 𝑓 𝑥 =1

𝑛; 𝑥 = 1, 2, … , 𝑛; 𝑛 > 0 𝑖𝑠 𝑎𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟

𝑐 𝑓 𝑥 =3

2(𝑥 − 1)2; 0 < 𝑥 < 2

[5] Find the mean and variance of the Weibull random variable having the p. d. f.

f(x)= 𝑐

𝑎 𝑥−𝜇

𝑎 𝑐−1

exp − 𝑥−𝜇

𝑎 𝑐 𝑖𝑓 𝑥 > 𝜇


𝑤𝑕𝑒𝑟𝑒, 𝑐 > 0, 𝑎 > 0 𝑎𝑛𝑑 𝜇 ∈ −∞, ∞ .

[6] A median of a distribution is a value m such that P(X≥ m)≥ ½ and P(X≤ m)≥ ½ , with equality for a

continuous distribution. Find the median of the distribution with p. d. f. f(x)= 3𝑥2 , 0 < 𝑥 < 1; =

0, otherwise.

[7] Let X be a continuous, nonnegative random variable with d. f. F(x). Show that

E(X)= ∫ (1 − 𝑓(𝑥))𝑑𝑥∞

0.

[8] A target is made of three concentric circles of radii 1

3, 1, 3 feet. Shots within the inner circle give 4

points, within the next ring 3 points and within the third ring 2 points. Shots outside the target give 0. Let

X be the distance of the hit from the centre (in feet) and let the p. d. f. of X be

f(x)= 2

𝜋 1+𝑥2 𝑥 > 0


What is the expected vale of the score in a single shot?

[9] Find the moment generating function (m. g. f.) for the following distributions

(a) X is a (Binomial r. v.) discrete random variable with p. m. f.

𝑃 𝑋 = 𝑥 = 𝑛

𝑥 𝑝𝑥(1 − 𝑝)𝑛−𝑥 𝑥 = 0, 1, 2, … , 𝑛


n is a positive integer.

(b)X is a (Poisson r. v.) continuous random variable with p. d. f.

𝑓𝑥 𝑥 =

𝑒−𝑥/𝛽𝑥𝛼−1

⎾𝛼𝛽𝛼, 𝑥 > 0

0, 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒

Find E(X) and V(X) from the m. g. f. s.

[10] The m. g. f. of a random variable X is given by

𝑀𝑋 𝑡 =1

2𝑒−5𝑡 +

1

6𝑒4𝑡 +

1

8𝑒5𝑡 +

5

24𝑒25𝑡

Find the distribution function of the random variable.

[11] Let X be a random variable with P(X ≤ 0)= 0 and let 𝜇= E(X) exists. Show that P(X≥ 2𝜇) ≤ 0.5.

[12] Let X be a random variable with E(X)= 3 and E(𝑋2)= 13, determine a lower bound for P(-2 < X < 8).

[13] Let x be a random variable with p. m. f.

P(X= x)=

1

8 𝑥 = −1, 1

6

8 𝑥 = 0


Using the p. m. f. , show that the bound for Chebychev’ s inequality cannot be improved.

[14] A communication system consists of n components, each of which will independently function with

probability p. The system will be able to operate effectively if al least one half of its components function.

(a) For what value of a p a 5-component system is more likely to operate effectively than a 3-component

system?

(b) In general, when is a (2k + 1) –component system better than a (2k- 1) –component system?

[15] An interviewer is given a list of 8 people whom he can attempt to interview. He is required to

interview exactly 5 people. If each person(independently) agrees to be interviewed with probability 2/3,

what is the probability that his list will enable him to complete his task?

[16] A pipe-smoking mathematician carries at all times 2 match boxes, 1 in his left-hand pocket and 1 in

his right- hand pocket. Each time he needs a match he is equally likely to take it from either pocket.

Consider the moment when the mathematician first discovers that one of his matchboxes is empty. If it is

assumed that both matchboxes initially contained N matches, what is the probability that there are exactly

k matches in the other box, k= 0, 1, …, N?

Solution Key

(1) Let X denote the # of throws reqd. to get a 6

𝔛 = 1, 2, 3, …

𝑃 𝑋 = 𝑥 = 5

6

𝑥−1 1

6 𝑥 ∈ 𝔛

= 0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒

𝐸 𝑋 = 𝑥

∞

1

5

6

𝑥−1 1

6=

1

6 𝑥

∞

1

5

6

𝑥−1

= 1 + 2 5

6 + 3

5

6

2

+ ⋯ 1

6

=1

6 1 +

5

6+

5

6

2

+ ⋯ +5

6 1 +

5

6+

5

6

2

+ ⋯ + 5

6

2

1 +5

6+

5

6

2

+ ⋯ + ⋯

=1

6

1

1 −56

+5

6

1

1 −56

+ 5

6

2 1

1 −56

+ …

=1

6 6 1 +

5

6+

5

6

2+ ⋯ = 6

(2) X: length of run of heads or tails starting with trial 1

𝔛= {1, 2, …}

𝑃 𝑋 = 𝑥 = (1 − 𝑝)𝑥𝑝 + 𝑝𝑥(1 − 𝑝)

↑ ↑

Run of x 𝑇𝑠 run of x H

𝐸 𝑋 = 𝑥 1 − 𝑝 𝑥𝑝 + 𝑝𝑥 1 − 𝑝

∞

1

= 𝑝 1 − 𝑝 𝑥

∞

1

1 − 𝑝 𝑥−1 + 𝑥

∞

1

𝑝𝑥−1

= 𝑝 1 − 𝑝 1

𝑝2+

1

1 − 𝑝 2 ← 𝑎𝑠 𝑖𝑛 (1)

=1 − 2𝑝 + 2𝑝2

𝑝(1 − 𝑝) .

(3) (a) 𝐸 𝑋 = 𝑥 ∞1

1

𝑥 𝑥+1 =

1

𝑥+1∞1 𝑛𝑜𝑡 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑡

⟹ 𝐸 𝑋 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡

𝑏 𝐸 𝑋 = 𝑥 𝑥 >1

1

2𝑥2𝑑𝑥 = ∞ ⟹ 𝐸 𝑥 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡

(d) 𝐸 𝑋 = ∫ 𝑥

𝜋

∞

−∞

1

1+𝑥2 𝑑𝑥 =2

𝜋 ∫

𝑥

1+𝑥2

∞

0𝑑𝑥 =

1

𝜋 log 1 + 𝑥2 | ∞

0= ∞

⟹ 𝐸 𝑋 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡.

(4) Trial calculations

(a) e. g. 𝐸 𝑋 = 𝑎 ∫ 𝑥𝑎1

0𝑑𝑥 =

𝑎

𝑎+1; 𝐸 𝑋2 =

𝑎

𝑎+2

𝑉 𝑋 = 𝐸 𝑋2 − (𝐸(𝑋))2 =𝑎

𝑎 + 2−

𝑎

𝑎 + 1

2

= ⋯

(5) 𝐸 𝑋 =𝑐

𝑎∫ 𝑥

∞

𝜇 𝑥−𝜇

𝑎 𝑐−1

𝑒−

𝑥−𝜇

𝑎 𝑐

𝑑𝑥

𝑦 = 𝑥 − 𝜇

𝑎 𝑐

𝑑𝑦 =𝑐

𝑎 𝑥 − 𝜇

𝑎 𝑐−1

⟹ 𝐸 𝑋 = 𝑎𝑦1𝑐 + 𝜇 𝑒−𝑦

∞

0

𝑑𝑦 = 𝑎⎾1

𝑐+ 1 + 𝜇

𝐸 𝑋2 =𝑐

𝑎 𝑥2

𝑥 − 𝜇

𝑎 𝑐−1

𝑒−

𝑥−𝜇𝑎

𝑐∞

𝜇

𝑑𝑥

𝑎𝑠 𝑖𝑛𝐸 𝑋 = 𝑎𝑦1𝑐 + 𝜇

2∞

0

𝑒−𝑦 𝑑𝑦

= 𝑎2 2

𝑐+ 1 + 2𝑎𝜇

1

𝑐+ 1 + 𝜇2

𝑉 𝑋 = 𝐸𝑋2 − 𝐸𝑋 2

= 𝑎2 2

𝑐+ 1 + 2𝑎𝜇

1

𝑐+ 1 + 𝜇2 − 𝑎

1

𝑐+ 1 + 𝜇

2

= ⋯

(6)∫ 3𝑥2𝑚

0𝑑𝑥= ∫ 3

1

0𝑥2𝑑𝑥 =

1

2 ⟹ 𝑚 = ⋯

(7) ∫ 1 − 𝐹 𝑥 ∞

0𝑑𝑥 = ∫ ∫ 𝑓𝑥

∞

𝑥

∞

0 𝑦 𝑑𝑦 𝑑𝑥

0 < 𝑥 < 𝑦 < ∞

= 𝑓𝑥

∞

0

𝑦 𝑑𝑥𝑦

0

𝑑𝑦 = 𝑦∞

0

𝑓𝑥 𝑦 𝑑𝑦 = 𝐸(𝑋)

(8) Let Z denote the r. v. ∋

Z: Score in a shot 𝔛𝑧 = 0, 1, 2, 3 4

𝑃 𝑍 = = > 3) =2

𝜋

1

1 + 𝑥2

∞

3

𝑑𝑥 = 2

𝜋tan−1 𝑥 |

∞

3=

1

3

𝑃 𝑍 = 2 = 𝑃 1 < 𝑋 < 3 2

𝜋

1

1 + 𝑥2

3

1

𝑑𝑥 =1

6

P(Z=3) = P 1

3< 𝑥 < 1 =

2

𝜋∫

1

1+𝑥2

1

0𝑑𝑥 =

1

6

𝑃 𝑧 = 4 = 𝑃 0 < 𝑥 <1

3 =

2

𝜋

1

1 + 𝑥2

1

3

0

=1

3

Expendent score,

E(Z)=0 ×1

3+ 2.

1

6+ 3.

1

6+ 4 .

1

3=

1

3+

1

2+

4

3= ⋯

(9)(a) 𝑀𝑋 𝑡 = 𝐸 𝑒𝑡𝑋 = 𝑒𝑡𝑥𝑛0 𝑛

𝑥 𝑝𝑥 1 − 𝑝 𝑛−𝑥 = 𝑛

𝑥 𝑛

0 (𝑝𝑒𝑡)𝑥 1 − 𝑝 𝑛−𝑥

= (1 − 𝑝 + 𝑝𝑒𝑡)𝑛

𝑞 = 1 − 𝑝

𝑑

𝑑𝑡 𝑀𝑋 𝑡

𝑡 = 0𝑛 𝑞 + 𝑝𝑒𝑡 𝑛−1𝑝𝑒𝑡 𝑡 = 0

= 𝑛𝑝 = 𝐸(𝑥)

𝑑2

𝑑𝑡2𝑀𝑋 𝑡 = 𝑛 𝑛 − 1 𝑞 + 𝑝𝑒𝑡 𝑛−2 𝑝𝑒𝑡 2 + 𝑛 𝑞 + 𝑝𝑒𝑡 𝑛−1𝑝𝑒𝑡

𝑑2𝑀𝑋 𝑡

𝑑𝑡2|𝑡 = 0

= 𝑛 𝑛 − 1 𝑝2 + 𝑛𝑝 = 𝜇21 = 𝐸(𝑋2)

𝑉 𝑋 = 𝐸𝑋2 − (𝐸𝑋)2 = 𝑛 𝑛 − 1 𝑝2 + 𝑛𝑝 − 𝑛2𝑝2 = 𝑛𝑝 1 − 𝑝 = 𝑛𝑝𝑞

Sly (b)

(10) (i) 𝑋 ∼ 𝐺 𝛼, 𝛽

𝑀𝑋 𝑡 = 𝐸 𝑒𝑡𝑋 =1

⎾𝛼𝛽𝛼 𝑒𝑡𝑋

∞

0

𝑥𝛼−1𝑒−

𝑥𝛽𝑑𝑥

=1

⎾𝛼𝛽𝛼 𝑥𝛼−1𝑒

−𝑥 1𝛽

−𝑡 ∞

0

𝑑𝑥 𝑟𝑒𝑔𝑖𝑜𝑛 𝑜𝑓 𝑒𝑥𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡 <1

𝛽

=⎾𝛼

⎾𝛼𝛽𝛼.

1

1𝛽

− 𝑡 𝛼 =

1

𝛽− 𝑡

𝛼

𝐸 𝑋 =𝑑

𝑑𝑡𝑀𝑋 𝑡 |𝑡 = 0 = −𝛼 1 − 𝛽𝑡 −𝛼−1(−𝛽)|𝑡=0

= 𝛼𝛽

𝐸 𝑋2 =𝑑2𝑀𝑋 𝑡

𝑑𝑡2|𝑡=0

= 𝛼𝛽(− 𝛼 + 1 1 − 𝛽𝑡 −𝛼−2(−𝛽))|𝑡=0

= 𝛼 𝛼 + 1 𝛽2

𝑉 𝑋 = 𝐸 𝑋2 − ((𝐸𝑋))2 = 𝛼2𝛽2 + 𝛼𝛽2 − 𝛼2𝛽2 = 𝛼𝛽2

Sly (e)

(ii)𝑀𝑋 𝑡 = 𝑒−5𝑡 1

2+ 𝑒4𝑡 1

6+ 𝑒5𝑡 1

8+

5

24𝑒25𝑡

A 4 pt distn.

p. m. f.

𝑋 = 𝑥 − 5 4 5 25

𝑃 𝑋 = 𝑥 1

2

1

6

1

8

5

24

d. f. 𝐹𝑋 𝑥 =

0 𝑥 < −51

2 − 5 ≤ 𝑥 < 4

1

2+

1

6 4 ≤ 𝑥 < 5

1

2+

1

6+

1

8 5 ≤ 𝑥 < 25

1

2+

1

6+

1

8+

5

24= 1 𝑥 ≥ 25

(11) X is (+) ve values r. v. , by Markov’s inequality

𝑃 𝑋 ≥ 2𝜇 = 𝑃 𝑋 ≥ 2𝜇 ≤𝐸((𝑋))

2𝜇=

1

2

(12) 𝑃 −2 < 𝑋 < 8 = 𝑃 −2−3

2<

𝑋−𝐸 𝑋

𝑉 𝑋 <

8−3

2

= 𝑃 −5

2<

𝑋 − 𝐸 𝑋

𝑉 𝑋 <

5

2 = 𝑃 𝑋 − 𝜇 ≤

5

2 𝑉 𝑋

= 1 − 𝑃 𝑋 − 𝜇 ≥5

2 𝑉 𝑋

≥ 1 −𝑉 𝑋

254 . 𝑉 𝑋

𝑐𝑕𝑒𝑏𝑦𝑠𝑕𝑒𝑣 ′𝑠 𝑠𝑖𝑛𝑒𝑞𝑢𝑎𝑙

= 1 −4

25=

21

25

(13) 𝐸 𝑋 = −1

8+

1

8= 0 = 𝜇; 𝑉 𝑋 = 𝐸𝑋2 =

1

8+

1

8=

1

4= 𝜍2

By chebyshev’s inequality

∀ 𝑡 > 0 𝑃 𝑋 − 𝜇 ≥ 𝑡 ≤ 𝜍2

𝑡2

𝑖. 𝑒. 𝑃 𝑋 ≥ 𝑡 ≤1

4𝑡2

𝐴𝑙𝑠𝑜, 𝑃 𝑋 ≥ 𝑡 = 1

8+

1

8=

1

4 0 < 𝑡 ≤ 1

0 𝑡 > 1

𝑖. 𝑒. 𝑃 𝑋 ≥ 𝑡 =1

4∀ 𝑡 ∋ 0 < 𝑡 ≤ 1

⟹ for t= 1, bound from chebyshev’s inequality is attained exactly and hence cannot be improved

(14) Let X be the r. v. denoting the # of components functioning X∼ B(n, p)

P(system works effectively)= P(X≥ [n/2] +1).

(a) 𝑃 5 𝑐𝑜𝑚𝑝. 𝑠𝑦𝑠𝑡𝑒𝑚 𝑤𝑜𝑟𝑘𝑠 = 𝑝5

𝑖. 𝑒. 𝑝5 = 𝑝5 𝑋 ≥ 3 = 5

3 𝑝3 1 − 𝑝 2 +

5

4 𝑝4 1 − 𝑝 + 𝑝5

& 𝑝3 = 3

2 𝑝2 1 − 𝑝 + 𝑝3 = 𝑃 3 𝑐𝑜𝑚𝑝. 𝑠𝑦𝑠𝑡𝑒𝑚 𝑤𝑜𝑟𝑘𝑠

𝑝5 > 𝑝3

𝑖𝑓 5

3 𝑝3 1 − 𝑝 2 +

5

4 𝑝4 1 − 𝑝 + 𝑝5 >

3

2 𝑝2 1 − 𝑝 + 𝑝3

𝑠𝑖𝑚𝑝𝑙𝑦𝑓𝑦 𝑡𝑜 𝑔𝑒𝑡 𝑡𝑕𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑎𝑠 𝑝 >1

2

𝑏 𝑃2𝑘+1 𝑋 ≥ 𝑘 + 1 = 𝑝2𝑘+1

= 𝑃2𝑘−1 𝑋 ≥ 𝑘 + 1 + 𝑃2𝑘−1 𝑋 = 𝑘 𝑃2 𝑋 ≥ 1 + 𝑃2𝑘−1 𝑋 = 𝑘 − 1 𝑃2 𝑋 = 2

𝑖. 𝑒. 𝑝2𝑘+1 = 𝑃2𝑘+1 𝑋 ≥ 𝑘 + 1 + 𝑃2𝑘+1 𝑋 = 𝑘 1 − 1 − 𝑝 2 + 𝑃2𝑘−1 𝑋 = 𝑘 − 1 𝑝2

𝐹𝑢𝑟𝑡𝑕𝑒𝑟 𝑝2𝑘−1 = 𝑃2𝑘−1 𝑋 ≥ 𝑘 = 𝑃2𝑘−1 𝑋 = 𝑘 + 𝑃2𝑘−1 𝑋 ≥ 𝑘 + 1

𝑠𝑖𝑛𝑐𝑒 𝑝2𝑘−1 = 𝑃2𝑘−1 + 𝑃2𝑘−1 𝑋 = 𝑘 ͍ − 𝑃2𝑘−1 𝑋 = 𝑘 1 − 𝑝 2 + 𝑃2𝑘−1 𝑋 = 𝑘 − 1 𝑝2

⟹ 𝑝2𝑘+1 = 𝑝2𝑘−1 − 𝑃2𝑘−1 𝑋 = 𝑘 1 − 𝑝 2 + 𝑃2𝑘−1 𝑋 = 𝑘 − 1 𝑝2

⟹ 𝑝2𝑘+1 > 𝑝2𝑘−1 𝑖𝑓

𝑃2𝑘−1 𝑋 = 𝑘 − 1 − 𝑝 2 + 𝑃2𝑘−1 𝑋 = 𝑘 − 1 𝑝2 > 0

𝑖. 𝑒. 2𝑘 − 1

𝑘

𝑝𝑘 1 − 𝑝 𝑘−1 −1 − 𝑝2 + 2𝑝 +

2𝑘 − 1

𝑘 − 1 𝑝𝑘−1 1 − 𝑝 𝑘𝑝2 > 0

𝑖. 𝑒. 𝑝𝑘 1 − 𝑝 𝑘−1 −1 − 𝑝2 + 2𝑝 + 𝑝 − 𝑝2 > 0

𝑖. 𝑒. −1 − 2𝑝2 + 3𝑝 > 0

𝑖. 𝑒. 2𝑝 − 1 1 − 𝑝 > 0

𝑖. 𝑒. 𝑝 >1

2 𝑟𝑒𝑞𝑑 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛.

15 𝑋: # of intervals attempts to get 5 interviews,

𝑃 𝑋 = 𝑥 = 𝑥 − 1

4

2

3

4

1

3

𝑥−5

×2

3; 𝑥 = 5, 6, …

𝑟𝑒𝑞𝑑 𝑝𝑟𝑜𝑏 𝑃 𝑋 ≤ 8 = 𝑃 𝑋 = 5 + 𝑃 𝑋 = 6 + 𝑃 𝑋 = 7 + 𝑃 𝑋 = 8

= 4

4

2

3

5

+ 5

4

2

3

5

1

3

1

+ ⋯ + = ⋯

(16) P(selecting Box 1)= P(selecting box 2)= ½

Suppose Box 2 is found empty, then Box 2 has been chosen (n+1)th times, at it is time Box 1 contains

k matches if it has been chosen n-k times

Chosing Box 2≡ success.

Chosing Box 1≡ failure} Bernoulli trial p= ½

⟹ Box 2 found empty with k matches left in Box 1

≡ N- k failures preceding (N+ 1)th success

𝑝𝑟𝑜𝑏 = 𝑁+(𝑁−𝑘)𝑁

1

2 𝑁

1

2 𝑁−𝑘

×1

2= 2𝑁−𝑘

𝑁

1

2

2𝑁−𝑘+1

𝑠𝑙𝑦 Box 1 found empty with k matches in Box 2

𝑝𝑟𝑜𝑏 = 2𝑁 − 𝑘

𝑁

1

2

2𝑁−𝑘+1

⟹ 𝑟𝑒𝑞𝑑 𝑝𝑟𝑜𝑏 = 2𝑁 − 𝑘

𝑁

1

2

2𝑁−𝑘

Problem Set-5

[1] A machine contains two belts of different lengths. These have times to failure which are exponentially

distributed, with means 𝛼 and 2𝛼. The machine will stop if either belt fails. The failure of the belts are

assumed to be independent. What is the probability that the system performs after time 𝛼 from the start?

[2] Let X be a normal random variable with parameters 𝜇= 10 and 𝜍2 = 36. 𝐶𝑜𝑚𝑝𝑢𝑡𝑒

𝑎 𝑃 𝑋 > 5 , 𝑏 𝑃 4 < 𝑋 < 16 , 𝑐 𝑃 𝑋 < 8 .

[3] Let 𝑋 ∼ 𝑁 𝜇, 𝜍2 . 𝐼𝑓 𝑃 𝑋 ≤ 0 = 0.5 𝑎𝑛𝑑 𝑃 −1.96 ≤ 𝑋 ≤ 1.96 = 0.95, 𝑓𝑖𝑛𝑑 𝜇 𝑎𝑛𝑑 𝜍2 .

[4] It is assumed that the lifetime of computer chips produced by a certain semiconductor manufacturer

are normally distributed with parameters 𝜇= 1.4 × 106 𝑎𝑛𝑑 𝜍2 = 3 × 105 hours. What is the

approximate probability that a batch of 10 chips will contain at least 2 chips whose lifetime are less than

1.8 × 106 hours?

[5] Let X be a normal random variable with mean 0 and variance 1 i.e. N (0, 1). Prove that

𝑃 𝑋 > 𝑡 ≤ 2

𝜋 𝑒−𝑡2/2

𝑡; ∀ 𝑡 > 0

[6] Show that if X is a discrete random variable with values 0, 1, 2, … then

𝐸 𝑋 = (1 − 𝐹(𝑘))∞𝑘=0 , where F(x) is the distribution function of the random variable X.

[7] The cumulative distribution function of a random variable X defined over 0 ≤ 𝑥 < ∞ 𝑖𝑠 𝐹 𝑥 = 1 −

𝑒−𝛽𝑥2, 𝑤𝑕𝑒𝑟𝑒 𝛽 > 0. Find the mean, median and variance of X.

[8] Show that for any x > 0,1 − 𝜙 𝑥 ≤𝜙(𝑥)

𝑥, where 𝜙(x) is the c. d. f. and 𝜙(x) is the p. d. f. of standard

normal distribution.

[9] A point 𝑚0 is said to mode of a random variable X, if the p. m. f. or the p. d. f. of X has a maximum at

𝑚0. For the distribution given in problem [7], if 𝑚0 denotes the mode; 𝜇 and 𝜍2, the variance of the

corresponding random variable, then show that

𝑚0 = 𝜇 2

𝜋 𝑎𝑛𝑑 2𝑚0

2 − 𝜇2 = 𝜍2 .

[10] Let X be a poison random variable with parameter 𝜆. Find the probability mass function of Y=

𝑋2 − 5.

[11] Let X be a Binomial random variable with parameters n and p. Find the probability mass function of

Y= n- X.

[12] Consider the discrete random variable X with the probability mass function

𝑃 𝑋 = −2 =1

5, 𝑃 𝑋 = −1 =

1

6, 𝑃 𝑋 = 0 =

1

5,

𝑃 𝑋 = 1 =1

15, 𝑃 𝑋 = 2 =

10

30, 𝑃 𝑋 = 3 =

1

30.

Find the probability mass function of Y= 𝑋2.

[13] The probability mass function of the random variable X is given by

𝑃 𝑋 = 𝑥 = 1

3

2

3

𝑥

𝑥 = 0, 1, 2, …


Find the distribution of Y= X/ (X +1).

[14] Let X be a random variable with probability mass function

𝑃 𝑋 = 𝑥 =

𝑒−1, 𝑥 = 0

𝑒−1

2 𝑋 !, 𝑥 ∈ {±1, ±2, … }


Find the p. m. f. and distribution of the random variable Y= |X|.

……………………………………………………………………………………………………..

Useful data

𝜙 (1/3)= 0.6293, 𝜙(5/6)= 0.7967, 𝜙(1)= 0.8413, 𝜙(4/3) 0.918

……………………………………………………………………………………………………..

Solution Key

(1) Belt 1

X∼ Exp with mean 𝛼 ∼ 1

𝛼𝑒−𝑥/𝛼 ; 𝑥 > 0

Belt 2

Y∼ Exp with mean 2𝛼 ∼1

2𝛼𝑒−𝑥/2𝛼 ; 𝑥 > 0

P(system works beyond 𝛼)

P(X >𝛼∩ Y> 𝛼)= P(X > 𝛼) P(Y > 𝛼)

= ∫1

𝛼𝑒−

𝑥

𝛼∞

𝛼𝑑𝑥 ∗ ∫

1

2𝛼𝑒−

𝑥

2𝛼∞

𝛼𝑑𝑥 = 𝑒−1 × 𝑒−1/2 = 𝑒−3/2

(1) (a) 𝑃 𝑋 > 5 = 𝑃 𝑋−10

6>

5−10

6 = 𝑃 𝑍 > −

5

6 ; 𝑍 ∼ 𝑁 0, 1

= 1 − 𝜙 −5

6 = 1 − 1 − 𝜙

5

6

= 𝜙 5

6 = 0.7967

𝑏 𝑃 4 < 𝑋 < 16 = 𝑃 4 − 10

6< 𝑍 <

16 − 10

6 = 𝑃 −1 < 𝑍 < 1

= 𝜙 1 − 𝜙 −1 = 2𝜙 1 − 1

= ⋯

𝑐 𝑃 𝑋 < 8 = 𝑃 𝑍 <8 − 10

6 = 𝜙 −

1

3 = 1 − 𝜙

1

3 = ⋯

3 𝑃 𝑋 ≤ 0 =1

2= 𝑃 𝑋 ≥ 0 ⟹ 𝜇 = 0

𝑃 −1.96 ≤ 𝑋 ≤ 1.96 = 0.95

𝑃 −1.96

𝜍≤

𝑋

𝜍≤

1.96

𝜍 = 0.95

𝑃 −1.96

𝜍≤ 𝑍 ≤

1.96

𝜍 = 0.95; 𝑍 ∼ 𝑁(0, 1)

2𝜙 1.96

𝜍 − 1 = 0.5

𝜙 1.96

𝜍 = 0.975

⟹1.96

𝜍= 𝜙−1 0.975 = 1.96 ⟹ 𝜍 = 1

( 4) X : lifetime r. v.

X∼ 𝑁 𝜇, 𝜍2

= 1.4 × 106 𝑕𝑟𝑠

𝜍2 = 3 × 105𝑕𝑟𝑠

𝑃 𝑋 < 1.8 × 106

= 𝑋 − 1.4 × 106

3 × 105<

0.4 × 106

3 × 105

= 𝑃 𝑍 <4

3 𝑍 ∼ 𝑁 0, 1

= 𝜙 4

3 = 0.918

Y: r. v. denoting # of chips that have lifetime < 1.8 × 106𝑕𝑟

Y∼ Bin(10, 0.918)

⟹ P(Y≥ 2)= 1- P(Y< 2)

= 1-P(Y= 0) – P(Y = 1)

= 1- 100 . 918 ° 1 − .918 10 − 10

1 0.918 1 1 − .918 9

= ⋯

(5) 𝑋 ∼ N 0, 1

∀ t > 0 𝑃 𝑋 ≥ 𝑡 = 1 − P 𝑋 < 𝑡

= 1 − P −t < 𝑋 < 𝑡

= 1 − ϕ t − ϕ −t

= 1 − 2ϕ t − 1

= 1 − 2 1 − p X > 𝑡 − 1

= 2 − 2 + 2 P X > 𝑡 = 2 P X > 𝑡

P X > 𝑡 =1

2π e

−x2

2

∞

t

dx ≤1

2π

x

t

∞

t

e−x2

2 dx t < 𝑥 < ∞

y =x2

2

=1

2π 1

t e−y

∞

t2

2

dy = 1

2π e−t2

2

𝑡

⟹ P X ≥ t ≤ 2 1

2π e−t2

2

𝑡=

2

π e−t2

2

𝑡

(6) 𝑋 = 𝑥 0 1 2 …… ..

𝑃 𝑋 = 𝑥 𝑝0 𝑝1 𝑝2 …… . ..

1 − 𝐹 𝑘

∞

𝑘=0

= 𝑃 𝑋 > 𝑘

∞

𝑘=0

= 𝑃 𝑋 > 0 + 𝑃 𝑋 > 1 + 𝑃 𝑋 > 2 + ⋯

= 𝑝1 + 𝑝2 + 𝑝3 + ⋯ + 𝑝2 + 𝑝3 + ⋯ + 𝑝3 + 𝑝4 + ⋯

= 𝑝1 + 2𝑝2 + 3𝑝3 + ⋯

= 𝑖 𝑝𝑖

∞

𝑖=1

= 𝑖 𝑃(𝑋 = 𝑖)

𝑖=0

= 𝐸(𝑋)

(7) d. f.

F(x)= 0 , 𝑥 < 0

1 − 𝑒−𝛽𝑥2, 𝑥 ≥ 0

𝛽 > 0

𝑝. 𝑑. 𝑓. 𝑓 𝑥 = 2𝛽 𝑥𝑒−𝛽𝑥2

, 𝑥 ≥ 0

0 𝑜𝑤

𝐸 𝑋 = 2𝛽 𝑥2𝑒−𝛽𝑥2∞

0

𝑑𝑥

𝑦 = 𝑥2

= 𝛽 𝑦12𝑒−𝛽𝑦

∞

0

= 𝛽.⎾

32

𝛽32

=1

2

𝜋

3= 𝜇

𝐸𝑋2 = 2𝛽 𝑥3𝑒−𝛽𝑥2∞

0

𝑑𝑥 = 𝛽 𝑦 𝑒−𝛽𝑦∞

0

𝑑𝑦 = 𝛽⎾2

𝛽2=

1

𝛽

𝑉 𝑋 = 𝐸 𝑋2 − 𝐸𝑋 2 =1

𝛽− 𝜇2 =

1

𝛽−

𝜋

4𝛽

𝑚𝑒𝑑𝑖𝑎𝑛 ∶ 𝑚0

𝑚0 ⟹ 𝐹 𝑚0 =1

2= 1 − 𝐹 𝑚0

𝑖. 𝑒. 2𝛽 𝑥𝑚0

0

𝑒−𝛽𝑥2𝑑𝑥 = 2𝛽 𝑥𝑒−𝛽𝑥2

∞

𝑚0

𝑑𝑥 =1

2

𝑖. 𝑒. 1 − 𝑒−𝛽𝑚02

=1

2

⟹ 𝑚0 = ⋯

(8) 1 − 𝜙 𝑥 =1

2𝜋∫ 𝑒

−𝑦2

2∞

𝑥𝑑𝑦

= 1

2𝜋

1

𝑦

∞

𝑥

𝑦𝑒−𝑦2

2 𝑑𝑦

=1

2𝜋 1

𝑦. −𝑒

−𝑦2

2 ∞

𝑥− −

1

𝑦2 −𝑒

−𝑦2

2 ∞

𝑥

𝑑𝑦

=1

2𝜋 1

𝑥𝑒

−𝑥2

2 − 1

𝑦2

∞

0

𝑒−𝑦2

2 𝑑𝑦

≥ 0

⟹ 1 − 𝜙 𝑥 ≤1

𝑥

1

2𝜋𝑒

−𝑥2

2 =𝜙(𝑥)

𝑥.

(9)

Mode- pt at which f(x) is maximum.

𝑓 ′ 𝑥 = 2𝛽 𝑥𝑒−𝛽𝑥 2 −2𝛽𝑥 + 𝑒−𝛽𝑥 2

𝑓 ′ 𝑥 = 0 ⟹ 2𝛽𝑥2 = 1 ⟹ 𝑥 =1

2𝛽

𝑓 ′′ 𝑥 = 2𝛽𝑑

𝑑𝑥 𝑒−𝛽𝑥 2

(1−2𝛽𝑥2 )

= 2𝛽 𝑒−𝛽𝑥 2 −4𝛽𝑥 + 1 − 2𝛽𝑥2 𝑒−𝛽𝑥 2

−2𝛽𝑥

𝑓 ′′ (𝑥)|𝑥=

1

2𝛽

= 2𝛽 𝑒−12 −4

𝛽

2 < 0

𝛽 > 0

⟹ 𝑚∗, 𝑡𝑕𝑒 𝑚𝑜𝑑𝑒 𝑜𝑓 𝑡𝑕𝑒 𝑑𝑖𝑠𝑡𝑛 𝑖𝑠 𝑎𝑡 1

2𝛽.

𝑚∗ =1

2𝛽

𝜇 = 𝐸 𝑋 = 𝜋

2.

1

𝛽=

𝜋

2 2 𝑚∗

𝑖. 𝑒. 𝜇 = 𝜋

2 𝑚∗

& 2𝑚∗2 − 𝜇2 = 2 2

𝜋𝜇2 − 𝜇2

=4

𝜋𝜇2 − 𝜇2 =

4

𝜋 .𝜋

4 .1

𝛽− 𝜇2

𝑖. 𝑒. 2𝑚∗2 − 𝜇2 = 𝜍2 =1

𝛽− 𝜇2

= 𝐸𝑋2 − 𝜇2 = 𝑉 𝑋 .

(10) X∼ P(𝜆)

p. m. f. 𝑃 𝑋 = 𝑥 = 𝑒−𝜆𝜆𝑥

𝑥!, 𝑥 = 0, 1, 2, … .


𝑌 = 𝑋2 − 5 ⟹ 𝑟𝑎𝑛𝑔𝑒 𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 𝑌 = −5, −4, −1, 4, 11, … = 𝓎

𝑃 𝑌 = 𝑦 = 𝑃 𝑋2 − 5 = 𝑦 = 𝑃 𝑋2 = 𝑦 + 5

𝑝. 𝑚. 𝑓. 𝑜𝑓 𝑌: 𝑃 𝑌 = 𝑦 = 𝑃 𝑋 = 𝑦 + 5 =

𝑒−𝜆𝜆 𝑦+5

𝑦 + 5 ! , 𝑦 ∈ 𝓎


(11) X∼ B(n, p)

p. m. f. 𝑃 𝑋 = 𝑥 = 𝑛𝑥 𝑝𝑥 1 − 𝑝 𝑛−𝑥 , 𝑥 = 0, 1, … , 𝑛


𝑌 = 𝑛 − 𝑥 ⟹ 𝓎 = 0, 1, … , 𝑛 .

𝑃 𝑌 = 𝑦 = 𝑃 𝑛 − 𝑋 = 𝑦 = 𝑃 𝑋 = 𝑛 − 𝑦

⟹ 𝑝. 𝑚. 𝑓. 𝑜𝑓 𝑌

𝑃 𝑌 = 𝑦 =

𝑛

𝑛 − 𝑦 𝑝𝑛−𝑦 1 − 𝑝 𝑛− 𝑛−𝑦 ; 𝑦 = 0, 1, … , 𝑛


= 𝑛

𝑦 1 − 𝑝 𝑦𝑝𝑛−𝑦 , 𝑦 = 0, 1, … , 𝑛


⟹ 𝑌 ∼ 𝐵 𝑛, 1 − 𝑝 .

(12) 𝑌 = 𝑋2 → 𝑟𝑎𝑛𝑔𝑒 𝑟𝑝 = 0, 1, 4, 9

𝑝. 𝑚. 𝑓. 𝑃 𝑌 = 𝑦 =

𝑃 𝑋 = 0 𝑦 = 0

𝑃 𝑋 = −1 + 𝑃 𝑋 = 1 𝑦 = 1

𝑃 𝑋 = −2 + 𝑃 𝑋 = 2 𝑦 = 4

𝑃 𝑋 = 3 𝑦 = 9

=

1

5, 𝑦 = 0

1

6+

1

15, 𝑦 = 1

1

5+

1

3, 𝑦 = 4

1

30, 𝑦 = 9

(13) 𝑃 𝑋 = 𝑥 = 1

3

2

3

𝑥, 𝑥 = 0, 1, 2, … . .


𝑌 =𝑋

𝑋 + 1⟹ 𝑋 =

𝑌

1 − 𝑌

𝑟𝑎𝑛𝑔𝑒 𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 𝑌 = 0,1

2,2

3,3

4, …

𝑃 𝑌 = 𝑦 = 𝑃 𝑋

𝑋 + 1= 𝑦 = 𝑃 𝑋 =

𝑦

1 − 𝑦

= 1

3

2

3

𝑦1−𝑦

, 𝑦 = 0,1

2,2

3, …


(14) 𝑝. 𝑚. 𝑓 𝑜𝑓 𝑋

𝑃 𝑋 = 𝑥 =

𝑒−1, 𝑥 = 0

𝑒−1

2 𝑋 !, 𝑥 ∈ ±1, ±2, …


𝑌 = 𝑋 𝓎 = 0, 1, 2, …

𝑃 𝑌 = 0 = 𝑃 𝑋 = 0 = 𝑒−1

𝑃 𝑌 = 1 = 𝑃 𝑋 = −1 + 𝑃 𝑋 = 1

=𝑒−1

2+

𝑒−1

2= 𝑒−1

𝑃 𝑌 = 2 = 𝑃 𝑋 = −2 + 𝑃 𝑋 = 2

=𝑒−1

2.2!+

𝑒−1

2.2!=

𝑒−1

2

𝑠𝑙𝑦 𝑓𝑜𝑟 𝑘 = 1, 2, ….

𝑃 𝑌 = 𝑘 = 𝑃 𝑋 = −𝑘 + 𝑃 𝑋 = 𝑘

=𝑒−1

2. 𝑘!+

𝑒−1

2. 𝑘!=

𝑒−1

𝑘!

𝑝. 𝑚. 𝑓. 𝑜𝑓 𝑌

𝑃 𝑌 = 𝑦 =

𝑒−1

𝑦!, 𝑦 = 0, 1, 2, … .

0, 𝑜𝑡𝑒𝑟𝑤𝑖𝑠𝑒

𝑌 ∼ 𝑃 1 .

Problem Set-6

[1] The probability density function of the random variable X is

𝑓𝑋 𝑥 = 1 0 < 𝑥 < 10 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒

𝑖. 𝑒. 𝑋 ∼ 𝑈(0, 1). Find the distribution of the following functions of X

(a) 𝑌 = 𝑋 ; 𝑏 𝑌 = 𝑋2; 𝑐 𝑌 = 2𝑋 + 3; 𝑑 𝑌 = −𝜆 log 𝑋; 𝜆 > 0.

[2] Let X be a random variable with 𝑈 0, 𝜃 , 𝜃 > 0 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛. 𝐹𝑖𝑛𝑑 𝑡𝑕𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑌 =

min 𝑋,𝜃

2 .

[3] The probability density function of X is given by

𝑓𝑋 𝑥 = 1

2 −

1

2≤ 𝑥 ≤

3

2 0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒

Find the distribution of Y=𝑋2 .

[4] The probability density function of X is by

𝑓𝑋 𝑥 = 𝑘𝑥𝑝−1

1 + 𝑥 𝑝+𝑞 𝑥 > 0

0 𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒,

𝑝, 𝑞 > 0. Derive the distribution of Y= (1 + 𝑋)−1 .

[5] The probability density function of X is by

𝑓𝑋 𝑥 = 𝑘 𝑥𝛽−1 exp −𝛼𝑥𝛽 𝑥 > 0

0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒,

𝛼, 𝛽 > 0. 𝐷𝑒𝑟𝑖𝑣𝑒 𝑡𝑕𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑌 = 𝑋𝛽

[6] According to the Maxwell-Boltzmann law of theoretical physics, the probability density function of V,

the velocity of a gas molecule, is

𝑓𝑉 𝑣 = 𝑘 𝑣2 exp −𝛽𝑣2 𝑣 > 0


𝑤𝑕𝑒𝑟𝑒 𝛽> 0 is a constant which depends on the mass and absolute temperature of the molecule and k > 0

is a normalizing constant. Derive the distribution of the kinetic energy E= 𝑚𝑉2/2.

[7] The probability density function of the random variable X is

𝑓𝑋 𝑥 = 3

8(𝑥 + 1)2 − 1 < 𝑥 < 1


Find the distribution of the following functions of Y= 1- 𝑋2.

[8] Let X be a random variable with U(0, 1) distribution. Find the distribution function of Y= min (X, 1-

X) and the probability density function of Z= (1- Y)/ Y.

[9] Suppose X∼ N 𝜇, 𝜍2 , 𝜇 ∈ ℜ, 𝜍 ∈ ℜ+. Find the distribution of 2X – 6.

[10] Let X be a continuous random variable on (a, b) with p. d. f. f and c. d. f. F. Find the p. d. f. of Z= -

log (F(X)).

[11] Let X be a continuous r. v. having the following p. d. f.

f(x)= 6𝑥 1 − 𝑥 𝑖𝑓 0 ≤ 𝑥 ≤ 1

0 𝑖𝑓 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒.

Derive the distribution function of X and hence find the p. d. f. of Y= 𝑋2(3 − 2𝑋).

[12] Let X be a distributed as double exponential with p. d. f. f(x)

=1

2𝑒−|𝑥|; 𝑥 ∈ ℜ. 𝐹𝑖𝑛𝑑 𝑡𝑕𝑒 𝑝. 𝑑. 𝑓. 𝑜𝑓 𝑌 = |𝑋|

[13] 3 balls are placed randomly in 3 boxes𝐵1 , 𝐵2 𝑎𝑛𝑑𝐵3. Let N be the total number of boxes which are

occupied an 𝑋𝑖 be the total number of balls in the box𝐵𝑖 , i= 1, 2, 3. Find the joint p. m. f. of (N, 𝑋1) and

(𝑋1 , 𝑋2). Obtain the marginal distributions of N, 𝑋1𝑎𝑛𝑑 𝑋2 from the joint p. m. f. s.

[14] The joint p. m. f. of X and Y is given by

p(x, y)= 𝑐 𝑥𝑦 𝑖𝑓 𝑥, 𝑦 ∈ { 1, 1 , 2, 1 , 2, 2 , (3, 1)}


Find the constant c, the marginal p. m. f. of X and Y and the conditional p. m. f. of X given Y= 2.

[15] The joint p. m. f. of X and Y is given by

p(x, y)= (𝑥+2𝑦)

18 𝑖𝑓 𝑥, 𝑦 ∈ { 1, 1 , 1, 2 , 2, 1 , (2, 2)}


(a) Find the marginal distributions.

(b) Verify whether X and Y are independent random variables.

(c) Find P(X< Y), P(X+ Y > 2).

(d) Find the conditional p. m. f. of Y given X= x, x= 1, 2.

[16] 5 cards are drawn at random without replacement from a deck of 52 playing cards. Let the random

variables 𝑋1 , 𝑋2 , 𝑋3 denote the number of spades, the number of hearts, the number of diamonds,

respectively, that appear among the five cards. Find the joint p. m. f. of 𝑋1 , 𝑋2 , 𝑋3. Also determine

whether the 3 random variables are independent.

[17] Consider a sample of size 3 drawn with replacement from an urn containing 3 white , 2 black and 3

red balls. Let the random variables𝑋1 , 𝑎𝑛𝑑 𝑋2 denote the number of white balls and number of black

balls in the sample, respectively. Determine whether the two random variables are independent.

[18] Let 𝑋 = 𝑋1 , 𝑋2 , 𝑋3 𝑇 𝑏𝑒 𝑎 𝑟𝑎𝑛𝑑𝑜𝑚 𝑣𝑒𝑐𝑡𝑜𝑟 𝑤𝑖𝑡𝑕 𝑗𝑜𝑖𝑛𝑡 𝑝. 𝑚. 𝑓.

𝑓𝑋1 ,𝑋2 ,𝑋3 𝑥1, 𝑥2 , 𝑥3 =

1

4 𝑥1 , 𝑥2 , 𝑥3 ∈


X= {(1, 0, 0), (0, 1, 0), (0, 0, 1),(1, 1, 1)}. Show that 𝑋1 , 𝑋2 , 𝑋3 are pair wise independent but are not

mutually independent.

Solution Key

(1) X∼ U(0, 1)

𝑓𝑋 𝑥 = 0 𝑥 < 𝑈

𝑥 0 ≤ 𝑥 ≤ 11 𝑥 > 1

𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑢𝑠𝑖𝑛𝑠 𝑑. 𝑓. 𝑚𝑒𝑡𝑕𝑜𝑑

𝑎 𝑌 = 𝑋

𝑑. 𝑓. 𝑜𝑓 𝑌: 𝑓𝑌 𝑦 = 𝑃 𝑋 ≤ 𝑦

= 𝑃 𝑋 ≤ 𝑦2 =

0 𝑦 < 0

𝑦2 0 ≤ 𝑦 ≤ 11 𝑦 > 1

𝑝. 𝑑. 𝑓. 𝑜𝑓 𝑌𝑖𝑠 𝑓𝑌 𝑦 =

0, 𝑦 < 02𝑦, 0 ≤ 𝑦 ≤ 1

1, 𝑦 > 1

= 2𝑦, 0 ≤ 𝑦 ≤ 1


𝑏 𝑌 = 𝑋2

𝑑. 𝑓. 𝑜𝑓 𝑌: 𝑓𝑌 𝑦 = 𝑃 𝑋2 ≤ 𝑦 =

0, 𝑦 < 0

𝑃 − 𝑦 ≤ 𝑥 ≤ 𝑦 , 0 ≤ 𝑦 ≤ 1

1, 𝑦 > 1

𝑓𝑜𝑟 0 ≤ 𝑦 ≤ 1

𝑃 − 𝑦 ≤ 𝑋 ≤ 𝑦 = 𝑃 0 ≤ 𝑋 ≤ 𝑦 = 𝐹𝑋 𝑦 = 𝑦

⟹ 𝐹𝑌 𝑦 =

0, 𝑦 < 0

𝑦, 0 ≤ 𝑦 ≤ 1

1, 𝑦 > 1

𝑝. 𝑑. 𝑓 𝑜𝑓 𝑌: 𝑓𝑌 𝑦 =

1

2 𝑦 0 ≤ 𝑦 ≤ 1


𝑐 𝑌 = 2𝑋 + 3 → 3, 5

𝑑. 𝑓. 𝑜𝑓 𝑌: 𝐹𝑌 𝑦 = 𝑃 2𝑋 + 3 ≤ 𝑦

= 𝑃 𝑋 ≤𝑦 − 3

2 =

0, 𝑦 < 3𝑦 − 3

2, 3 ≤ 𝑦 ≤ 5

1, 𝑦 > 5

𝑝. 𝑑. 𝑓. 𝑓𝑌 𝑦 = 1

2, 3 ≤ 𝑦 ≤ 5


⟹ 𝑌 ∼ 𝑈 3, 5

𝑑 𝑌 = −𝜆 log 𝑋 → 0, ∞

𝐹𝑌 𝑦 = 𝑃 𝑌 ≤ 𝑦 = 𝑃 −𝜆 log 𝑋 ≤ 𝑦 = 𝑃 𝑋 > 𝑒−

𝑦𝜆

= 1 − 𝑃 𝑋 ≤ 𝑒−

𝑦𝜆

𝑖. 𝑒. 𝐹𝑌 𝑦 = 0 𝑦 < 0

1 − 𝑒−

𝑦𝜆 𝑦 ≥ 0

𝑝. 𝑑. 𝑓. 𝑜𝑓 𝑌: 𝑓𝑌 𝑦 = 1

𝜆𝑒

−𝑦𝜆 , 𝑦 ≥ 0


𝑖. 𝑒. 𝑌 ∼ 𝐸𝑥𝑝 𝜆 (𝑠𝑐𝑎𝑙𝑒 𝜆)

(2) X∼𝑈 0, 𝜃

𝑌 = min 𝑋,𝜃

2 → 0,

𝜃

2 ← 𝑟𝑎𝑛𝑔𝑒 𝑠𝑝 𝑜𝑓 𝑌

𝐹𝑌 𝑦 = 𝑃 𝑌 ≤ 𝑦 = 𝑃 min 𝑋,𝜃

2 ≤ 𝑦

= 1 − 𝑝 min 𝑋,𝜃

2 > 𝑦

= 1 − 𝑝 𝑋 > 𝑦 ∩𝜃

2> 𝑦

𝑛𝑜𝑤 𝑃 𝑋 > 𝑦,𝜃

2> 𝑦 = 1 𝑖𝑓 𝑦 < 0

= 0 𝑖𝑓 𝑦 ≥𝜃

2

𝑓𝑜𝑟 0 ≤ 𝑦 <𝜃

2; 𝑃 𝑋 > 𝑦,

𝜃

2> 𝑦 = 𝑃 𝑋 > 𝑦 =

1

𝜃 𝑑𝑥

𝜃

𝑦

=𝜃 − 𝑦

𝜃

⟹ 𝐹𝑌 𝑦 =

0, 𝑦 < 0𝑦

𝜃, 0 ≤ 𝑦 <

𝜃

2

1, 𝑦 ≥𝜃

2

𝑁𝑜𝑡𝑒: − 𝐹𝑌 𝑦 𝑕𝑎𝑠 𝑎 𝑗𝑢𝑚𝑝 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑢𝑖𝑡𝑦 𝑎𝑡 𝜃

2

(3) 𝑓𝑋 𝑥 = 1

2,

1

2≤ 𝑥 ≤

3

2


𝑋 ∼ 𝑈 −1

2,

3

2

𝑌 = 𝑋2 → 𝑦 ∈ 0,9

4

𝐹𝑌 𝑦 = 𝑃 𝑋2 ≤ 𝑦 = 𝑃 − 𝑦 ≤ 𝑋 ≤ 𝑦

𝑓𝑜𝑟 𝑢 < 0; 𝐹𝑌 𝑦 = 0

& 𝑦 >9

4; 𝐹𝑌 𝑦 = 1

𝑓𝑜𝑟, 0 ≤ 𝑦 ≤1

4; 𝐹𝑌 𝑦 =

1

2

𝑦

− 𝑦

𝑑𝑥 = 𝑦

𝑓𝑜𝑟,1

4< 𝑦 <

9

4; 𝐹𝑌 𝑦 = 0.

−12

− 𝑦

𝑑𝑥 + 1

2

𝑦

−12

𝑑𝑥 =1

2 𝑦 +

1

2

= 1

4+

𝑦

2

⟹ 𝐹𝑌 𝑦 =

0 𝑦 < 0

𝑦 0 ≤ 𝑦 ≤1

41

2 𝑦 +

1

2 ,

1

4≤ 𝑦 ≤

9

4

1, 𝑦 ≥9

4

𝑝. 𝑑. 𝑓. 𝑓𝑌 𝑦 =

1

2 𝑦, 0 ≤ 𝑦 ≤

1

4

1

4 𝑦,

1

4< 𝑦 ≤

9

4

(4) 𝑓𝑋 𝑥 = 𝑘

𝑥𝑝−1

1+𝑥 𝑝+𝑞 , 𝑥 > 0


𝑌 =1

1 + 𝑋⟹ 𝑋 =

1 − 𝑌

𝑌= 𝑔−1 𝑌 ; 𝑌 ∈ 0, 1

𝐽 =𝑑𝑥

𝑑𝑦= −

1

𝑦2

𝑓𝑌 𝑦 = 𝑓𝑋 𝑔−1 𝑦 𝐽 , 0 ≤ 𝑦 ≤ 1


= 𝑘. 1 − 𝑦

𝑦 𝑝−1

1

𝑦−1 𝑝+𝑞

.1

𝑦2, 0 ≤ 𝑦 ≤ 1


𝑖. 𝑒. 𝑓𝑌 𝑦 = 𝑘. 𝑦𝑞−1 1 − 𝑦 𝑝−1 , 0 ≤ 𝑥 ≤ 1


𝑘 = 𝐵𝑒𝑡𝑎 𝑞, 𝑝 −1

⟹ 𝑌 ∼ 𝐵𝑒𝑡𝑎 (𝑞, 𝑝)

(5) 𝑓𝑋 𝑥 = 𝑘 𝑥𝛽−1𝑒−𝛼𝑥𝛽, 𝑥 > 0


𝑌 = 𝑥𝛽 𝐽 =𝑑𝑥

𝑑𝑦=

1

𝛽𝑦

1𝛽

−1 𝑥 = 𝑦

1𝛽 = 𝑔−1 𝑦

𝑓𝑌 𝑦 = 𝑓𝑋𝑔−1 𝑦 𝐽 , 𝑦 > 0


= 𝑘. 𝑦1𝛽

𝛽−1

𝑒−𝛼𝑦 1

𝛽. 𝑦

1𝛽

−1 , 𝑦 > 0


= 𝑘𝑦1−

1𝛽𝑒−𝛼𝑦

1

𝛽. 𝑦

1𝛽

−1, 𝑦 > 0


= 𝑘

𝑒−𝛼𝑦

𝛽, 𝑦 > 0


𝑘 = 𝛼𝛽 ⟹ 𝑌 ∼ 𝐸𝑥𝑝 1

𝛼 .

(6) 𝑓𝑉 𝑣 = 𝑘𝑣2𝑒−𝛽𝑣 , 𝑣 > 00 , 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒

𝐸 =1

2𝑚𝑉2; 𝑉2 =

2𝐸

𝑚.

𝜕𝑒

𝜕𝑣= 𝑚𝑣

𝐽 = 𝜕𝑣

𝜕𝑒=

1

2𝑚𝑒.

𝑓𝐸 𝑒 = 𝑘.

2𝑒

𝑚 𝑒

−𝛽 2𝑒𝑚

.

1

2𝑚𝑒; 𝑒 > 0


= 𝑐. 𝑒12 exp −

2𝛽

𝑚𝑒 , 𝑒 > 0


𝑐 𝑖𝑠 ∋ 𝐶.∞

0

𝑒12 exp −

2𝛽

𝑚𝑒 𝑑𝑒 = 1

𝑖. 𝑒. 𝐶.⎾

32

2𝛽𝑚

32

= 1 ⟹ 𝐶 =

2𝛽𝑚

32

⎾32

⟹ 𝐸 ∼ 𝐺𝑎𝑚𝑚𝑎 (… )

(7) 𝑓𝑋 𝑥 = 3

8 𝑥+1 2 , −1<𝑥<1


𝑌 = 1 − 𝑋2 , 𝑦 ∈ 0, 1

𝑥2 = 1 − 𝑦 ⟹ 𝑥 = ± 1 − 𝑦

𝑥 ∈ −1, 0 → 𝑥 = − 1 − 𝑦 = 𝑔−11 𝑦 →

𝑑𝑥

𝑑𝑦 =

1

2 1 − 𝑦

𝑥 ∈ 0, 1 → 𝑥 = 1 − 𝑦 = 𝑔2−1 𝑦 →

𝑑𝑥

𝑑𝑦 =

1

2 1 − 𝑦

−1, 0 −1, 0 0, 1 (−1, 0)

↓ ↓ ↓ ↓

𝑓𝑌 𝑦 = 𝑓𝑋 𝑔1−1 𝑦

𝑑𝑥

𝑑𝑦 + 𝑓𝑋 𝑔2

−1 𝑦 𝑑𝑥

𝑑𝑦 . 0 < 𝑦 < 1

=3

8 1 − 1 − 𝑦

2.

1

2 1 − 𝑦+

3

8 1 + 1 − 𝑦

2.

1

2 1 − 𝑦

=3

16 1 − 𝑦 1 − 1 − 𝑦

2+ 1 + 1 − 𝑦

2

=3

16 1 − 𝑦 2 1 + 1 − 𝑦

𝑖. 𝑒. 𝑓𝑌 𝑦 =

3

8 1 − 𝑦 −

12 + 1 − 𝑦

12 , 0 < 𝑦 < 1


(8) Y= min (X, 1- X) → range of Y(0, ½ )

𝐹𝑌 𝑦 = 𝑃 𝑌 ≤ 𝑦 = 𝑃 min 𝑋, 1 − 𝑋 ≤ 𝑦 = 1 − 𝑃 min 𝑋, 1 − 𝑋 > 𝑦

= 1 − 𝑃 𝑋 > 𝑦, 1 − 𝑋 > 𝑦

= 1 − 𝑃 𝑋 > 𝑦, 1 − 𝑦 > 𝑋

= 1 − 𝑃 𝑦 < 𝑋 < 1 − 𝑦

𝑃 𝑦 < 𝑥 < 1 − 𝑦 =

1 𝑦 ≤ 0

𝑑𝑥1−𝑦

𝑦

𝑖𝑓 0 < 𝑦 <1

2

0 𝑦 ≥1

2

⟹ 𝐹𝑌 𝑦 =

0 𝑦 ≤ 0

2𝑦 0 < 𝑦 <1

2

1 𝑦 ≥1

2

⟹ 𝑝. 𝑑. 𝑓. 𝑓𝑌 𝑦 = 2, 0 < 𝑦 <1


𝑍 =1 − 𝑌

𝑌=

1

𝑌− 1 → 𝑟𝑎𝑛𝑔𝑒 𝑜𝑓 𝑍 𝑖𝑠 1, ∞

𝐹𝑍 Ʒ = 𝑃 𝑍 ≤ Ʒ

𝑖𝑓 Ʒ ≤ 1, 𝑡𝑕𝑒𝑛 𝐹𝑍 Ʒ = 0

𝑖𝑓 Ʒ > 1, 𝑡𝑕𝑒𝑛 𝑃 𝑍 ≤ Ʒ = 𝑃 1

𝑌− 1 ≤ Ʒ = 𝑃

1

𝑌≤ Ʒ + 1

= 𝑃 𝑌 ≥1

Ʒ + 1 = 1 − 𝑃 𝑦 <

1

Ʒ + 1

= 1 −2

Ʒ + 1 𝑢𝑠𝑖𝑛𝑔 𝑑. 𝑓. 𝑜𝑓 𝑌

⟹ 𝐹𝑍 Ʒ =

0, 𝑖𝑓Ʒ ≤ 1

1 −2

Ʒ + 1, 𝑖𝑓 Ʒ > 1

⟹ 𝑝. 𝑑. 𝑓. 𝑜𝑓 𝑍 𝑖𝑠 𝑓𝑍 Ʒ =

2

Ʒ + 1 2, Ʒ > 1


(9) X∼ N 𝜇, 𝜍2

𝑌 = 2𝑋 − 6; 𝑦 ∈ (−∞, ∞)

𝐹𝑌 𝑦 = 𝑃 𝑌 ≤ 𝑦 = 𝑃 2𝑋 − 6 ≤ 𝑦

= 𝑃 𝑋 ≤𝑦 + 6

2

= 𝑃 𝑋 − 𝜇

𝜍≤

𝑦 + 62

− 𝜇

𝜍 = 𝜙

𝑦 + 6 − 2𝜆

2𝜍

𝑓𝑌 𝑦 = 𝜙 𝑦 + 6 − 2𝜆

2𝜍 .

1

2𝜍. 𝑦 ∈ −∞, ∞

=1

2𝜋𝑒

12

𝑦 − 2𝜇 − 6

2𝜍

2

.1

2𝜍=

1

2𝜋 2𝜍 exp −

1

2 4𝜍2 𝑦 − 2𝜇 − 6

2

⟹ 𝑌 ∼ 𝑁(2𝜇 − 6, 4𝜍2)

(10) X∼𝑓𝑋 𝑥

𝑍 = − log 𝐹 𝑋 ; Ʒ ∈ 0, ∞

Ʒ = − log 𝐹 𝑋 ⟹ 𝑥 = 𝐹−1 𝑒−𝑧 𝜕𝑧

𝜕𝑥 =

𝑓 𝑥

𝐹 𝑥 ⟹ 𝐽 =

𝐹 𝑥

𝑓 𝑥

𝑝. 𝑑. 𝑓. 𝑜𝑓 𝑍 = 𝑓 𝐹−1 𝑒−𝑧 .𝐹 𝐹−1 𝑒−𝑧

𝑓 𝐹−1 𝑒−𝑧

= 𝐹 𝐹−1 𝑒−𝑧 = 𝑒−𝑧 ;

⟹ 𝑓𝑍 𝑧 = 𝑒−Ʒ , Ʒ > 0


(11) 𝑓𝑋 𝑥 = 6𝑥 1 − 𝑥 , 0 ≤ 𝑥 ≤ 1

0 𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒

𝐹𝑋 𝑥 = 𝑃 𝑋 ≤ 𝑥 =

0, 𝑥 < 0

6𝑦 − 6𝑦2 𝑥

0

𝑑𝑥, 0 ≤ 𝑥 ≤ 1

1, 𝑥 > 1

=

0, 𝑥 < 0

𝑥2 3 − 2𝑥 , 0 ≤ 𝑥 ≤ 11, 𝑥 > 1

𝐹𝑋 𝑋 = 𝑋2 3 − 2𝑋

If X ∼𝐹𝑋 𝑥 𝑑𝑖𝑠𝑡𝑕 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛, 𝑡𝑕𝑒𝑚 𝑌 = 𝐹 𝑋 ∼ 𝑈 0, 1

↓

[X ∼𝑓𝑋 𝑥 𝑝. 𝑑. 𝑓. & 𝑑. 𝑓. 𝐹.]

↑

Cont r. v. Y= F(X) →y ∈ (0, 1)[ General result as in prob. 10]

X= 𝐹−1 𝑦

𝑑𝑦

𝑑𝑥= 𝑓 𝑥

𝑑𝑥

𝑑𝑦 =

1

𝑓 𝑥

⟹ 𝑝. 𝑑. 𝑓. 𝑜𝑓 𝑌 ∶ 𝑓𝑌 𝑦 =

𝑓𝑋 𝐹−1 𝑦

𝑓𝑋 𝐹−1 𝑦 = 1 𝑖𝑓 0 < 𝑦 < 1


⟹ 𝑌 ∼ 𝑈 0, 1

⟹ 𝐹𝑜𝑟 𝑡𝑕𝑒 𝑔𝑖𝑣𝑒𝑛 𝑑𝑖𝑠𝑡𝑛 𝑋2 3 − 2𝑋 = 𝐹 𝑋 ∼ 𝑈(0, 1)

(12) X∼ Double exponential

𝑓𝑋 𝑥 =1

2𝑒− 𝑥 ; −∞ < 𝑥 < ∞

𝑌 = 𝑋 𝑟𝑎𝑛𝑔𝑒 𝑜𝑓 𝑌 ∶ 0, ∞

𝑥 ∈ −∞, 0 → 𝑥 = −𝑦 → 𝑑𝑥

𝑑𝑦 = 1

𝑥 ∈ 0, ∞ → 𝑥 = 𝑦 → 𝑑𝑥

𝑑𝑦 = 1

⟹ 𝑓𝑌 𝑦 = 𝑓𝑋 𝑔1−1 𝑦 𝐽 + 𝑓𝑋 𝑔2

−1 𝑦 𝐽

↑

In (-∞, 0)

𝑖. 𝑒. 𝑓𝑌 𝑦 = 1

2𝑒−𝑦 +

1

2𝑒−𝑦 , 0 < 𝑦 < ∞


∴ 𝑝. 𝑑. 𝑓. 𝑜𝑓 𝑌 ∶ 𝑓𝑌 𝑦 = 𝑒−𝑦 , 0 < 𝑦 < ∞


(13) 𝑋𝑖 = 0, 1, 2, 3 𝑓𝑜𝑟 𝑖 = 1, 2, 3

𝑁 = 1, 2, 3

Possible configurations with 3 boxes and 3 balls

𝐵1 𝐵2 𝐵3

3 0 0 →

0 3 0

0 0 3

2 1 0 → each with prob. = 1

3+3−13

=1

10

2 0 1

1 2 0

↓N 𝑋1 𝑋2 𝑋3

1

1

1

2

2

2

2

2

2

3

3 0 0

0 3 0

0 0 3

2 1 0

2 0 1

1 2 0

0 2 1

1 0 2

0 1 2

1 1 1

0 2 1

1 0 2

0 1 2

1 1 1

𝑗𝑡 𝑝. 𝑚. 𝑓. 𝑜𝑓 (𝑁, 𝑋1) 𝑗𝑡 𝑝. 𝑚. 𝑓. 𝑜𝑓 (𝑁, 𝑋1)

𝑋1 N

0 1 2 3

1

2

3

2

10 0 0

1

10

2

10

2

10

2

10 0

0 1

10 0 0

4

10

3

10

2

10

1

10

Marginal of 𝑋1 4

10

3

10

2

10

1

10

Marginal of 𝑋2

(14) 𝑝 𝑥, 𝑦 = 𝐶 𝑥𝑦 = 1 𝑥 ,𝑦 𝑥 ,𝑦

⟹ 𝐶 1, 1 + 2, 1 + 2, 2 + 3, 1 = 1

⟹ 𝐶 =1

10

𝑗𝑡 𝑝. 𝑚. 𝑓.

Marginal of Y

𝑋1

𝑋2 0 1 2 3

1

2

3

2

10 0 0

1

10

2

10

2

10

2

10 0

0 1

10 0 0

𝑌 X

1 2

1

2

3

1

10 0

2

10

4

10

3

10 0

1

10

6

10 } marginal of

X 3

10

6

10

4

10

Conditional p. m. f. of X given Y= 2, 𝑝 𝑥 ,2

𝑝𝑌 2 = 1 𝑖𝑓 𝑥 = 2

= 0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒.

(15) 𝑗𝑡 𝑝. 𝑚. 𝑓.

↔

Marg of Y

(b)P(X= 1, Y=1)= 3

18≠ 𝑃 𝑋 = 1 , 𝑃 𝑌 = 1 =

8

18.

7

18

⟹ 𝑋 & 𝑌 𝑚𝑎𝑔𝑖𝑛𝑎𝑙

𝑐 𝑃 𝑋 < 𝑌 = 𝑃 𝑋 = 1, 𝑌 = 2 =5

18

𝑃 𝑋 + 𝑌 > 2 = 𝑃 𝑋 = 1, 𝑌 = 2 + 𝑃 𝑋 = 2, 𝑌 = 1 + 𝑃 𝑋 = 2, 𝑌 = 2 =15

18

𝑑 𝑚𝑎𝑟𝑔 𝑜𝑓 𝑋: 𝑝𝑋 𝑥 = 𝑃 𝑋 = 𝑥 =𝑥 + 3

9; 𝑥 = 1, 2

𝑝𝑌|𝑋=𝑥 =

118

(𝑥 + 2𝑦)

118 (2𝑥 + 6)

=𝑥 + 2𝑦

2𝑥 + 6; 𝑦 = 1, 2

(16) 𝑖𝑓 𝑝. 𝑚. 𝑓. 𝑜𝑓 𝑋1 , 𝑋2 , 𝑋3

𝑝𝑋1 ,𝑋2 ,𝑋3 𝑥1 , 𝑥2 , 𝑥3 =

13𝑥1

13𝑥2

13𝑥3

135−𝑥1−𝑥2−𝑥3

525

; 𝑥𝑖 ≥ 0 & 𝑥𝑖 ≤ 5

3

1

𝑝𝑋𝑖 𝑥 =

13𝑥 39

5−𝑥

525

𝑥 = 0, 1, 2, 3, 4, 5

𝑝𝑋1 𝑥1 𝑝𝑋2

𝑥2 𝑝𝑋3 𝑥3 ≠ 𝑝𝑋1 ,𝑋2 ,𝑋3

𝑥1, 𝑥2 , 𝑥3

⟹(𝑋1 , 𝑋2 , 𝑋3) 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑖𝑛𝑑𝑒𝑝.

(17) 𝑋1: # 𝑜𝑓 𝑤𝑕𝑖𝑡𝑒 𝑏𝑎𝑙𝑙𝑠

𝑋2: # 𝑜𝑓 𝑏𝑙𝑎𝑐𝑘 𝑏𝑎𝑙𝑙𝑠.

𝑊, 2 𝐵 , 1 𝑅 − 7

𝑌 X

1 2

1

2

3

18

5

18

4

18

6

18

8

18

10

18 } marginal of

X

7

18

11

18

= 𝑝𝑋1 ,𝑋2 𝑥1 , 𝑥2 =

3!

𝑥1! 𝑥2! 3 − 𝑥1 − 𝑥2 !

3

8

𝑥1

2

8

𝑥2

3

8

3−𝑥1−𝑥2

; 𝑥𝑖 ≥ 0, 𝑥1 + 𝑥2

≤ 3

𝑋1 , 𝑋2 ∼ 𝑀𝑢𝑙𝑡 3,3

8,2

8

𝑝𝑋1 𝑥1 =

3

𝑥1

3

8

𝑥1

5

8

3−𝑥1

; 𝑥1 = 0, 1, 2, 3

𝑝𝑋2 𝑥2 =

3

𝑥2

2

8

𝑥1

6

8

3−𝑥2

; 𝑥2 = 0, 1, 2, 3

𝑖. 𝑒. 𝑋1 ∼ 𝐵 3,3

8 ; 𝑋2 ∼ 𝐵 3,

2

8

𝑝𝑋1 𝑥1 𝑝𝑋2

𝑥2 ≠ 𝑝𝑋1 ,𝑋2 𝑥1, 𝑥2

⟹ 𝑋1& 𝑋2 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑖𝑛𝑑𝑒𝑝.

(18) From the jt p. m. f. of 𝑋1 , 𝑋2

𝑃 𝑋1 = 0, 𝑋2 = 0 = 𝑃 𝑋1 = 0, 𝑋2 = 1 = 𝑃 𝑋1 = 1, 𝑋2 = 0 = 𝑃 𝑋1 = 1, 𝑋2 = 1 =1

4

𝐹𝑢𝑟𝑡𝑕𝑒𝑟 𝑋1 , 𝑋2 ≡ 𝑋1 , 𝑋3 ≡ 𝑋2 , 𝑋3

& 𝑃 𝑋𝑖 = 0 =1

2= 𝑃 𝑋𝑖 = 1 ; 𝑖 = 1, 2, 3

⟹ 𝑋1 , 𝑋2 , 𝑋3 𝑎𝑟𝑒 𝑝𝑎𝑖𝑟 𝑤𝑖𝑠𝑒 𝑖𝑛𝑑𝑒𝑝.

𝐵𝑢𝑡 𝑃 𝑋1 = 0, 𝑋2 = 0, 𝑋3 = 0 =1

4≠ 𝑃 𝑋1 = 0 𝑃 𝑋2 = 0 𝑃 𝑋3 = 0 =

1

8

⟹ 𝑋1 , 𝑋2 , 𝑋3 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑖𝑛𝑑𝑒𝑝.

Problem Set-7

[1] The joint p. d. f. of (X, Y) is given by f(x, y)= 4𝑥𝑦 0 < 𝑥 < 1, 0 < 𝑦 < 1


Find the probability p. d. f. s and verify whether the random variables are independent. Also find

P(0<X< ½ , ¼ <Y < 1), P(X+ Y <1)

[2] If the joint p. d. f. of (X, Y) f(x, y)= 𝑒−(𝑥+𝑦) 0 < 𝑥, 𝑦 < ∞


Show that X and Y are independent.

[3] If the joint p. d. f. of (X, Y) is f(x, y)= 2𝑒−(𝑥+𝑦) 0 < 𝑥 < 𝑦 < ∞


Show that X and Y are not dependent.

[4] Show that the random variables X and Y with joint p. d. f.

f(x, y)= 12𝑥𝑦 1 − 𝑦 0 < 𝑥 < 1, 0 < 𝑦 < 1


Are independent.

[5] Suppose the joint p. d. f. of (X, Y) is f(x, y)= 𝑐𝑥2𝑦 0 < 𝑥 < 𝑦 < 1


Find (a) the value of the constant c, (b) the marginal p. d. f. s of X and Y and (c) P(X+ Y ≤ 1).

[6] The joint p. d. f. of (X, Y) is given by f(x, y)= 6 1 − 𝑥 − 𝑦 𝑥 > 0, 𝑦 > 0, 𝑥 + 𝑦 < 1


Find the marginal p. d. f. of X and Y and P(2X + 3Y < 1).

[7] The joint p. d. f. of (X, Y) is f(x, y)= 𝑥 + 𝑦 0 < 𝑥 < 1, 0 < 𝑦 < 1


Find the conditional distribution of Y given X= x, 0< x<1; the conditional mean and conditional varience

of the conditional distribution.

[8] Suppose the conditional p. d. f. of X given Y = y is f(x|y)= 𝑐𝑥

𝑦2 0 < 𝑥 < 𝑦


Further, the marginal distribution of Y is g(y)= 𝑑𝑦4 0 < 𝑦 < 10 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒.

(a) Find the constants c and d.

(b) The p. d. f. of (X, Y).

(c) P(0.25 < X < 0.5) and P(0.25 < X < 0.5 |Y= 0.625)

[9] Let f(x) and g(y) be two arbitrary p. d. f. s with corresponding distribution functions F(x) and G(y)

respectively. Suppose the joint p. d. f. of X and Y is given by

h(x, y)= f(x) g(y)[1+𝛼 2𝐹 𝑥 − 1 {2𝐺 𝑦 − 1}], |𝛼|≤ 1.

Show that the marginal p. d. f. of X and Y are f(x) and g(y), respectively. Does there exist a value of 𝛼 for

which the random variables X and Y are independent?

[10] Suppose the marginal density of the random variable is 𝑓𝑋 𝑥 = 4𝑥 1 − 𝑥2 , 0 < 𝑥 < 1


And the conditional density of the random variable Y given X = x is

𝑓𝑌|𝑋=𝑥 𝑦|𝑥 =

2𝑦

1 − 𝑥2 , 𝑥 < 𝑦 < 1, 0 < 𝑥 < 1


Find the conditional p. d. f. of X given Y= y, E(X|Y= ½ ) and Var (X| Y = ½ ).

[11] The joint p. d. f. of (X, Y) f(x, y)= 𝑒− 𝑥+𝑦 0 < 𝑥, 𝑦 < ∞


Find the joint m. g. f. of (X, Y) and the m. g. f. of Z= X+ Y are have hence V(Z).

[12] Derive the joint m. g. f. of

𝑋1 , 𝑋2 ∼ 𝑁2 𝜇1 , 𝜇2 , 𝜍12 , 𝜍2

2, 𝜌 𝑎𝑛𝑑 𝑢𝑠𝑖𝑛𝑔 𝑡𝑕𝑒 𝑗𝑜𝑖𝑛𝑡 𝑚. 𝑔. 𝑓. 𝑓𝑖𝑛𝑑 𝜌 𝑋1 , 𝑋2 .

[13] Let the joint p. d. f. of (X, Y) be f(x, y)= 2 0 < 𝑥 < 𝑦 < 1


Find the conditional mean and conditional variance of X given Y= y and that of Y given X= x. Compute

further 𝜌 𝑋, 𝑌 .

[14] Let X, Y and Z be three random variables and a and b be two scalar constants. Prove that (a) Cov(X,

b)= Cov(Y, b)= 𝐶𝑜𝑣 𝑍, 𝑏 = 0; 𝑏 𝐶𝑜𝑣 𝑋, 𝑎𝑌 + 𝑏 = 𝑎 𝐶𝑜𝑣 𝑋, 𝑌 ; 𝑐 𝐶𝑜𝑣 𝑋, 𝑌 + 𝑍 = 𝐶𝑜𝑣 𝑋, 𝑌 +𝐶𝑜𝑣 𝑋, 𝑍 ; 𝑑 𝜌 𝑋, 𝑎𝑌 + 𝑏 = 𝜌(𝑋, 𝑌) for a>0.

[15] Let 𝑋1 , 𝑋2 , 𝑎𝑛𝑑𝑋3 be three independent random variables each with a variance

𝜍2 . 𝐷𝑒𝑓𝑖𝑛𝑒 𝑡𝑕𝑒 𝑛𝑒𝑤 𝑟𝑎𝑛𝑑𝑜𝑚 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠

𝑊1 = 𝑋1 , 𝑊2 = 3 − 1

2𝑋1 +

3 − 3

2𝑋2 𝑎𝑛𝑑 𝑊3

= 2 − 1 𝑋2 + 2 − 2 𝑋3 . 𝐹𝑖𝑛𝑑 𝜌 𝑊1 , 𝑊2 , 𝜌 𝑊1 , 𝑊3 𝑎𝑛𝑑 𝜌 𝑊2 , 𝑊3 .

[16] Let (X, Y)∼𝑁2 3, 1, 16, 25, 0.6 . Find (a)P(3< Y< 8); (b) P(3 < y< 8| X= 7); (c) P(-3 <X <3) and (d)

P(-3 <X < 3| Y= 4).

[17] Let (X, Y) ∼ 𝑁2 5, 10, 1, 25, 𝜌 𝑤𝑖𝑡𝑕 𝜌 > 0. 𝐼𝑓 𝑖𝑡 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑡𝑕𝑎𝑡 𝑃 4 < 𝑦 < 16 𝑋 = 5) =0.954 𝑎𝑛𝑑 𝜙 2 = 0.977, find the value of 𝜌.

[18] Let 𝑋1 , 𝑋2 , … , 𝑋20 be independent random variables with identical distributions, each with a mean 2

and variance 3. Define Y= 𝑋𝑖15𝑖=1 𝑎𝑛𝑑 𝑍 = 𝑋𝑖

20𝑖=11 . Find E(Y), E(Z), V(Y), V(Z) and 𝜌(Y, Z).

[19] Let X and Y be a jointly distributed random variables with E(X)= 15, E(Y)= 20, V(X)= 25, V(Y)=

100 and 𝜌(X, Y)= -0.6. Find 𝜌(X- Y, 2X – 3Y).

[20] suppose that the lifetime of light bulbs of a certain kind follows exponential distribution with p. d. f.

𝑓𝑋 𝑥 = 1

50𝑒−

𝑥50 𝑥 > 0


Find the probability that among 8 such bulbs, 2 will last less that 40 hours, 3 will last anywhere between

40 and 60 hours, 2 will last anywhere 60 and 80 hours and 1 will last for more than 80 hours. Find the

expected number of bulbs in a lot of 8 bulbs with lifetime between 60 and 80 hours and also the expected

number of bulbs in a lot of 8 with lifetime 60 and 80 hours, given that the number of bulbs with lifetime

anywhere between 40 and 60 hours is 2.

[21] Let the random variables X and Y have the following joint p. m. f. s

(a) P(X= x, Y= y)= 1/3, if (x, y)∈{(0, 0), (1, 1), (2, 2)} and 0 otherwise.

(b) P(X= x, Y= y)= 1/3 , if (x, y)∈ {(0, 2), (1, 1), (2, 0)} and 0 otherwise.

(c) P(X= x, Y= y)= 1/3 , if (x, y) ∈ {(0, 0), (1, 1), (2, 0)} and 0 otherwise.

In each of the above cases find the coefficient of correlation between X and Y.

[22] The joint p. m. f. of (X, Y) is

P(X= x, Y= y)= xy/10, if (x, y)∈ {(1, 1), (2, 1), (2, 2), (3, 1)} and 0 otherwise.

Find the joint m. g. f. of X and Y and the coefficient of correlation between X and Y. Using the joint m. g.

f. , find the p. m. f. Z= X+ Y.

[23] Let 𝑀𝑋,𝑌 𝑢, 𝑣 𝑑𝑒𝑛𝑜𝑡𝑒 𝑡𝑕𝑒 𝑗𝑜𝑖𝑛𝑡 𝑚. 𝑔. 𝑓. 𝑋, 𝑌 𝑎𝑛𝑑 𝛹 𝑢, 𝑣 = log 𝑀𝑋,𝑌 𝑢, 𝑣 . 𝑆𝑕𝑜𝑤 𝑡𝑕𝑎𝑡

𝜕𝛹 (𝑢 ,𝑣)

𝜕𝑢|𝑢=𝑣=0 ,

𝜕𝛹 (𝑢 ,𝑣)

𝜕𝑣|𝑢=𝑣=0 ,

𝜕2𝛹(𝑢 ,𝑣)

𝜕𝑢2 | 𝑢=𝑣=0 ,𝜕2𝛹(𝑢 ,𝑣)

𝜕𝑣2 |𝑢=𝑣=0𝑎𝑛𝑑 𝜕2𝛹(𝑢 ,𝑣)

𝜕𝑢𝜕𝑣|𝑢=𝑣=0 yeilds the mean, the

variance and the covariance of the two random variables.

𝑓𝑋,𝑌 𝑥, 𝑦 =1

2 𝑓𝜌 𝑥, 𝑦 + 𝑓−𝜌 𝑥, 𝑦 ; −∞ < 𝑥, 𝑦 < ∞

Where, 𝑓𝜌 𝑥, 𝑦 is the probability density function of 𝑁2 0, 0, 1, 1, 𝜌 𝑎𝑛𝑑 𝑓−𝜌 𝑥, 𝑦 is the probability

density function of 𝑁2 0, 0, 1, 1, −𝜌 . Find the marginal p. d. f. s of X and Y, the correlation coefficient

between X and Y. Are the 2 variables independent?

[25] Let the joint p. d. f. of X and Y be given by

𝑓𝑋,𝑌 𝑥, 𝑦 = 𝑘, 𝑖𝑓 − 𝑥 < 𝑦 < 𝑥; 0 < 𝑥 < 1


Find the value of the constant k and obtain the conditional expectations E(X|Y= y) and E(Y| X= x). Verify

whether the 2 random variables are independent and / or uncorrelated.

[26] The joint moment generating function of X and Y is given by

𝑀𝑋,𝑌 𝑠, 𝑡 = 𝑎 𝑒𝑠+𝑡 + 1 + 𝑏 𝑒𝑠 + 𝑒𝑡 , 𝑎, 𝑏 > 0; 𝑎 + 𝑏 =1

2.

Find the correlation coefficient between X and Y.

[27] Let X and Y be jointly distributed random variables with

E(X)= E(Y)= 0, E 𝑋2 = 𝐸 𝑌2 = 2 𝑎𝑛𝑑 𝜌 𝑋, 𝑌 =1

3

𝐹𝑖𝑛𝑑 𝜌 𝑋

3+

2𝑌

3,2𝑋

3+

𝑌

3 .

Solution Key

(1) 𝑗𝑡 𝑝. 𝑑. 𝑓. 𝑜𝑓 𝑋, 𝑌

𝑓𝑋,𝑌 𝑥, 𝑦 = 4𝑥𝑦 0 < 𝑥 < 1, 0 < 𝑦 < 1


𝑀𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑜𝑓 𝑋 ∶

𝑓𝑋 𝑥 = 4𝑥𝑦 𝑑𝑦1

0

= 2𝑥 0 < 𝑥 < 1


𝑆𝑙𝑦 𝑓𝑌 𝑦 = 2𝑦, 0 < 𝑦 < 1


𝑜𝑏𝑠𝑒𝑟𝑣𝑒 𝑡𝑕𝑎𝑡𝑓𝑋,𝑌 𝑥, 𝑦 = 𝑓𝑋 𝑥 𝑓𝑌 𝑦

⟹ 𝑋 & 𝑌 𝑎𝑟𝑒𝑤 𝑖𝑛𝑑𝑒𝑝.

𝑃 0 < 𝑋 <1

2,1

4< 𝑌 < 1 = 𝑃 0 < 𝑋 <

1

2 𝑃

1

4< 𝑌 < 1

= 2𝑥 𝑑𝑥

12

0

2𝑦 𝑑𝑦1

14

= ⋯

𝑃 𝑋 + 𝑌 < 1 = 𝑃 𝑋 < 1 − 𝑦 1

0

𝑓𝑌 𝑦 𝑑𝑦 → 𝑋 & 𝑌 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝.

= 2𝑥 𝑑𝑥1−𝑦

0

1

0

2𝑦 𝑑𝑦

= ⋯

(2) 𝑓𝑋 𝑥 = ∫ 𝑒−𝑥𝑒−𝑦∞

0𝑑𝑦 = 𝑒−𝑥 𝑥 > 0


𝑓𝑌 𝑦 = 𝑒−𝑦 𝑦 > 0

= 0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒 𝑓𝑋,𝑌 𝑥, 𝑦 = 𝑓𝑋 𝑥 𝑓𝑌 𝑦

⟹ 𝑋 & 𝑌 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝.

(3) 𝑓𝑋 𝑥 = ∫ 2𝑒−𝑥𝑒−𝑦∞

𝑥𝑑𝑦

= 2𝑒−𝑥 𝑒−𝑥 = 2𝑒−2𝑥 𝑥 > 0


𝑠𝑙𝑦 𝑓𝑌 𝑦 = 2 𝑒−𝑦𝑦

0

𝑒−𝑥𝑑𝑥 = 2 𝑒−𝑦 1 − 𝑒−𝑦 𝑦 > 0

= 0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒 𝑓 𝑥, 𝑦 ≠ 𝑓 𝑥 𝑓 𝑦

⟹ 𝑋 & 𝑌 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑖𝑑𝑒𝑝.

(4) 𝑓𝑋 𝑥 = 12𝑥 ∫ 𝑦 − 𝑦2 1

0𝑑𝑦 = 12𝑥

𝑦2

2−

𝑦3

3 | 1

0

⟹ 𝑓𝑋 𝑥 = 2𝑥 0 < 𝑥 < 10 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒

𝑓𝑌 𝑦 = 12𝑦 1 − 𝑦 𝑥1

0

𝑑𝑥 = 6 𝑦 1 − 𝑦 0 < 𝑦 < 1

= 0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒. 𝑓 𝑥, 𝑦 = 𝑓 𝑥 𝑓 𝑦

⟹ 𝑋 & 𝑌 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝.

(5) ∫ ∫ 𝑓 𝑥, 𝑦 1

𝑥

1

0𝑑𝑦 𝑑𝑥 = 1, 𝑖. 𝑒. 𝐶 ∫ 𝑥21

0 ∫ 𝑦1

𝑥𝑑𝑦 𝑑𝑥 = 1

⟹ 𝐶 𝑥21

0

1

2 1 − 𝑥2 𝑑𝑥 = 1

⟹𝑐

2 𝑥3

3−

𝑥5

5

1

0= 1 ⟹ 𝐶 = 15

(b) 𝑓𝑋 𝑥 = 15𝑥2 ∫ 𝑦1

𝑥 𝑑𝑦 =

15

2𝑥2 1 − 𝑥2 , 0 < 𝑥 < 1


𝑓𝑌 𝑦 = 15𝑦 𝑥2𝑦

0

𝑑𝑥 = 5𝑦4 , 0 < 𝑦 < 10, 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒

𝑐 𝑃 𝑋 + 𝑌 ≤ 1 = 15

𝑥+𝑦≤1

𝑥2𝑦 𝑑𝑦 𝑑𝑥

𝑥 < 𝑦

= 15 𝑥2

12

0

𝑦 1−𝑥

𝑥

𝑑𝑦 𝑑𝑥 = 15 𝑥2

12

0

𝑦2

2|

1 − 𝑥

𝑥𝑑𝑥

= ⋯ =15

192.

𝐴𝑙𝑡 𝑃 𝑋 + 𝑌 ≤ 1 = 15

𝑥+𝑦≤1

𝑥2𝑦 𝑑𝑦 𝑑𝑥

𝑥 < 𝑦

= 15 𝑦

12

0

𝑥2𝑦

0

𝑑𝑥 𝑑𝑦 + 15 𝑦1

12

𝑥21−𝑦

0

𝑑𝑥 𝑑𝑦

= ⋯ =15

15 × 32+

15

10 × 32=

15

192

(6) 𝑓𝑋 𝑥 = ∫ 𝑓𝑋,𝑌 𝑥, 𝑦 1−𝑥

0𝑑𝑦 = 6 ∫ 1 − 𝑥 − 𝑦

1−𝑥

0𝑑𝑦 = 6 1 − 𝑥 𝑦 −

𝑦2

2 1−𝑥

0

= 3 1 − 𝑥 2 , 0 < 𝑥 < 1


𝑠𝑙𝑦 𝑏𝑦 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑦

𝑓𝑌 𝑦 = 3 1 − 𝑦 2 , 0 < 𝑦 < 1


𝑃 2𝑋 + 3𝑌 < 1 = 6 1 − 𝑥 − 𝑦

1−2𝑥3

0

12

0

𝑑𝑦 𝑑𝑥

= 6 1 − 𝑥 𝑦 −𝑦2

2

12

0

1 − 2𝑥30

𝑑𝑥

= 6 1 − 𝑥 1 − 2𝑥

3 −

1

2

1 − 2𝑥

3

2

12

0

𝑑𝑥

= 6 1 + 2𝑥2 − 3𝑥

3−

1 + 4𝑥2 − 4𝑥

18

12

0

𝑑𝑥

= 6 8𝑥2 − 14𝑥 + 5

18

12

0

𝑑𝑥

=6

18 8

𝑥3

3− 14

𝑥2

2+ 5𝑥

120

=6

18

8

3.1

8 − 7 .

1

4+

5

2 =

13

36

𝐴𝑙𝑡 𝑃 2𝑋 + 3𝑌 < 1 = 6 1 − 𝑦 − 𝑥

1−3𝑦2

0

13

0

𝑑𝑥 𝑑𝑦

= 6 1 − 𝑦 𝑥 −𝑥2

2

1 − 3𝑦20

13

0

𝑑𝑦

= 6 1 − 𝑦 1 − 3𝑦

2 −

1

2

1 − 3𝑦

2

2

13

0

𝑑𝑦

= ⋯ =13

36

(7) 𝑓𝑋 𝑥 = ∫ 𝑓 𝑥, 𝑦 1

0 𝑑𝑦 = ∫ 𝑥 + 𝑦

1

0𝑑𝑦

= 𝑥 +1

2 0 < 𝑥 < 1

0 𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒

𝑓𝑌|𝑋 =𝑓 𝑥, 𝑦

𝑓𝑋 𝑥 =

𝑥 + 𝑦

12

2𝑥 + 1 =

2 𝑥 + 𝑦

2𝑥 + 1 0 < 𝑦 < 1

𝐸 𝑌|𝑋 = 𝑦1

0

2 𝑥 + 𝑦

2𝑥 + 1𝑑𝑦 =

2

2𝑥 + 1 𝑥𝑦 + 𝑦2

1

0

𝑑𝑦

=2

2𝑥 + 1 𝑥

2+

1

3 =

2 3𝑥 + 2

6 2𝑥 + 1 =

3𝑥 + 2

6𝑥 + 3

𝐸 𝑌2 𝑋 = 𝑦21

0

2 𝑥 + 𝑦

2𝑥 + 1𝑑𝑦 =

2

2𝑥 + 1 𝑦2𝑥 + 𝑦3

1

0

𝑑𝑦

=2

2𝑥 + 1 𝑥

3+

1

4 =

2 4𝑥 + 3

12 2𝑥 + 1 =

4𝑥 + 3

6 2𝑥 + 1

𝑉 𝑌 𝑋 = 𝐸 𝑌2 𝑋 − 𝐸2 𝑌 𝑋

=4𝑥 + 3

6(2𝑥 + 1)−

3𝑥 + 2

3 2𝑥 + 1

2

= ⋯

(8) f(x, y)= f(x| y) g(y)= c d x 𝑦2; 0 < 𝑥 < 𝑦, 0 < 𝑦 < 1

𝑔 𝑦 1

0

𝑑𝑦 = 1 ⟹ 𝑑 𝑦41

0

𝑑𝑦 = 1 ⟹ 𝑑 = 5

⟹ 𝑓 𝑥, 𝑦 = 5 𝑐 𝑥𝑦2; 0 < 𝑥 < 𝑦 < 1

⟹ 5𝑐 𝑦21

0

𝑥𝑦

0

𝑑𝑥 𝑑𝑦 = 1 ⟹5𝑐

2 𝑦4

1

0

𝑑𝑦 = 1

⟹ 𝑐 = 2

⟹ 𝑓 𝑥, 𝑦 = 10𝑥 𝑦2; 0 < 𝑥 < 𝑦 < 1


𝑓𝑋 𝑥 = 10𝑥 𝑦21

0

𝑑𝑦 0 < 𝑥 < 1

𝑓𝑋 𝑥 = 10

3𝑥 1 − 𝑥3 0 < 𝑥 < 1


𝑃 0.25 < 𝑋 < 0.5 =10

3 𝑥 − 𝑥4

12

14

𝑑𝑥 = ⋯

𝑃 1

4< 𝑋 <

1

2 𝑌 = 0.625 = 𝑓𝑋|𝑌=𝑦

12

14

𝑑𝑥

= 2 𝑥

(0.625)2

12

14

𝑑𝑥 =2

0.625 2

1

2

1

2

2

− 1

4

2

= ⋯

(9) Marginal of X from h (x, y)

𝑓𝑋 𝑥 = 𝑕 𝑥, 𝑦 ∞

−∞

𝑑𝑦 = 𝑓 𝑥 ∞

−∞

𝑔 𝑦 1 + 𝛼 2 𝐹 𝑥 − 1 2𝐺 𝑦 − 1 𝑑𝑦

𝑓 𝑥 = 𝑔 𝑦 ∞

−∞

𝑑𝑦 + 𝑓 𝑥 𝛼 2𝐹 𝑥 − 1 𝑔 𝑦 2𝐺 𝑦 − 1 ∞

−∞

𝑑𝑦

= 𝑓 𝑥 × 1 + 𝑓 𝑥 𝛼 2𝐹 𝑥 − 1 𝑔 𝑦 2𝐺 𝑦 − 1 ∞

−∞

𝑑𝑦

𝑔 𝑦 ∞

−∞

2𝐺 𝑦 − 1 𝑑𝑦 ≟ 2𝑢 − 1 𝑑𝑢1

0

= 2 𝑢2

2 − 𝑢|

1

0= 0

⟹ 𝑓𝑋 𝑥 = 𝑓 𝑥 + 0

𝑠𝑙𝑦 𝑓𝑌 𝑦 = 𝑕 𝑥, 𝑦 ∞

−∞

𝑑𝑥 = 𝑔 𝑦

𝑕 𝑥, 𝑦 = 𝑓𝑋 𝑥 . 𝑓𝑌 𝑦 = 𝑓 𝑥 𝑔 𝑦 𝑖𝑓𝑓 𝛼 = 0

(10) 𝑓𝑋,𝑌 𝑥, 𝑦 = 𝑓𝑌|𝑋=𝑥 𝑦 𝑥 𝑓𝑋 𝑥

= 8𝑥𝑦, 0 < 𝑥 < 𝑦 < 1


𝑀𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑝. 𝑑. 𝑓. 𝑜𝑓 𝑌

𝑓𝑌 𝑦 = 8𝑦 𝑥 𝑑𝑥

𝑦

0

= 4𝑦3 , 0 < 𝑦 < 1


𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙 𝑝. 𝑑. 𝑓 𝑜𝑓 𝑋 𝑔𝑖𝑣𝑒𝑛 𝑌

𝑓𝑋|𝑌=𝑦 𝑥 𝑦 =

8𝑥𝑦

4𝑦3=

2𝑥

𝑦3, 0 < 𝑥 < 𝑦; 0 < 𝑦 < 1


𝐸 𝑋 𝑌 = 𝑦 =2

𝑦2 𝑥2

𝑦

0

𝑑𝑥 =2

𝑦2−

𝑦3

3=

2𝑦

3

⟹ 𝐸 𝑋 𝑌 =1

2 =

1

3

𝐸 𝑋2 𝑌 = 𝑦 =2

𝑦2 𝑥3

𝑦

0

𝑑𝑥 = 2

𝑦2.𝑦4

4=

𝑦2

2

⟹ 𝐸 𝑋2 𝑌 =1

2 =

1

8

𝑉 𝑋 𝑌 = 𝑦 = 𝐸 𝑋2 𝑌 = 𝑦 − 𝐸2 𝑋 𝑌 = 𝑦

=1

8−

1

9=

1

72.

(11) Jt . m. g. f.

𝑀𝑋1 ,𝑋2 𝑡1 , 𝑡2 = 𝐸 𝑒𝑡1𝑋1+𝑡2𝑋2 = 𝑒𝑡1𝑥1+𝑡2𝑥2

∞

0

∞

0

𝑒− 𝑥1+𝑥2 𝑑𝑥2𝑑𝑥1

= 𝑒−𝑥2 1−𝑡1 ∞

0

𝑑𝑥1 𝑒−𝑥2 1−𝑡2 ∞

0

𝑑𝑥2

∞

0

= 1 − 𝑡1 −1 1 − 𝑡2

−1 𝑖𝑓𝑡1 , 𝑡2 < 1

𝑁𝑜𝑡𝑒: 𝑠𝑖𝑛𝑐𝑒 𝑋1& 𝑋2 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝, 𝑤𝑒 𝑤𝑟𝑖𝑡𝑒

𝑀𝑋1 ,𝑋2 𝑡1, 𝑡2 = 𝑀𝑋1

𝑡1 𝑀𝑋2 𝑡2

𝑚. 𝑔. 𝑓. 𝑜𝑓 𝑍 = 𝑋1 + 𝑋2

𝑀𝑍 𝑡 = 𝐸 𝑒𝑡 𝑋1+ 𝑋2 = 1 − 𝑡 −2 , 𝑡 < 1

𝐸 𝑧 =𝜕𝑀𝑍 𝑡

𝑑𝑡|𝑡=0 = 2 1 − 𝑡 −3|𝑡=0 = 2

𝐸 𝑧2 =𝜕2𝑀𝑍 𝑡

𝑑𝑡2|𝑡=0 = 6(1 − 𝑡)−4| 𝑡=0 = 6 ⟹ 𝑉 𝑧 = 2

(12) 𝑀𝑋1 ,𝑋2 𝑡1, 𝑡2 = 𝐸 𝑒𝑡1𝑋1+𝑡2𝑋2

= 𝐸𝐸 𝑒𝑡1𝑋1+𝑡2𝑋2 |𝑋1 = 𝐸 𝑒𝑡1𝑋1𝐸 𝑒𝑡2𝑋2 |𝑋1

𝑠𝑖𝑛𝑐𝑒 𝑋2 𝑋1 ∼ 𝑁 𝜇2 + 𝜌𝜍2

𝜍1

𝑥1 − 𝜇1 , 𝜍22 1 − 𝜌2

𝐸 𝑒𝑡2𝑋2 |𝑋1 → 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑎𝑡 𝑚. 𝑔. 𝑓. 𝑜𝑓 𝑋2 𝑔𝑖𝑣𝑒𝑛 𝑋1

𝑀𝑋1 ,𝑋2 𝑡1, 𝑡2 = 𝑒𝑡1𝑋2 𝑒

𝑡2 𝜇2+𝜌𝜍2𝜍1

𝑥1−𝜇1 +

𝑡22

2𝜍2

2 1 − 𝜌2

= 𝑒𝑡2𝜇2+𝑡2

2

2𝜍2

2 1−𝜌2 𝐸 𝑒𝑡1𝑋1+𝑡2𝜌

𝜍2𝜍1

𝑋1 𝑒−𝑡2𝜌

𝜍2𝜍1

𝜇1

= 𝑒𝑡2𝜇2+

𝑡22

2𝜍2

2 1−𝜌2 −𝑡2𝜌𝜍2𝜍1

𝜇1𝑋𝐸 𝑒 𝑡1+ 𝑡2𝜌

𝜍2𝜍1

𝑋1

= 𝑒𝑡2𝜇2+

𝑡22

2𝜍2

2 1−𝜌2 −𝑡2𝜌𝜍2𝜍1

𝜇1𝑒 𝑡1+ 𝑡2𝜌

𝜍2𝜍1

𝜇1 +𝜍1

2

2 𝑡1 + 𝑡2𝜌

𝜍2

𝜍1

= exp 𝑡2𝜇2 +𝑡2

2

2𝜍2

2 1 − 𝜌 2 − 𝑡2𝜌𝜍2

𝜍1

𝜇1 + 𝑡1𝜇1 + 𝑡2𝜌

𝜍2

𝜍1

+𝜍1

2

2 𝑡1

2 + 𝑡22𝜌 2

𝜍22

𝜍12

+ 2𝑡1 𝑡2𝜌𝜍2

𝜍1

= exp 𝑡2𝜇2 +𝑡2

2

2𝜍2

2 + 𝑡1𝜇1 +𝑡1

2

2𝜍1

2 + 𝑡1 𝑡2𝜌𝜍1𝜍2

= exp 𝑡1𝜇1 + 𝑡2𝜇2 +1

2 𝑡1

2𝜍12 + 𝑡2

2𝜍22 + 2𝑡1 𝑡2𝜍1𝜍2𝜌

𝜕𝑀𝑋1 ,𝑋2 𝑡1, 𝑡2

𝜕𝑡1|𝑡1=0,𝑡2=0 = 𝜇1 𝑠𝑙𝑦

𝜕𝑀𝑋1 ,𝑋2

𝜕𝑡2|𝑡1=0,𝑡2=0 = 𝜇2& 𝑉 𝑋1 = 𝜍1

2 , 𝑉 𝑋2 = 𝜍22

𝐸 𝑋1 , 𝑋2 =𝜕2𝑀𝑋1 ,𝑋2

𝑡1, 𝑡2

𝜕𝑡1 𝜕𝑡2|𝑡1=0,𝑡2=0 = 𝜌𝜍1𝜍2 + 𝜇1𝜇2

⟹ 𝐶𝑜𝑣 𝑋1 , 𝑋2 = 𝜌𝜍1𝜍2 + 𝜇1𝜇2 − 𝜇1𝜇2 = 𝜌𝜍1𝜍2

⟹ 𝑐𝑜𝑟𝑣 𝑋1 , 𝑋2 = 𝜌

(13) f(x, y)= 2, 0 < 𝑥 < 𝑦 < 1


𝑓𝑋 𝑥 = 21

𝑥

𝑑𝑦 = 2(1 − 𝑥), 0 < 𝑥 < 1


𝑓𝑌 𝑦 = 2𝑦

0

𝑑𝑥 = 2𝑦, 0 < 𝑦 < 1


𝑓𝑌|𝑋=𝑥 =

2

2 1 − 𝑥 , 𝑥 < 𝑦 < 1


𝑓𝑋|𝑌=𝑦 =

2

2𝑦, 0 < 𝑥 < 𝑦


𝐸 𝑌 𝑋 = 𝑦1

𝑥

1

1 − 𝑥𝑑𝑦 =

1 − 𝑥2

2 1 − 𝑥 =

1 + 𝑥

2

𝐸 𝑌2 𝑋 =∫ 𝑦21

𝑥1

1 − 𝑥𝑑𝑦 =

1 − 𝑥3

3 1 − 𝑥 .

⟹ 𝑉 𝑌 𝑋 = 𝐸 𝑌2 𝑋 − 𝐸2 𝑌 𝑋 =1 − 𝑥3

3 1 − 𝑥 −

1 + 𝑥

2

𝑠𝑙𝑦 𝐸 𝑋 𝑌 , 𝐸 𝑋2 𝑌 𝑎𝑛𝑑 𝑕𝑒𝑛𝑐𝑒 𝑉 𝑋 𝑌 . (14) (a) 𝐶𝑜𝑣 𝑋, 𝑏 = 𝐸 𝑋 − 𝐸 𝑋 𝑏 − 𝐸 𝑏 = 0 = 𝐶𝑜𝑣 𝑋, 𝑏 =

𝑍𝐶𝑜𝑣 𝑍, 𝑏

𝑏 𝐶𝑜𝑣 𝑋, 𝑎𝑌 + 𝑏 = 𝐸 𝑋 − 𝐸 𝑋 𝑎𝑌 + 𝑏 − 𝐸 𝑎𝑌 + 𝑏

= 𝐸 𝑋 − 𝐸 𝑋 𝑎𝑌 + 𝑏 − 𝑎𝐸 𝑌 − 𝑏

= 𝑎 𝐶𝑜𝑣 𝑋, 𝑌

𝑐 𝐶𝑜𝑣 𝑋, 𝑌 + 𝑍 = 𝐸 𝑋 − 𝐸 𝑋 𝑌 + 𝑍 − 𝐸 𝑌 − 𝐸 𝑍

= 𝐸 𝑋 − 𝐸 𝑋 𝑌 − 𝐸 𝑌 + 𝑧 − 𝐸 𝑧

= 𝐶𝑜𝑣 𝑋, 𝑌 + 𝐶𝑜𝑣 𝑋, 𝑧 𝑑 𝐶𝑜𝑣 𝑋, 𝑎𝑌 + 𝑏 = 𝑎 𝑐𝑜𝑣 𝑋, 𝑌

𝑐𝑜𝑣 𝑋, 𝑎𝑌 + 𝑏 =𝑐𝑜𝑣(𝑋, 𝑎𝑌 + 𝑏)

𝑉 𝑋 𝑉 𝑎𝑌 + 𝑏 12

=𝑎 𝑐𝑜𝑣(𝑥, 𝑌)

[𝑉(𝑋)𝑎2𝑉(𝑌)]12

= 𝑐𝑜𝑣 𝑋, 𝑌

(15) 𝑐𝑜𝑣 𝑤1 , 𝑤2 = 𝐶𝑜𝑣 𝑋1 , 3−1

2𝑋1 +

3− 3

2𝑋2 =

3−1

2𝑉 𝑋1 +

3− 3

2𝑐𝑜𝑣 𝑋1 , 𝑋2 =

3−1

2𝜍2

2

𝑉 𝑤1 = 𝜍2& 𝑉 𝑤2 = 3 − 1

2

2

𝜍2 + 3 − 3

2

2

𝜍2 = 3 − 1 2𝜍2

⟹ 𝜌𝑤1 ,𝑤2=

1

2

𝑠𝑙𝑦 𝜌𝑤1 ,𝑤3& 𝜌𝑤2 ,𝑤3

↓

𝑐𝑜𝑣 𝑤1 , 𝑤3 = 𝑐𝑜𝑣 𝑋1 , 2 − 1 𝑋2 + 2 − 2 𝑋3 = 0

⟹ 𝜌𝑤1 ,𝑤3= 0

(16) (a) 𝑃 3 < 𝑌 < 8 𝑌 ∼ 𝑁 1, 25

= 𝑃 3 − 1

5<

𝑌 − 1

5<

8 − 1

5 = 𝜙

7

5 − 𝜙

2

5

= ⋯ 𝑓𝑟𝑜𝑚 𝑡𝑎𝑏𝑙𝑒

𝑏 𝑃 3 < 𝑌 < 8 𝑋 = 7 𝑌 𝑋 ∼ 𝑁 1 + 𝑃5

4 𝑥 − 3 , 25 1 − 𝑝2

= 𝑃 3 − 4

4<

𝑌 − 4

4<

8 − 4

4 𝑋 = 7

= 𝜙 1 − 𝜙 −0.25

= ⋯ 𝑐 𝑃 −3 < 𝑋 < 3 𝑋 ∼ 𝑁 3, 16

= 𝑃 −3 − 3

4<

𝑋 − 3

4<

3 − 3

4 = 𝜙 0 − 𝜙 −

6

4 = ⋯

𝑑 𝑃 −3 < 𝑋 < 3 𝑌 = 4 𝑋 𝑌 ∼ 𝑁 3 + 𝑝4

5 𝑦 − 1 , 16 1 − 𝑝2

= 𝑃 −3 − 4.44

3.2<

𝑋 − 4.44

3.2<

3 − 4.44

3.2 𝑌 = 4) = 𝜙 −

1.44

3.2 − 𝜙 −

7.44

3.2 = ⋯

(17) 𝑋, 𝑌 ∼ 𝑁2 5, 10, 1, 25, 𝜌 ; 𝜌 > 0 𝑌|𝑋 = 5 ∼ 𝑁2 10, 25 1 − 𝜌2

𝑃 4 < 𝑌 < 16 𝑋 = 5 = 𝑃 4 − 10

5 1 − 𝜌2<

𝑌 − 10

5 1 − 𝜌2<

16 − 10

5 1 − 𝜌2 𝑋 = 5

= 𝜙 6

5 1 − 𝜌2 − 𝜙 −

6

5 1 − 𝜌2

= 2𝜙 6

5 1 − 𝜌2 − 1 = 0.954 𝑔𝑖𝑣𝑒𝑛 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛

⟹ 𝜙 6

5 1 − 𝜌2 = 0.977 = 𝜙 2

⟹6

5 1 − 𝜌2= 2 ⟹ 1 − 𝜌2 = 0.36 ⟹ 𝜌 = 0.8 𝑎𝑠 𝜌 > 0

(18) 𝐸 𝑌 = 𝐸 𝑋𝑖 151 = 30

𝑉 𝑌 = 15 𝑉 𝑋𝑖 = 45 ; 𝑉 𝑍 = 10 × 3 = 30

𝑐𝑜𝑣 𝑌, 𝑍 = 𝑐𝑜𝑣 𝑋𝑖

15

1

, 𝑋𝑖

10

11

= 5𝑉 𝑋𝑖 = 15

𝜌𝑌,𝑍 =15

[45 × 30]12

(19) 𝑈 = 𝑋 − 𝑌; 𝑉 = 2𝑋 − 3𝑌

𝐸 𝑈 = −5

𝐸 𝑉 = 2 × 15 − 3 × 20 = −30

𝑉 𝑈 = 𝑉 𝑋 + 𝑉 𝑌 − 2𝑐𝑜𝑣 𝑋, 𝑌 𝑉 𝑣 = 4𝑉 𝑋 + 9𝑉 𝑌 − 12 𝑐𝑜𝑣 𝑋, 𝑌

𝑁𝑜𝑤𝜌𝑋,𝑌 = −0.6 =𝑐𝑜𝑣 𝑋, 𝑌

25 × 100=

𝑐𝑜𝑣 𝑋, 𝑌

50⟹ 𝑐𝑜𝑣 𝑋, 𝑌 = −30

⟹ 𝑉 𝑈 = 185 & 𝑉 𝑣 = 100 + 900 + 360 = 1360

⟹ 𝑐𝑜𝑣 𝑈, 𝑉 = 𝑐𝑜𝑣 𝑋 − 𝑌, 2𝑋 − 3𝑌

= 2𝑉 𝑋 − 3 𝑐𝑜𝑣 𝑋, 𝑌 − 2𝑐𝑜𝑣 𝑌, 𝑋 + 3𝑉 𝑌 = 2 × 25 − 5 −30 + 30 × 100 = 500

⟹ 𝜌𝑈,𝑉 =500

185 × 1360

(20) X : r. v. denoting life time

𝑋 ∼ 𝐸𝑥𝑝 50 𝑝. 𝑑. 𝑓. 𝑓 𝑥 = 1

50𝑒−

𝑥50 , 𝑥 > 0


𝐹𝑋 𝑥 = 1 − 𝑒−𝑥

50 , 𝑥 > 0

𝑌1: # 𝑜𝑓 𝑏𝑢𝑖𝑏𝑠 𝑜𝑢𝑡 𝑜𝑓 8 𝑡𝑜 𝑕𝑎𝑣𝑒 𝑙𝑖𝑓𝑒𝑡𝑖𝑚𝑒 < 40

𝑌2 ∶ ………………… . . ≥ 40& < 60 𝑌3: ………………………… ≥ 60 & ≤ 80

𝑌4 ∶ ………………… . > 80

𝑃 𝑋 < 40 = 𝐹𝑋 40 = 1 − 𝑒−4050 = 𝑝1 , 𝑠𝑎𝑦

𝑃 40 ≤ 𝑋 < 60 = 𝐹𝑋 60 − 𝐹𝑋 40 = 𝑒−4050 − 𝑒−

6050 == 𝑝2 , 𝑠𝑎𝑦

𝑃 60 ≤ 𝑋 ≤ 80 = 𝐹𝑋 80 − 𝐹𝑋 60 = 𝑒−6050 − 𝑒−

8050 = 𝑝3 , 𝑠𝑎𝑦

𝑃 𝑋 > 80 = 1 − 𝑝1 − 𝑝2 − 𝑝1 = 𝑒−8050

𝑗𝑡. 𝑝. 𝑚. 𝑓. 𝑜𝑓 𝑌1 , 𝑌2 , 𝑌3 𝑖𝑠 𝑚𝑎𝑙𝑡𝑖𝑛𝑜𝑚𝑖𝑎𝑙 8, 𝑝1 , 𝑝2 , 𝑝3

⟹ 𝑃 𝑌1 = 2, 𝑌2 = 3, 𝑌3 = 2 =8!

2! 3! 22! 1!𝑝1

2𝑝23𝑝3

2 1 − 𝑝1 − 𝑝2 − 𝑝3

𝑚𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛𝑌3 ∼ 𝐵𝑖𝑛 8, 𝑝3 = 𝑒−6050 − 𝑒−

8050

𝐸 𝑌3 = 8 𝑒−6050 − 𝑒−

8050

𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙 𝑌3|𝑌2 = 𝑦2 ∼ 𝐵𝑖𝑛 8 − 𝑦2 ,𝑝3

1 − 𝑝2

⟹ 𝐸 𝑌3 𝑌2 = 1 = 8 − 1 𝑒−

6050 − 𝑒−

8050

1 − 𝑒−4050 − 𝑒−

6050

.

(21) (a)

Y

X

0 1 2

0

1

2

1

3 0 0

0 1

3 0

0 0 1

3

1

3

1

3

1

3

1

3

1

3

1

3

𝐸 𝑋 = 1 = 𝐸 𝑌

𝑉 𝑋 = 𝐸 𝑋2 − 1

=5

3− 1 =

2

3= 𝑉 𝑌

𝐸 𝑋𝑌 = 0 × 0 1

3+ 1 × 1

1

3+ 2 × 2 ×

1

3=

5

3

𝑐𝑜𝑣 𝑋, 𝑌 =5

3− 1 =

2

3

𝜌𝑋,𝑌 = 1

𝑏

Y

X

0 1 2

0

1

2

0 0 1

3

0 1

3 0

1

3 0 0

𝑠𝑙𝑦 ⟹ 𝜌𝑋 ,𝑌 = −1

(c)

Y

X

0 1 2

0

1

2

0 0 1

3

0 1

3 0

1

3 0 0

𝜌𝑋,𝑌 = 0.

(22)

Y

X

1 2

1

2

3

1

10 0

2

10

4

10

3

10 0

1

10

6

10

3

10

6

10

4

10

𝐸 𝑋 =1

10+ 2

6

10+ 3

3

10=

22

10

𝐸 𝑌 =6

10+ 2

4

10=

14

10

𝐸 𝑋2 =1

10+ 4

6

10+ 9

3

10=

52

10

𝐸 𝑌2 =6

10+ 4

4

10=

22

10

𝑉 𝑋 =52

10−

22

10

2

= ⋯

𝑉 𝑌 = 22

10

2

− 14

10

2

= ⋯

𝐸 𝑋𝑌 = 1 × 1 1

10+ 2 × 1

2

10+ 2 × 2

4

10+ 3 × 1

3

10=

30

10= 3

𝑐𝑜𝑣 𝑋, 𝑌 = 𝐸 𝑋𝑌 − 𝐸 𝑋 𝐸 𝑌 = 3 −22

10.14

10= ⋯

𝑐𝑜𝑣 𝑋, 𝑌 =𝑐𝑜𝑣 𝑋, 𝑌

𝑉 𝑋 𝑉 𝑌 12

= ⋯

𝑗𝑡 𝑚. 𝑔. 𝑓. 𝑜𝑓 𝑋, 𝑌

𝑀𝑋,𝑌 𝑡1 , 𝑡2 = 𝑒𝑡1𝑥+𝑡2𝑦

𝑥 ,𝑦

𝜌 𝑋 = 𝑥, 𝑌 = 𝑦

= 𝑒𝑡1+𝑡2 ×1

10+ 𝑒2𝑡1+𝑡2

2

10+ 𝑒2(𝑡1+𝑡2)

4

10+ 𝑒3𝑡1+𝑡2 .

3

10.

(23) 𝑀𝑋,𝑌 𝑢, 𝑣 = 𝐸 𝑒𝑢𝑋+𝑣𝑌

= 𝛹 𝑢, 𝑣 = 𝑙0𝑔𝑀𝑋 ,𝑌 𝑢, 𝑣 𝜕𝛹 𝑢, 𝑣

𝜕𝑢=

1

𝑀𝑋 ,𝑌 𝑢, 𝑣 .𝜕𝑀𝑋,𝑌 𝑢, 𝑣

𝜕𝑢

𝜕𝛹 𝑢, 𝑣

𝜕𝑢|𝑢=0,𝑣=0 =

1

𝑀 0, 0 . 𝐸 𝑋 = 𝐸 𝑋

𝑠𝑙𝑦 𝜕𝛹 0, 0

𝜕𝑣=

𝜕𝛹 𝑢, 𝑣

𝜕𝑣|𝑢=0,𝑣=0 = 𝐸 𝑌

𝜕2𝛹 𝑢, 𝑣

𝜕𝑢2=

1

𝑀 𝑢, 𝑣 𝜕2𝑀 𝑢, 𝑣

𝜕𝑢2+

−1

𝑀 𝑢, 𝑣 2

𝜕𝑀 𝑢, 𝑣

𝜕𝑢

𝜕𝑀 𝑢, 𝑣

𝜕𝑢

= 1

𝑀 𝑢, 𝑣 𝜕2𝑀 𝑢, 𝑣

𝜕𝑢2−

𝜕𝑀 𝑢, 𝑣

𝜕𝑢.

1

𝑀 𝑢, 𝑣

2

𝜕2𝛹 𝑢, 𝑣

𝜕𝑢2|𝑢=0,𝑣=0 = 𝐸 𝑋2 − 𝐸2 𝑋 = 𝑉 𝑋 =

𝜕2𝛹 0, 0

𝜕𝑢2

𝑠𝑙𝑦𝜕2𝛹 𝑢, 𝑣

𝜕𝑢2|𝑢=0,𝑣=0 = 𝑉 𝑌

𝜕2𝛹 𝑢, 𝑣

𝜕𝑣 𝜕𝑢=

1

𝑀 𝑢, 𝑣 .𝜕2𝑀 𝑢, 𝑣

𝜕𝑣 𝜕𝑢−

1

𝑀 𝑢, 𝑣 2 .

𝜕𝑀 𝑢, 𝑣

𝜕𝑣.𝜕𝑀 𝑢, 𝑣

𝜕𝑢

𝜕2𝛹 𝑢, 𝑣

𝜕𝑣 𝜕𝑢|𝑢=0,𝑣=0 = 𝐸 𝑋𝑌 − 𝐸 𝑋 𝐸 𝑌

𝑖. 𝑒.𝜕2𝛹 𝑢, 𝑣

𝜕𝑣 𝜕𝑢|𝑢=0,𝑣=0 = 𝑐𝑜𝑣 𝑋, 𝑌 .

(24) 𝑀𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑝. 𝑑. 𝑓. 𝑜𝑓 𝑋

𝑓𝑋 𝑥 = 𝑓𝑋,𝑌

∞

−∞

𝑥, 𝑦 𝑑𝑦

=1

2 𝑓𝜌

∞

−∞

𝑥, 𝑦 𝑑𝑦 +1

2 𝑓−𝜌

∞

−∞

𝑥, 𝑦 𝑑𝑦

=1

2𝜙 𝑥 +

1

2𝜙 𝑥 𝜙 𝑥 𝑝. 𝑑. 𝑓. 𝑜𝑓 𝑁 0, 1

= 𝜙 𝑥 ⟹ 𝑋 ∼ 𝑁 0, 1

𝑠𝑙𝑦 𝑓𝑌 𝑦 = 𝜙 𝑦 ⟹ 𝑌 ∼ 𝑁 0, 1

𝐸 𝑋𝑌 = 𝑥𝑦∞

−∞

∞

−∞

𝑓𝑋,𝑌 𝑥, 𝑦 𝑑𝑥 𝑑𝑦

=1

2 𝑥𝑦

∞

−∞

∞

−∞

𝑓𝜌 𝑥, 𝑦 𝑑𝑥 𝑑𝑦 +1

2 𝑥𝑦

∞

−∞

∞

−∞

𝑓−𝜌 𝑥, 𝑦 𝑑𝑥 𝑑𝑦

=1

2 𝜌 +

1

2 −𝜌 = 0

𝑐𝑜𝑣 𝑋, 𝑌 = 𝐸 𝑋𝑌 − 𝐸 𝑋 𝐸 𝑌 = 0 − 0.0 = 0

𝜌 𝑋, 𝑌 = 0 ⟹ 𝑋 & 𝑌 𝑎𝑟𝑒 𝑢𝑛𝑐𝑟𝑟𝑒𝑙𝑎𝑡𝑒𝑑

𝑠𝑖𝑛𝑐𝑒, 𝑓𝑋,𝑌 𝑥, 𝑦 ≠ 𝑓𝑋 𝑥 𝑓𝑌 𝑦 .

𝑋 & 𝑌 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 .

(25) ∫ ∫ 𝑘𝑥

−𝑥

1

0 𝑑𝑦 𝑑𝑥 = 1 ⟹ 𝑘 ∫ 2𝑥 𝑑𝑥

1

0= 1 ⟹ 𝑘 = 1

𝑀𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑜𝑓 𝑋, 𝑓𝑋 𝑥 = 𝑘 𝑑𝑦𝑥

−𝑥

= 2𝑥, 0 < 𝑥 < 10 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒

𝑀𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑜𝑓 𝑌, 𝑓𝑌 𝑦 = 𝑑𝑥1

𝑦

= 1 − 𝑦 , −1 < 𝑦 < 1


𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑌|𝑋 = 𝑥; 𝑓𝑌|𝑋=𝑥 = 1

2𝑥, −𝑥 < 𝑦 < 𝑥


𝐸 𝑌 𝑋 = 𝑥 = 𝑦 1

2𝑥𝑑𝑦

𝑥

−𝑥

= 0

𝑠𝑙𝑦 𝐸 𝑋 𝑌 = 𝑦 = 𝑥1

𝑦

1

1 − 𝑦 𝑑𝑥 =

1 − 𝑦2

2 1 − 𝑦 .

𝑓𝑋|𝑌=𝑦 = 1 − 𝑦 −1 , 𝑦 < 𝑥 < 1


𝑓𝑋,𝑌 𝑥, 𝑦 = 1 ≠ 𝑓𝑋 𝑥 𝑓𝑌 𝑦

⟹ 𝑋 & 𝑌 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑖𝑛𝑑𝑒𝑝.

𝐸 𝑋𝑌 = 𝑥𝑦∞

−∞

1

0

𝑑𝑦 𝑑𝑥 = 0

𝐸 𝑌 = 𝐸. 𝐸 𝑌 𝑋 = 0

⟹ 𝑐𝑜𝑣 𝑋, 𝑦 = 𝜌𝑋 ,𝑌 = 0

⟹ 𝑋 & 𝑌 𝑎𝑟𝑒 𝑢𝑛𝑐𝑜𝑟𝑟𝑒𝑙𝑎𝑡𝑒𝑑.

(26) 𝑀𝑋,𝑌 𝑠, 𝑡 = 𝑎 𝑒𝑠+𝑡 + 1 + 𝑏 𝑒𝑠 + 𝑒𝑡 , 𝑎, 𝑏 > 0, 𝑎 + 𝑏 =1

2

𝐸 𝑋 =𝜕

𝜕𝑠(𝑎 𝑒𝑠+𝑡 + 1 + 𝑏 𝑒𝑠 + 𝑒𝑡 )|𝑠=𝑡=0

= 𝑎 𝑒𝑡 𝑒𝑠 + 𝑏 𝑒𝑠|𝑡=𝑠=0 = 𝑎 + 𝑏 =1

2= 𝐸 𝑌

𝐸 𝑋2 =𝜕2

𝜕𝑠2 𝑎 𝑒𝑠+𝑡 + 1 + 𝑏 𝑒𝑠 + 𝑒𝑡 |𝑠=𝑡=0

= 𝑎 𝑒𝑡 𝑒𝑠 + 𝑏 𝑒𝑠|𝑠=𝑡=0 = 𝑎 + 𝑏 =1

2= 𝐸 𝑌2

𝑉 𝑋 = 𝑉 𝑌 =1

2−

1

4=

1

4

𝐸 𝑋𝑌 =𝜕2

𝜕𝑡 𝜕𝑠 𝑎 𝑒𝑠+𝑡 + 1 + 𝑏 𝑒𝑠 + 𝑒𝑡 |𝑠=𝑡=0

= 𝑎 𝑒𝑡 𝑒𝑠|𝑠=𝑡=0 = 𝑎

∴ 𝑐𝑜𝑣 𝑋, 𝑌 = 𝑎 −1

4⟹ 𝜌𝑋 ,𝑌 =

𝑎 −14

14

= 4𝑎 − 1.

(27) 𝑣𝑎𝑟 𝑋

3+

2𝑌

3 = 𝑣𝑎𝑟

2𝑋

3+

𝑌

3 ←∵ 𝑉 𝑋 = 𝑉 𝑌

=1

9𝑉 𝑋 +

4

9𝑉 𝑌 + 2 𝑐𝑜𝑣

𝑋

3,2𝑌

3

=2

9+

8

9+

4

9×

2

3=

2

9+

8

9+

8

27=

38

27

𝑐𝑜𝑣 𝑋

3+

2𝑌

3,2𝑋

3+

𝑌

3

=2

9𝑉 𝑋 +

1

9𝑐𝑜𝑣 𝑋, 𝑌 +

4

9𝑐𝑜𝑣 𝑋, 𝑌 +

2

9𝑉 𝑌

=4

9+

2

27+

8

27+

4

9=

34

27

𝑐𝑜𝑣 𝑋

3+

2𝑌

3,2𝑋

3+

𝑌

3

=

34273827

=34

38.

Problem Set-8

[1] The joint probability mass function of the random variables 𝑋1 𝑎𝑛𝑑 𝑋2 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦

𝑃 𝑋1 = 𝑥1 , 𝑋2 = 𝑥2 = 2

3

𝑥1+𝑥2

1

3

2−𝑥1−𝑥2

𝑖𝑓 𝑥1 , 𝑥2 = 0, 0 , 0, 1 , 1, 0 , (1, 1)


(a) Find the joint probability mass function of𝑌1 = 𝑋1 − 𝑋2 𝑎𝑛𝑑 𝑌2 = 𝑋1 + 𝑋2.

(b) Find the marginal probability mass functions of 𝑌1 𝑎𝑛𝑑 𝑌2.

(c) Verify whether 𝑌1 𝑎𝑛𝑑 𝑌2 are independent.

[2] Let the joint probability mass function of 𝑋1 𝑎𝑛𝑑 𝑋2 𝑏𝑒

𝑃 𝑋1 = 𝑥1 , 𝑋2 = 𝑥2 =

𝑥1𝑥2

36 𝑖𝑓 𝑥1 = 1, 2, 3; 𝑥2 = 1, 2, 3;


(a) Find the joint probability mass function of𝑌1 = 𝑋1 𝑋2 𝑎𝑛𝑑 𝑌2 = 𝑋2 .

(b) Find the marginal probability mass functions of 𝑌1

(c) Find the probability mass function of Z=𝑋1 + 𝑋2 .

[3] (a) Let X∼

𝐵𝑖𝑛 𝑛1 , 𝑝 𝑎𝑛𝑑 𝑌 ∼

𝐵𝑖𝑛 𝑛2 , 𝑝 𝑏𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑟𝑎𝑛𝑑𝑜𝑚 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠. 𝐹𝑖𝑛𝑑 𝑡𝑕𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑋 𝑔𝑖𝑣𝑒𝑛 𝑋 +

𝑌 = 𝑡, 𝑡 ∈ 0, 1, … , min 𝑛1 , 𝑛2 .

(b) Let X∼𝐵𝑖𝑛 𝑛1,1

2 𝑎𝑛𝑑𝑌 ∼ 𝐵𝑖𝑛 𝑛2 ,

1

2 be independent random variables.

Find the distribution of Y=𝑋1 − 𝑋2 + 𝑛2.

[4] Let X∼ Poisson

𝜆1 𝑎𝑛𝑑 𝑌 ∼

𝑃𝑜𝑖𝑠𝑠𝑜𝑛 𝜆2 𝑏𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑟𝑎𝑛𝑑𝑜𝑚 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠. 𝐹𝑖𝑛𝑑 𝑡𝑕𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑋 𝑔𝑖𝑣𝑒𝑛 𝑋 +

𝑌 = 𝑡, 𝑡 ∈ 0, 1, … .

[5] Let 𝑋1 , 𝑋2 , 𝑋3 𝑎𝑛𝑑 𝑋4 be four mutually independent random variables each having probability

density function

f(x)= 3 1 − 𝑥 2 0 < 𝑥 < 10 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒.

Find the probability density functions of Y= min(𝑋1 , 𝑋2 , 𝑋3 , 𝑋4)𝑎𝑛𝑑 𝑍 = 𝑋1 , 𝑋2 , 𝑋3 , 𝑋4 .

[6] Suppose 𝑋1 , … . , 𝑋𝑛 are n independent random variables, where 𝑋𝑖 𝑖 = 1, … , 𝑛 has the exponential

distribution Exp 𝛼1 , 𝑤𝑖𝑡𝑕 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛

𝑓𝑋𝑖 𝑥 =

𝛼𝑖𝑒−𝛼𝑖𝑥 𝑥 > 0

0 𝑜𝑡𝑗𝑕𝑒𝑟𝑤𝑖𝑠𝑒.

Find the probability density functions of Y= min 𝑋1 , … . , 𝑋𝑛 𝑎𝑛𝑑 𝑍 = max(𝑋1 , … . , 𝑋𝑛).

[7] Let X and Y be the respective arrival times of two friends A and B who agree to meet at a spot and

wait for the other only for t minutes. Supposing that X and Y are i. i. d. Exp (𝜆). Show that the probability

of A and B meeting each other is 1 − 𝑒−𝜆𝑡 .

[8] Let 𝑋1 𝑎𝑛𝑑 𝑋2 𝑏𝑒 𝑖. 𝑖. 𝑑. 𝑈 0, 1 . Define two new random variables as 𝑌1 = 𝑋1 + 𝑋1𝑎𝑛𝑑 𝑌2 = 𝑋2 −

𝑋1. Find the joint probability density function of 𝑌1 𝑎𝑛𝑑 𝑌2 are also the marginal probability density

functions of 𝑌1 𝑎𝑛𝑑 𝑌2.

[9] Let X and Y be i. i. d. N(0, 1). Find the probability density function of Z= X/ Y.

[10] Let X and Y be independent random variables with probability density functions

𝑓𝑋 𝑥 = 𝑥𝛼1−1

⎾𝛼2𝜃𝛼2

𝑒−𝑦/𝜃 𝑦 > 0


Find the distributions of 𝑈 = 𝑋 + 𝑌 𝑎𝑛𝑑 𝑉 =𝑋

(𝑋+𝑌) and also that they are independently distributed.

[11] Let X and Y be i.i.d. random variables with common probability density function

f(x)= 𝑐

1+𝑥4 − ∞ < 𝑥 < ∞


Where, c is a normalizing constant. Find the probability density function of Z= X / Y.

[12] Let X and Y be i. i. d. N(0, 1), define the random variables R and𝛩 𝑏𝑦 𝑋 = 𝑅 cos 𝛩 , 𝑌 = 𝑅 𝑠𝑖𝑛𝛩,

(a) show that R and 𝛩 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑤𝑖𝑡𝑕𝑅2

2∼ 𝐸𝑥𝑝 1 𝑎𝑛𝑑 𝛩 ∼ 𝑈 0, 2𝜋

𝑏 𝑠𝑕𝑜𝑤 𝑡𝑕𝑎𝑡 𝑋2 + 𝑌2 𝑎𝑛𝑑𝑋

𝑌 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡𝑙𝑦 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑.

[13] Let 𝑈1𝑎𝑛𝑑𝑈2 be i.i.d. U(0, 1) random variables. Show that

𝑋1 = −2𝑙𝑛𝑈1 cos(2𝜋𝑈2) 𝑎𝑛𝑑 𝑋2 = −2𝑙𝑛𝑈1 sin(2𝜋𝑈2) 𝑎𝑟𝑒 𝑖. 𝑖. 𝑑. 𝑁(0, 1) random variables.

[14] Let 𝑋1 , 𝑋2 , 𝑎𝑛𝑑 𝑋3 be i. i. d. with probability density function

f(x)= 𝑒−𝑥 𝑥 > 0


Find the probability density function of 𝑌1 , 𝑌2 , 𝑌3; where

𝑌1 =𝑋1

𝑋1 + 𝑋2; 𝑌2 =

𝑋1 + 𝑋2

𝑋1 + 𝑋2 + 𝑋3; 𝑌3 = 𝑋1 + 𝑋2 + 𝑋3

[15] Let 𝑋1 , 𝑋2 , 𝑎𝑛𝑑 𝑋3 be three mutually independent chi-square random variables with 𝑛1 , 𝑛2 , 𝑛3

degrees of freedom respectively; i.e. 𝑋1 ∼ 𝜆𝑛1

2, 𝑋2 ∼ 𝜆𝑛2

2 𝑎𝑛𝑑 𝑋3 ∼ 𝜆𝑛3

2 and they are independent.

(a) Show that 𝑌1 =𝑋1

𝑋2 𝑎𝑛𝑑𝑌2 = 𝑋1 + 𝑋2𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑎𝑛𝑑 𝑡𝑕𝑎𝑡 𝑌2 is chi-square random variable

with 𝑛1 + 𝑛2 degree of freedom.

(b) Find the probability density functions of

𝑍1 =𝑋1/𝑛1

𝑋2/𝑛2 𝑎𝑛𝑑 𝑍2 =

𝑋3/𝑛3

(𝑋1 + 𝑋2)/(𝑛1 + 𝑛2)

[16] Let X and Y be independent random variables such that X∼ N(0, 1) and Y∼𝜆𝑛2

2.

Find the probability density function of 𝑇 =𝑋

𝑌/𝑛.

[17] Let 𝑋1 , … , 𝑋𝑛 be a random sample from N(0, 1) distribution. Find the m.g. f. of Y= 𝑋𝑖2𝑛

𝑖=1 and

identify its distribution. Further, suppose 𝑋𝑛+1 is another random sample from N(0, 1) independent

of 𝑋1 , … , 𝑋𝑛 . Derive the distribution of 𝑋𝑛+1

𝑌

𝑛

.

[18] X and Y are i. i. d. random variables each having geometric distribution with the following p. m.

f.

P(X= x)= 1 − 𝑝 𝑥𝑝, 𝑥 = 0, 1, …


𝐼𝑑𝑒𝑛𝑡𝑖𝑓𝑦 𝑡𝑕𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑜𝑓𝑋

𝑋+𝑌. 𝐹𝑢𝑟𝑡𝑕𝑒𝑟 𝑓𝑖𝑛𝑑 𝑡𝑕𝑒 𝑝. 𝑚. 𝑓. 𝑜𝑓 𝑍 = min(𝑋, 𝑌).

Solution Key

(1) 𝑃 𝑌1 = 𝑦1 , 𝑌2 = 𝑦2 = 𝑃 𝑋1 − 𝑋2 = 𝑦1 , 𝑋1 + 𝑋2 = 𝑦2

= 𝑃 𝑋1 =𝑦1 + 𝑦2

2, 𝑋2 =

𝑦1 − 𝑦2

2

=

2

3

0

1

3

2−0

𝑖𝑓 𝑦1 + 𝑦2

2= 0,

𝑦1 − 𝑦2

2= 0, 𝑖. 𝑒. 𝑦1 = 0, 𝑦2 = 0

2

3

1

1

3

2−1


2= 1 ,

𝑦1 − 𝑦2

2= 0, 𝑖. 𝑒. 𝑦1 = 1, 𝑦2 = 1

2

3

1

1

3

2−1


2= 0,

𝑦1 − 𝑦2

2= 1, 𝑖. 𝑒. 𝑦1 = −1, 𝑦2 = 1

2

3

2

1

3

2−2


2= 1 ,

𝑦1 − 𝑦2

2= 1, 𝑖. 𝑒. 𝑦1 = 0, 𝑦2 = 2

𝑖. 𝑒.

𝑃 𝑌1 = 𝑦1 , 𝑌2 = 𝑦2 =

1

9 𝑖𝑓 𝑦1 , 𝑦2 = (0, 0)

2

9𝑖𝑓 𝑦1 , 𝑦2 = −1, 1 , (1, 1)

4

9 𝑖𝑓 𝑦1 , 𝑦2 = (0, 2)


𝑌2 𝑌1

0 1 2

0 -1 1

1

9 0

4

9

0 2

9 0

0 2

9 0

𝑃 𝑌1 = 𝑦1 =

5

9 𝑦1 = 0

2

9 𝑦1 = −1

2

9 𝑦1 = 1


𝑃 𝑌2 = 𝑦2 =

1

9 𝑦2 = 0

4

9 𝑦2 = 1

4

9 𝑦2 = 2


𝑠𝑖𝑛𝑐𝑒 𝑃 𝑌1 = 𝑦1 , 𝑌2 = 𝑦2 ≠ 𝑃 𝑌1 = 𝑦1 𝑃 𝑌2 = 𝑦2 ∀ 𝑦1 , 𝑦2

𝑌1 & 𝑌2 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑖𝑛𝑑𝑒𝑝.

(2) 𝑌1 = 𝑋1𝑋2 ; 𝑌2 = 𝑋2

𝑗𝑡 𝑝. 𝑚. 𝑓.

𝑃 𝑌1 = 𝑦1 , 𝑌2 = 𝑦2 = 𝑃 𝑋1𝑋2 = 𝑦1 , 𝑋2 = 𝑦2 = 𝑃 𝑋1 =𝑦1

𝑦2, 𝑋2 = 𝑦2

=

𝑦1

36 𝑖𝑓

𝑦1

𝑦2= 1, 2, 3; 𝑦2 = 1, 2, 3


𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑌1 𝑖𝑛 1, 2, 3, 4, 6, 9 .

𝑃 𝑌1 = 𝑦1 =

𝑃 𝑋1 = 1, 𝑋2 = 1 =

1

36 𝑦1 = 1

𝑃 𝑋1 = 1, 𝑋2 = 2 + 𝑃 𝑋1 = 2, 𝑋2 = 1 =2

36+

2

36=

4

36 𝑦1 = 2

𝑃 𝑋1 = 1, 𝑋2 = 3 + 𝑃 𝑋1 = 3, 𝑋2 = 1 =3

36+

3

36=

6

36 𝑦1 = 3

𝑃 𝑋1 = 2, 𝑋2 = 2 =4

36 𝑦1 = 4

𝑃 𝑋1 = 2, 𝑋2 = 3 + 𝑃 𝑋1 = 3, 𝑋2 = 2 =6

36+

6

36=

12

36 𝑦1 = 6

𝑃 𝑋1 = 3, 𝑋2 = 3 =9

36 𝑦1 = 9


𝑍 = 𝑋1 + 𝑋2 → 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑍 𝑖𝑛 2, 3, 4, 5, 6

𝑃 𝑧 = 3

= 𝑃 𝑋1 + 𝑋2 = 3

=

𝑃 𝑋1 = 1, 𝑋2 = 1 =

1

36 𝔍 = 2

𝑃 𝑋1 = 1, 𝑋2 = 2 + 𝑃 𝑋1 = 2, 𝑋2 = 1 =4

36 𝔍 = 3

𝑃 𝑋1 = 1, 𝑋2 = 3 + 𝑃 𝑋1 = 2, 𝑋2 = 2 + 𝑃 𝑋1 = 3, 𝑋2 = 1 =10

36 𝔍 = 4

𝑃 𝑋1 = 2, 𝑋2 = 3 + 𝑃 𝑋1 = 3, 𝑋2 = 2 =12

36 𝔍 = 5

𝑃 𝑋1 = 3, 𝑋2 = 2 =9

36 𝔍 = 6


(3) 𝑃 𝑋 = 𝑥 𝑋 + 𝑌 = 𝑡 =𝑃 𝑋=𝑥 ,𝑋+𝑌=𝑡

𝑃 𝑋+𝑌=𝑡 =

𝑃 𝑋=𝑥 ,𝑌=𝑡−𝑥

𝑃 𝑋+𝑌=𝑡

=𝑃 𝑋 = 𝑥 𝑃 𝑌 = 𝑡 − 𝑥

𝑃 𝑋 + 𝑌 = 𝑡 =

𝑛1𝑥 𝑝𝑥 1 − 𝑝 𝑛−𝑥 𝑛2

𝑡−𝑥 𝑝𝑡−𝑥 1 − 𝑝 𝑛2− 𝑡−𝑥

𝑛1+𝑛2𝑡 𝑝𝑡 1 − 𝑝 𝑛1+𝑛2−𝑡

= 𝑛1

𝑥 𝑛2

𝑡−𝑥

𝑛1+𝑛2𝑡

; 0 ≤ 𝑥 ≤ 𝑛1 0 ≤ 𝑡 − 𝑥 ≤ 𝑛2

↑ 𝑕𝑦𝑝𝑒𝑟 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑛1 , 𝑛2

(4) 𝑃 𝑋 = 𝑥 𝑋 + 𝑌 = 𝑡 =𝑃 𝑋=𝑥 ,𝑌=𝑡−𝑥

𝑃 𝑋+𝑌=𝑡 [𝑋 ∼ 𝑃 𝜆1 , 𝑌 ∼ 𝑃 𝜆2 ; 𝑋 + 𝑌 ∼ 𝑃 𝜆1 + 𝜆2 ]

= 𝑃 𝑋 = 𝑥 𝑃 𝑌 = 𝑡 − 𝑥

𝑃 𝑋 + 𝑌 = 𝑡 =

𝑒−𝜆1𝜆1

𝑥

𝑥! 𝑒−𝜆2𝜆2

𝑡−𝑥

𝑡 − 𝑥 !

𝑒− 𝜆1+𝜆2 𝜆1 + 𝜆2 𝑡

𝑡!

= 𝑡

𝑥

𝜆1

𝜆1 + 𝜆2

𝑥

1 −𝜆1

𝜆1 + 𝜆2

1

𝑖. 𝑒. 𝑋| 𝑋 + 𝑌 = 𝑡 𝐵𝑖𝑛 𝑡,𝜆1

𝜆1 + 𝜆2

(5) 𝑓𝑋 𝑥 = 3 1 − 𝑥 2 0 < 𝑥 < 10 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒.

𝐹𝑋 𝑥 =

0 𝑥 < 0

3 ∫ 1 − 𝑡 2𝑑𝑡 0 ≤ 𝑥 < 1𝑥

0= 1 − 1 − 𝑥 3

1 𝑥 ≥ 1

𝑌 = 𝑀𝑖𝑛 𝑋1 , 𝑋2 , 𝑋3 , 𝑋4 ; 𝑍 = 𝑀𝑎𝑥 𝑋1 , 𝑋2 , 𝑋3 , 𝑋4

𝑋1 , 𝑋2 , 𝑋3 , 𝑋4 𝑖. 𝑖. 𝑑. 𝑓𝑟𝑜𝑚 𝑓𝑋 𝑥

𝑑. 𝑓. 𝑜𝑓 𝑌 = 𝐹𝑌 𝑦 = 𝑃 𝑌 ≤ 𝑦 = 1 − 𝑃 𝑌 > 𝑦

= 1 − 𝑃 𝑋𝑖 > 𝑦

4

1

= 1 − 1 − 𝑃 𝑋 ≤ 𝑦 4

= 1 − 1 − 1 − 1 − 𝑦 2 4 = 1 − 1 − 𝑦 12 0 < 𝑦 < 1

𝑓𝑌 𝑦 = 12 1 − 𝑦 11 0 < 𝑦 < 1


𝑑. 𝑓. 𝑜𝑔 𝑍: 𝑓𝑍 Ʒ = 𝑃 𝑍 ≤ Ʒ = 𝑃 𝑋𝑖 ≤ Ʒ

4

1

= [𝑃 𝑋 ≤ Ʒ ]4

= 1 − 1 − Ʒ 3 4 0 < Ʒ < 1

𝑓𝑍 Ʒ = 12(1 − Ʒ)2(1 − (1 − Ʒ)3)3 0 < Ʒ < 1


(6) Similar to (5) .

(7) X: arrival time of A

Y : arrival time of B

X & Y i. i. d. Exp(𝜆)- p. d. f.

𝑓 𝑥 = 𝜆𝑒−𝜆𝑥 𝑥 > 0

𝑟𝑒𝑞𝑑 𝑝𝑟𝑜𝑏 = 𝑃 𝑋 < 𝑌, 𝑌 − 𝑋 ≤ 𝑡 + 𝑃 𝑌 < 𝑋, 𝑋 − 𝑌 ≤ 𝑡

= 𝑃 𝑌 − 𝑡 ≤ 𝑋 ≤ 𝑌 + 𝑃 𝑋 − 𝑡 ≤ 𝑌 ≤ 𝑋

= 𝑃 𝑋 ≤ 𝑌 ≤ 𝑋 + 𝑡 + 𝑃 𝑌 ≤ 𝑋 ≤ 𝑌 + 𝑡 𝑗𝑡 𝑝. 𝑑. 𝑓 𝑜𝑓 𝑋, 𝑌 → 𝜆2𝑒−𝜆 𝑥+𝑦 𝑥 > 0, 𝑦 > 0

= 𝜆2𝑒−𝜆 𝑥+𝑦 𝑥+𝑡

𝑥

∞

0

𝑑𝑦 𝑑𝑥 + 𝜆2𝑒−𝜆 𝑥+𝑦 𝑦+𝑡

𝑥

∞

0

𝑑𝑥 𝑑𝑦

= 2𝜆2 𝑒−𝜆𝑥∞

0

𝑒−𝜆𝑡𝑥+𝑡

𝑥

𝑑𝑦 𝑑𝑥

= 2𝜆2 1

𝜆 1 − 𝑒−𝜆𝑡 𝑒−2𝜆𝑥

∞

0

𝑑𝑥 = 1 − 𝑒−𝜆𝑡 .

(8) 𝑋1 , 𝑋2 ∼ 𝑈 0, 1

𝑌1 = 𝑋1 + 𝑋2 ⟹1

𝐽 =

𝜕𝑦1

𝜕𝑥1

𝜕𝑦1

𝜕𝑥2

𝜕𝑦2

𝜕𝑥1

𝜕𝑦2

𝜕𝑥2

= 1 1

−1 1 = 2

𝑌2 = 𝑋2 − 𝑋1 𝐽 =1

2

𝑓𝑋1 ,𝑋2 𝑥1, 𝑥2 = 1; 0 < 𝑥1 < 1, 0 < 𝑥2 < 1

⟹ 𝑓𝑌1 ,𝑌2 𝑦1 , 𝑦2 =

1

2; 0 < 𝑦1 + 𝑦2 < 2, 0 < 𝑦1 − 𝑦2 < 2

𝑅𝑎𝑛𝑔𝑒 𝑢𝑛𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙𝑙𝑦 0 < 𝑦1 < 2 & − 1 < 𝑦2 < 1

𝑖2 =𝑦1 + 𝑦2

2, 𝑖1 =

𝑦1 − 𝑦2

2 𝐴𝑙𝑠𝑜 0 < 𝑥1 < 1 ; ⟹ 0 <

𝑦1 − 𝑦2

2< 1

⟹ 0 < 𝑦1 − 𝑦2 < 2

𝑦2 < 𝑦1 < 2 + 𝑦2 & 𝑦1 − 2 < 𝑦2 < 𝑦1} ____(1)

𝐴𝑙𝑠𝑜 0 < 𝑥2 < 1 ; 0 <𝑦1 + 𝑦2

2< 1

0 < 𝑦1 + 𝑦1 < 2

−𝑦2 < 𝑦1 < 2 − 𝑦2 & − 𝑦1 < 𝑦2 < 2 − 𝑦1} ____(2)

𝐶𝑜𝑚𝑏𝑖𝑛𝑖𝑛𝑔 1 & 2

max 𝑦2 , −𝑦2 < 𝑦1 < min 2 + 𝑦2 , 2 − 𝑦2

& max 𝑦1 − 2, −𝑦1 < 𝑦2 < min 𝑦1 , 2 − 𝑦1 ) ____(3)

𝐼𝑓 − 1 < 𝑦2 < 0 𝑡𝑕𝑒𝑛 𝑓𝑟𝑜𝑚 3 − 𝑦2 < 𝑦1 < 2 + 𝑦2 & 𝑖𝑓 0 < 𝑦2 < 1 𝑡𝑕𝑒𝑛 𝑓𝑟𝑜𝑚 3 𝑦2

< 𝑦1 < 2 − 𝑦2} _____(4)

𝐴𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑣𝑒𝑙𝑦 𝑖𝑓 0 < 𝑦1 < 1 𝑡𝑕𝑒𝑛 𝑓𝑟𝑜𝑚 3 − 𝑦1 < 𝑦2 < 𝑦1 & 𝑖𝑓 1 < 𝑦1

< 2 𝑡𝑕𝑒𝑛 𝑓𝑟𝑜𝑚 3 𝑦1 − 2 < 𝑦2 < 2 − 𝑦1}____(5)

⟹ 𝑀𝑎𝑟𝑔 𝑜𝑓 𝑌1

𝑓𝑌1 𝑦1 =

1

2 𝑑𝑦2

𝑦1

−𝑦1

= 𝑦1 𝑖𝑓 0 < 𝑦1 < 1

𝑈𝑠𝑖𝑛𝑔 5 →=1

2 𝑑𝑦2

2−𝑦1

𝑦1−2

= 2 − 𝑦1 𝑖𝑓 1 < 𝑦1 < 2

& 𝑀𝑎𝑟𝑔 𝑜𝑓 𝑌2

𝑓𝑌2 𝑦2 =

1

2 𝑑𝑦1

2+𝑦2

−𝑦2

= 1 + 𝑦2 𝑖𝑓 − 1 < 𝑦2 < 0

𝑢𝑠𝑖𝑛𝑔 4 → =1

2 𝑑𝑦1

2−𝑦2

𝑦2

= 1 − 𝑦2 𝑖𝑓 0 < 𝑦2 < 1

(9) X ∼N(0, 1)

Y∼ N(0, 1) > ind

𝑓𝑋,𝑌 𝑥, 𝑦 =1

2𝜋exp −

1

2 𝑥2 + 𝑦2

𝑈 = 𝑌} 𝑋 = 𝑈𝑍 𝐽 = 𝑧 𝑢1 0

= |𝑢|

𝑓𝑈,𝑍 𝑢, 𝑧 = 1

2𝜋exp −

1

2 𝑢2Ʒ2 + 𝑢2 𝑢 ; −∞ < 𝑢 < ∞, −∞ < 𝑧 < ∞

𝑓𝑍 Ʒ =1

2𝜋 𝑢

∞

−∞

exp −1

2𝑢2 1 + Ʒ2 𝑑𝑢

=1

𝜋 𝑢

∞

0

exp −𝑢2

2 1 + Ʒ2 𝑑𝑢 =

1

𝜋.

1

1 + Ʒ2; −∞ < Ʒ < ∞

𝑖. 𝑒. 𝑍 ∼ 𝐶𝑎𝑢𝑐𝑕𝑦 𝑑𝑖𝑠𝑡𝑛 0, 1

𝐼𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑋 ∼ 𝐶𝑎𝑢𝑐𝑕𝑦 𝜇, 𝜃 → 𝑓𝑋 𝑥 =𝜃

𝜋

1

1 + (𝑥 − 𝜇)2; −∞ < 𝑥 < ∞

(10) 𝑓𝑋,𝑌 𝑥, 𝑦 =1

⎾𝛼1⎾𝛼2𝜃𝛼1+𝛼2𝑥𝛼1−1𝑦𝛼2−1𝑒−

𝑥+𝑦

𝜃 , 𝑥 > 0, 𝑦 > 0


𝑈 = 𝑋 + 𝑌} 𝑋 = 𝑈𝑉

𝑉 =𝑋

𝑋 + 𝑌} 𝑌 = 𝑈(1 − 𝑉)

𝐽 = 𝑣 𝑢

1 − 𝑣 −𝑢 = −𝑢

𝑅𝑎𝑛𝑔𝑒 𝑢 > 0, 0 < 𝑣 < 1

𝑓𝑈,𝑉 𝑢, 𝑣 =1

⎾𝛼1⎾𝛼2𝜃𝛼1+𝛼2

𝑢𝑣 𝛼1−1 𝑢 1 − 𝑣 𝛼2−1

𝑒−𝑢𝜃 . 𝑢 𝑢 > 0, 0 < 𝑣 < 1


𝑖. 𝑒. 𝑓𝑈,𝑉 𝑢, 𝑣

=

1

⎾𝛼1 + 𝛼2 𝜃𝛼1+𝛼2𝑢𝛼1+𝛼2−1𝑒

−𝑢𝜃 ×

1

𝐵 𝛼1 , 𝛼2 𝑣𝛼1−1 1 − 𝑣 𝑢𝛼2−1

𝑢 > 0, 0 < 𝑣 < 1


⟹ 𝑓𝑈 𝑢 =1

⎾𝛼1 + 𝛼2 𝜃𝛼1+𝛼2𝑢𝛼1+𝛼2−1𝑒−

𝑢𝜃 𝑢 > 0

𝑈 ∼ 𝐺𝑎𝑚𝑚𝑎. = 0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒

𝑓𝑉 𝑣 =1

𝐵 𝛼1 , 𝛼2 𝑣𝛼1−1 1 − 𝑣 𝑢𝛼2−1

0 < 𝑣 < 1

𝑉 ∼ 𝐵𝑒𝑡𝑎. = 0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒.

⟹ 𝑈 & 𝑉 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝.

(11) 𝑓𝑋,𝑌 =𝐶2

1+𝑥4 1+𝑦4 − ∞ < 𝑥 < ∞, −∞ < 𝑦 < ∞

𝑈1 =𝑋

𝑌, 𝑈2 = 𝑌}

𝑋 = 𝑈1𝑈2

𝑌 = 𝑈2} 𝐽 =

𝑢2 𝑢1

0 1 = 𝑢2

𝑅𝑎𝑛𝑔𝑒 − ∞ < 𝑢1 < ∞, −∞ < 𝑢2 < ∞

𝑓𝑈1 ,𝑈2 𝑢1 , 𝑢2 =

𝐶2 𝑢2

1 + 𝑢14𝑢2

4 1 + 𝑢24

− ∞ < 𝑢1 < ∞, −∞ < 𝑢2 < ∞

𝑓𝑈1 𝑢1 = 𝑓𝑈1 ,𝑈2

𝑢1 , 𝑢2 ∞

−∞

𝑑𝑢2 = 2𝐶 𝑢2

1 + 𝑢14𝑢2

4 1 + 𝑢24

∞

0

𝑑𝑢2

=𝐶𝜋

2.

1

1 + 𝑢12 𝑎𝑛 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑛𝑔 .

𝑓𝑈1

∞

−∞

𝑢1 𝑑𝑢1 = 1 ⟹ 𝐶 =2

𝜋2

⟹ 𝑓𝑈1 𝑢1 =

2

𝜋.

1

1 + 𝑢12

− ∞ < 𝑢1 < ∞.

↑ 𝐶𝑎𝑢𝑐𝑕𝑦 𝑑𝑖𝑠𝑡𝑛.

(12) 𝑓𝑋,𝑌 𝑥, 𝑦 =1

2𝜋𝑒−

1

2 𝑥2+𝑦2 − ∞ < 𝑥 < ∞, −∞ < 𝑦 < ∞

𝑋 = 𝑅 cos 𝛩

𝑌 = 𝑅 sin 𝛩

𝐽 = 𝑐𝑜𝑠𝜃 −𝑟 𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝜃 𝑟 𝑐𝑜𝑠𝜃

= 𝑟

𝑅𝑎𝑛𝑔𝑒 𝑟 ≥ 0, 0 < 𝜃 < 2𝜋

𝑓𝑅,𝛩 𝑟, 𝜃 =1

2𝜋𝑒−

𝑟2

2 𝑟, 𝑟 > 0, 0 < 𝜃 < 2𝜋


𝑓𝑅 𝑟 = 𝑟𝑒−𝑟2

2 𝑟 > 0


𝑓 𝛩 𝜃 =1

2𝜋 0 < 𝜃 < 2𝜋 𝛩 ∼ 𝑈 0, 2𝜋


⟹ 𝑅 &𝛩 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝.

𝐷𝑒𝑓𝑖𝑛𝑒 𝑌 =𝑅2

2 𝑦 > 0

𝑅 = 2 𝑦 𝑑𝑟

𝑑𝑦=

1

2𝑦

𝑓𝑌 𝑦 =1

2𝑦 2𝑦𝑒−𝑦 𝑦 > 0


𝑖. 𝑒. 𝑓𝑌 𝑦 = 𝑒−𝑦 𝑦 > 0


⟹𝑅2

2∼ 𝐸𝑥𝑝 1 .

𝑈 = 𝑋2 + 𝑌2 = 𝑅2 − 𝑓𝑛 𝑜𝑓 𝑟. 𝑣. 𝑘

𝑉 =𝑋

𝑌= 𝑐𝑜𝑡𝛩 − 𝑓𝑛 𝑜𝑓 𝑟. 𝑣. 𝛩

𝑠𝑖𝑛𝑐𝑒 𝑅 & 𝛩 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝, 𝑈 & 𝑉 𝑎𝑟𝑒 𝑎𝑙𝑠𝑜 𝑖𝑛𝑑𝑒𝑝.

𝑖. 𝑒. 𝑋2 + 𝑌2 &𝑋

𝑌 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝.

(13) 𝑈1 ∼ 𝑈 0, 1

−𝑙𝑛𝑈1 ∼ 𝐸𝑥𝑝 1 − 𝑠𝑡𝑟𝑎𝑖𝑔𝑕𝑡 𝑓𝑜𝑟𝑤𝑎𝑟𝑑

𝑈2 ∼ 𝑈 0, 1

2𝜋𝑈2 ∼ 𝑈 0, 2𝜋 − 𝑠𝑡𝑟𝑎𝑖𝑔𝑕𝑡 𝑓𝑜𝑟𝑤𝑎𝑟𝑑.

⟹ −𝑙𝑛𝑈1 ∼ 𝐸𝑥𝑝 1 & 2𝜋𝑈2 ∼ 𝑈 0, 2𝜋 𝑎𝑛𝑑 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝. 𝑏𝑦 𝑝𝑟𝑜𝑏𝑙𝑒𝑚 # 12

𝑗𝑡 𝑑𝑖𝑠𝑡𝑛 𝑜𝑓 – 𝑙𝑛𝑈1 , 2𝜋𝑈2 𝑖𝑠 𝑠𝑎𝑚𝑒 𝑎𝑠 𝑗𝑡 𝑑𝑖𝑠𝑡𝑛 𝑜𝑓 𝑅2

2, 𝛩

𝑖. 𝑒. – 𝑙𝑛𝑈1 , 2𝜋𝑈2 ≝ 𝑅2

2, 𝛩

𝑖. 𝑒. – 2𝑙𝑛𝑈1 , 2𝜋𝑈2 ≝ 𝑅2, 𝛩

𝑖. 𝑒. – 2𝑙𝑛𝑈1 cos 2𝜋𝑈2 , – 2𝑙𝑛𝑈1 sin 2𝜋𝑈2 ≝ 𝑅 𝑐𝑜𝑠𝛩, 𝑅 𝑠𝑖𝑛𝛩

𝑖. 𝑒. 𝑋1 , 𝑋2 ≝ 𝑅 𝑐𝑜𝑠𝛩, 𝑅 𝑠𝑖𝑛𝛩

⟹ 𝑋1𝑎𝑛𝑑 𝑋2 𝑎𝑟𝑒 𝑖. 𝑖. 𝑑. 𝑁 0, 1 𝑟. 𝑣. 𝑠.

𝐷𝑖𝑣𝑒𝑟𝑡 𝑚𝑒𝑡𝑕𝑜𝑑 𝑈1 , 𝑈2 𝑖. 𝑖. 𝑑. 𝑈 0, 1 .

𝑓𝑈1 ,𝑈2 𝑢1, 𝑢2 = 1 ; 0 < 𝑢1 < 1, 0 < 𝑢2 < 1


𝑋1 = – 2𝑙𝑛𝑈1 cos 2𝜋𝑈2

𝑋2 = – 2𝑙𝑛𝑈1 sin 2𝜋𝑈2

𝑅𝑎𝑛𝑔𝑒 𝑜𝑓 𝑋1; −∞ < 𝑥1 < ∞, 𝑠𝑙𝑦 − ∞ < 𝑥2 < ∞

𝑋12 + 𝑋2

2 = – 2𝑙𝑛𝑈1 𝑋2

𝑋1= 𝑡𝑎𝑛 2𝜋𝑈2

𝑈1 = exp −1

2 𝑋1

2 + 𝑋22

𝑈2 =1

2𝜋tan−1

𝑋2

𝑋1

𝐽 = exp −

1

2 𝑋1

2 + 𝑋22 −𝑋1 exp −

1

2 𝑋1

2 + 𝑋22 −𝑋2

−𝑋2

2𝜋 𝑋12 + 𝑋2

2

𝑋1

2𝜋 𝑋12 + 𝑋2

2

𝐽 = exp −1

2 𝑋1

2 + 𝑋22 −

1

2𝜋

𝐽 =

exp −12 𝑋1

2 + 𝑋22

2𝜋

⟹ 𝑓𝑋1 ,𝑋2 𝑥1, 𝑥2 =

1

2𝜋 exp −

1

2 𝑥1

2 + 𝑥22 ; −∞ < 𝑥1 < ∞, −∞ < 𝑥2 < ∞ ↓

= 1

2𝜋𝑒−

12𝑥1

2

1

2𝜋𝑒−

12𝑥2

2

⟹ 𝑋1& 𝑋2 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝 𝑁 0, 1 𝑟. 𝑣. 𝑠.

(14) 𝑓𝑋1 ,𝑋2 ,𝑋3 𝑥1 , 𝑥2 , 𝑥3 = 𝑒− 𝑥1 ,𝑥2 ,𝑥3 ; 𝑥1 > 0, 𝑥2 > 0, 𝑥3 > 0

𝑌1 =𝑋1

𝑋1 + 𝑋2; 𝑌2 =

𝑋1

𝑋1 + 𝑋2 + 𝑋3; 𝑌3 = 𝑋1 + 𝑋2 + 𝑋3

𝑖. 𝑒. 𝑋1 = 𝑌1 𝑌2 𝑌3

𝑋2 = 𝑌2 𝑌3 1 − 𝑌1

𝑋3 = 𝑌3 1 − 𝑌2

𝑋1 + 𝑋2 = 𝑌2 𝑌3 , 𝑋1 = 𝑌1 𝑌2 𝑌3 , 𝑋2 = 𝑌2 𝑌3

𝐽 =

𝑦2𝑦3 𝑦1𝑦3 𝑦1𝑦2

−𝑦2𝑦3 𝑦3 1 − 𝑦1 𝑦2 1 − 𝑦1

0 −𝑦3 1 − 𝑦2 = 𝑦2𝑦3

2

𝑓𝑌1 ,𝑌2 ,𝑌3 𝑦1 , 𝑦2 , 𝑦3 = 𝑦2𝑦3

2 𝑒−𝑦3 ; 0 < 𝑦1 < 1, 𝑦3 > 0

𝑓𝑌1 𝑦1 = 𝑦2

1

0

𝑑𝑦2 𝑦32 𝑒−𝑦3

∞

0

𝑑𝑦3 = 1 0 < 𝑦1 < 1

𝑖. 𝑒. 𝑌1 ∼ 𝑈 0, 1

𝑓𝑌2 𝑦2 = 𝑦2 × 1 × 2 0 < 𝑦2 < 1

𝑖. 𝑒. 𝑌2 ∼ 𝐵𝑒𝑡𝑎 2, 1 𝑋 ∼ 𝐵𝑒𝑡𝑎 𝑚, 𝑛 𝑓𝑋 𝑥 =⎾𝑚 + 𝑛

⎾𝑚⎾𝑛𝑥𝑚−1 1 − 𝑥 𝑛−1

&𝑓𝑌3 𝑦3 = 𝑑𝑦1

1

0

𝑑𝑦2

1

0

𝑦32 𝑒−𝑦3

=1

2𝑒−𝑦3𝑦3

2 0 < 𝑦3 < ∞

𝑦2𝑦3 𝑦3 1 − 𝑦1 1 − 𝑦2 + 𝑦2𝑦3 1 − 𝑦1

𝑦1𝑦3 𝑦2𝑦3 𝑦1𝑦3 𝑦1𝑦2

0 𝑦3 𝑦2

0 − 𝑦3 1 − 𝑦2

(15) (a) 𝑓𝑋1 ,𝑋2 𝑥1, 𝑥2 =

1

2𝑛𝑖2 ⎾

𝑛𝑖2

2𝑖=1 𝑒−

𝑥𝑖2 𝑥𝑖

𝑛𝑖2

−1; 𝑥𝑖 > 0

= 𝐶 𝑒−𝑥𝑖2 𝑥𝑖

𝑛 𝑖2

−1

2

𝑖=1

; 𝑥𝑖 > 0

𝑌1 =𝑋1

𝑋2; 𝑌2 = 𝑋1 + 𝑋2 |

𝑋1 =𝑌1 𝑌2𝑌1 + 1

𝑋2 =𝑌2

𝑌1 + 1

1

𝐽=

1

𝑥2−

𝑥

𝑥22

1 1

=1

𝑥2+

𝑥

𝑥22

=𝑥1 + 𝑥2

𝑥22

=𝑦2

𝑦2

1 + 𝑦1

2 = 1 + 𝑦1

2

𝑦2

𝐽 =𝑦2

1 + 𝑦1 2

𝑓𝑌1 ,𝑌2 𝑦1 , 𝑦2 = 𝐶 𝑒−

𝑦22

𝑦1 𝑦2

𝑦1+1 𝑛1

2−1

𝑦2

1+𝑦1 𝑛2

2−1 𝑦2

𝑦1+1 2 𝑦1 > 0, 𝑦2 > 0

𝑖. 𝑒. 𝑓𝑌1 ,𝑌2 𝑦1 , 𝑦2 = 𝐶1𝑒

−𝑦22 𝑦2

𝑛1+𝑛22

−1 ↓

𝑓𝑌2 𝑋

𝑦1

𝑛22

−1

1 + 𝑦1 𝑛1+𝑛2

2

𝑦1 > 0, 𝑦2 > 0

↓

𝑓𝑌1

⟹ 𝑌1 & 𝑌2 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝

& 𝑓𝑌2 𝑦2 = 𝐶1𝑒

−𝑦22 𝑦2

𝑛1+𝑛22

−1𝑦2 > 0

𝑓𝑌2 𝑦2

∞

0

𝑑𝑦2 = 1 ⟹ 𝐶1 = ⎾𝑛1 + 𝑛2

2. 2

𝑛1+𝑛22

−1

⟹ 𝑌2 ∼ 𝜆2𝑤𝑖𝑡𝑕 𝑛1 + 𝑛2 𝑑. 𝑓.

𝑏 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑎 𝑎

𝑍1 =

𝑋1𝑛1

𝑋2𝑛2

∼ 𝐹𝑛1 ,𝑛2→ 𝐹 𝑑𝑖𝑠𝑡𝑛 𝑤𝑖𝑡𝑕 𝑛1 , 𝑛2 𝑑. 𝑓. &

𝑍2 =𝑋3/𝑛3

𝑋1 + 𝑋2/ 𝑛1 + 𝑛2 ∼ 𝐹𝑛3 ,𝑛1+ 𝑛2

(16) X∼ N(0, 1)

𝑓𝑋,𝑌 =1

2𝑥𝑒−

𝑥2

2 1

2𝑛2 ⎾

𝑛2

𝑒−𝑦2 𝑦

𝑛2−1

𝑇 =𝑋

𝑌𝑛

𝑑𝑒𝑓𝑖𝑛𝑒 𝑑𝑢𝑚𝑚𝑦 𝑈 = 𝑌 𝑋

𝑌 →

𝑇

𝑈 = 𝑌

⟹ 𝑋 = 𝑇 𝑈

𝑛

𝑌 = 𝑈

𝐽 =

𝑢

𝑛

𝑡

2 𝑛 𝑢0 1

= 𝑢

𝑛

⟹ 𝑓𝑇,𝑈 𝑡, 𝑢 =1

2𝜋 2𝑛2

𝑛2 𝑛

exp −1

2

𝑡2𝑢

𝑛 exp −

𝑢

2 , 𝑢

𝑛2−

12 − ∞ < 𝑡 < ∞ 𝑢 > 0

𝑓𝑇 𝑡 = 𝑓𝑇,𝑈 𝑡, 𝑢 ∞

0

𝑑𝑢

=1

2𝜋 2𝑛2

𝑛2 𝑛

. 𝑢𝑛2−

12

∞

0

exp −𝑢

2 1 +

𝑡2

𝑛 𝑑𝑢

=⎾

𝑛 + 12

2𝜋 2𝑛2

𝑛2 𝑛

.1

12

1 +𝑡2

𝑛

𝑛+12

− ∞ < 𝑡 < ∞

= ⋯

(17) 𝑀𝑌 𝑡 = 𝐸 𝑒𝑡𝑌 = 𝐸 𝑒𝑡 𝑋𝑖2𝑛

1

= 𝐸 𝑒𝑡𝑋𝑖2

𝑛

𝑖=1

= 𝑀𝑋𝑖2

𝑛

𝑖=1

𝑡

𝑋𝑖2 ∼ 𝜆1

2 → = 1 − 2𝑡 −12

𝑛

𝑖=1

= 1 − 2𝑡 −𝑛2

⟹ 𝑌 ∼ 𝜆𝑛2

𝑋𝑛+1 ∼ 𝑁 0, 1

𝑌 ∼ 𝜆𝑛2 > 𝑖𝑛𝑑𝑒𝑝.

𝑗𝑡 𝑝. 𝑑. 𝑓. 𝑜𝑓 𝑌 & 𝑋𝑛+1

𝑓𝑌,𝑋𝑛+1 𝑦, 𝑥 =

1

2𝑛2⎾

𝑛2

𝑒−𝑦2𝑦

𝑛2−1 ×

1

2𝜋 𝑒−

𝑥2

2

𝑇 =𝑋𝑛+1

𝑌𝑛

𝑈 = 𝑌

} ⟹ 𝑋𝑛+1 = 𝑇 𝑢

𝑛 𝑌 = 𝑈

𝐽 = 𝑢

𝑛

𝑡

2 𝑛 𝑢0 1

= 𝑢

𝑛.

𝑗𝑡 𝑝. 𝑑. 𝑓. 𝑜𝑓 𝑇 & 𝑈

𝑓𝑇,𝑈 𝑡, 𝑢 = 2𝑛2⎾

𝑛

2 2𝜋 𝑛

−1

exp −1

2 𝑡2𝑢

𝑛 exp −

𝑢

2 𝑢

𝑛2−1; −∞ < 𝑡 < ∞, 𝑢 > 0

𝑓𝑇 𝑡 = 2𝑛2⎾

𝑛

2 2𝜋 𝑛

−1

𝑢𝑛2−1

1

0

exp −𝑢

2 1 +

𝑡2

𝑛 𝑑𝑢

= 2𝑛2⎾

𝑛

2 2𝜋 𝑛

−1

⎾

𝑛 + 12

12

1 +𝑡2

𝑛

𝑛+12

; −∞ < 𝑡 < ∞

=⎾

𝑛 + 12

𝜋 ⎾𝑛2 𝑛

1 +𝑡2

𝑛

−𝑛+1

2

; −∞ < 𝑡 < ∞

(18) Z= X + Y ; Z∈ {0, 1, …}

𝑃 𝑧 = Ʒ = 𝑃 𝑋 + 𝑌 = Ʒ = 𝑃 𝑋 = 𝑥 ∩ 𝑌 = Ʒ − 𝑥

Ʒ

𝑥=0

= 𝑃 𝑋 = 𝑥 ∩ 𝑌 = Ʒ − 𝑥

Ʒ

𝑥=0

= 𝑃 𝑋 = 𝑥

Ʒ

𝑥=0

𝑃 𝑌 = Ʒ − 𝑥

= 𝑞𝑥𝑝

Ʒ

𝑥=0

𝑞Ʒ−𝑥𝑝 = 𝑝2 𝑞Ʒ

Ʒ

𝑥=0

𝑖. 𝑒. 𝑃 𝑧 = Ʒ = 𝑝2𝑞Ʒ Ʒ + 1 , Ʒ = 0, 1, …


𝑃 𝑋 = 𝑥, 𝑍 = Ʒ = 𝑃 𝑋 = 𝑥, 𝑌 = Ʒ − 𝑥

= 𝑝2𝑞Ʒ; 𝑥 = 0, , 1 … , Ʒ; Ʒ = 0, 1, … .


Problem Set-9

[1] Let 𝑋𝑛 be a sequence of 𝑁 1

𝑛, 1 −

1

𝑛 , 𝑠𝑕𝑜𝑤 𝑡𝑕𝑎𝑡 𝑋𝑛 → 𝑍, 𝑤𝑕𝑒𝑟𝑒 𝑍 ∼ 𝑁 0, 1 .

[2] Let 𝑋𝑛 be a sequence of i.i.d. random variables with E(𝑋𝑖)=𝜇, 𝑉𝑎𝑟 𝑋𝑖 = 𝜎2 𝑎𝑛𝑑 𝐸(𝑋𝑖 −

𝜇)4 = 𝜎4 + 1. 𝐹𝑖𝑛𝑑 lim𝑛→∞ 𝑃[𝜎2 −1

𝑛 ≤

𝑋1−𝜇 2+⋯+ 𝑋𝑛−𝜇 2

𝑛 ≤ 𝜎2 +

1

𝑛 ] .

[3] Let 𝑋1 , 𝑋2 , … , 𝑋𝑛 𝑏𝑒 𝑖. 𝑖. 𝑑. 𝐵 1, 𝑝 , 𝑆𝑛 = 𝑋𝑖𝑛𝑖=1 . 𝐹𝑖𝑛𝑑 𝑛 𝑤𝑕𝑖𝑐𝑕 𝑤𝑜𝑢𝑙𝑑 𝑔𝑢𝑎𝑟𝑎𝑛𝑡𝑒𝑒

𝑃 𝑆𝑛

𝑛− 𝑝 ≥ 0.01 ≤ 0.01, no matter whatever the unknown p may be.

[4] Let 𝑋1 , … , 𝑋𝑛 be i.i.d. from a distribution with mean

𝜇 𝑎𝑛𝑑 𝑓𝑖𝑛𝑖𝑡𝑒 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝜎2 . 𝑃𝑟𝑜𝑣𝑒 𝑡𝑕𝑎𝑡 𝑛 (𝑋𝑛 ͟−𝜇)

𝑆𝑛 → 𝑍, 𝑤𝑕𝑒𝑟𝑒 𝑍 ∼ 𝑁 0, 1 .

[5] The p. d. f. of a random variable X is f(x)= 1

𝑥2 𝑥 ≥ 1


Consider a random sample of size 72 from the distribution having the above p. d. f. compute,

approximately, the probability that more than 50 of these observations are less than 3.

[6] Let 𝑋1 , … , 𝑋100 be i. i. d. from poisson (3) distribution and let Y= 𝑋𝑖100𝑖=1 . Using CLT, find an

approximate value of P(100≤ Y ≤ 200).

[7] Let X∼ Bin (100, 0.6). Find an approximate value of P(10 ≤X ≤ 16).

[8] The p. d. f. of 𝑋𝑛 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 𝑓𝑛 𝑥 = 1

⎾𝑛 𝑒−𝑥𝑛−1 𝑥 > 0


Find the limiting distribution of 𝑌𝑛 =𝑋𝑛

𝑛.

[9] Let X̅ denote the mean of a random sample OF SIZE 64 FROM THE Gamma distribution with density

𝑓𝑛 𝑥 =

1

⎾𝑝𝛼𝑝 𝑒−

𝑥𝛼𝑥𝑛−1 𝑥 > 0


𝑊𝑖𝑡𝑕 𝛼 = 2, 𝑝 = 4. Compute the approximate value of P(7 < X̅ < 9).

[10] 𝑋1 , … , 𝑋𝑛 is a random sample from U(0, 2). Let 𝑌𝑛 = 𝑋𝑛 , 𝑠𝑕𝑜𝑤 𝑡𝑕𝑎𝑡 𝑛 𝑌𝑛 − 1 → 𝑁 0,

1

3 .

Solution Set

(1) 𝐹𝑋𝑛 𝑥 = 𝑃 𝑋𝑛 ≤ 𝑥 = 𝑃

𝑋𝑛−1

𝑛

1−1

𝑛

≤𝑥−

1

𝑛

1−1

𝑛

= 𝛷𝑥 −

1𝑛

1 −1𝑛

⟶ 𝛷 𝑥 𝑎𝑠 𝑛 → ∞

⟹ 𝑋𝑛 ⟶ 𝑋 ∼ 𝑁 0,1

𝐴𝑙𝑡 𝑚. 𝑔. 𝑓. 𝑜𝑓 𝑋𝑛

𝑀𝑋𝑛 𝑡 = exp

𝑡

𝑛+

𝑡2

2 1 −

1

𝑛

⟶ 𝑒𝑡2

2 ← 𝑚. 𝑔. 𝑓. 𝑜𝑓 𝑁 0, 1

⟹ 𝑋𝑛 ⟶ 𝑋 ∼ 𝑁 0, 1 .

(2) 𝑌𝑖 = 𝑋𝑖 − 𝜇 2

𝐸 𝑌𝑖 = 𝐸 𝑋𝑖 − 𝜇 2 = 𝜎2

𝑉 𝑌𝑖 = 𝐸 𝑋𝑖 − 𝜇 2 − 𝜎2 2

= 𝐸 𝑋𝑖 − 𝜇 4 + 𝜎4 − 2𝜎2𝐸 𝑋𝑖 − 𝜇 2

= 𝜎4 + 1 + 𝜎4 − 2𝜎4 = 1

𝑖. 𝑒. 𝐸 𝑌𝑖 = 𝜎2; 𝑉 𝑌𝑖 = 1 ∀ 𝑖 & 𝑌1 … 𝑌𝑛 𝑖. 𝑖. 𝑑.

𝑆𝑛 = 𝑌𝑖

𝐸𝑆𝑛 = 𝑛𝜎2

𝑉 𝑆𝑛 = 𝑛

𝐶𝐿𝑇 ⟹𝑆𝑛 − 𝐸 𝑆𝑛

𝑉 𝑆𝑛 → 𝑁 0, 1

𝑖. 𝑒. 𝑋1 − 𝜇 2 + … + 𝑋𝑛 − 𝜇 2 − 𝑛𝜎2

𝑛→ 𝑋 ∼ 𝑁 0, 1 .

lim𝑛→∞

𝑃 𝜎2 −1

𝑛≤

𝑋1 − 𝜇 2 + … + 𝑋𝑛 − 𝜇 2

𝑛≤ 𝜎2 +

1

𝑛

= lim𝑛→∞

𝑃 −1

𝑛≤

𝑋1 − 𝜇 2 + … + 𝑋𝑛 − 𝜇 2 − 𝑛𝜎2

𝑛≤

1

𝑛

= lim𝑛→∞

𝑃 −1 ≤ 𝑋1 − 𝜇 2 + … + 𝑋𝑛 − 𝜇 2 − 𝑛𝜎2

𝑛≤ 1

= 𝛷 1 − 𝛷 −1 = 2𝛷 1 − 1 = ⋯

(3) 𝐸𝑆𝑛 = 𝑛𝑝 ; 𝑉𝑆𝑛 = 𝑛𝑝𝑞

𝑃 𝑆𝑛

𝑛− 𝑝 ≥ 𝑡 ≤

𝐸 𝑆𝑛 − 𝑛𝑝 2

𝑡2𝑛2=

𝑛𝑝 1 − 𝑝

𝑛2𝑡2≤

1

4𝑛𝑡2≤ 0.01 𝑔𝑖𝑣𝑒𝑛

⟹ 𝑛 ≥1

0.04𝑡2 𝑓𝑜𝑟 𝑡 = 0.01

𝑛 ≥ ⋯

(4) 𝐶𝐿𝑇 ⟹ 𝑛 𝑋𝑛 −𝜇

𝜎⟶ 𝑍 ∼ 𝑁 0, 1

𝐴𝑙𝑠𝑜 𝑆𝑛2 =

1

𝑛 𝑋𝑖 − 𝑋 2

𝑛

𝑖=1

⟶ 𝜎2

(𝑆𝑛2 =

1

𝑛 𝑋𝑖

2 − 𝑋 2) ⟹ 𝑆𝑛 ⟶ 𝜎,

⎵

↓p ↓p

𝜎2 + 𝜇2 𝜇2 →

⎵

↓p

𝜎2

Using slutsky’s 𝑋𝑛 → 𝑋; 𝑌𝑛 → 𝑐𝑋𝑛

𝑌𝑛→

𝑋

𝑐

𝑛 𝑋𝑛 − 𝜇 𝜎

𝑆𝑛𝜎

→ 𝑋 ∼ 𝑁 0, 1

𝑖. 𝑒. 𝑛 𝑋𝑛

− 𝜇

𝑆𝑛⟶ 𝑋 ∼ 𝑁 0, 1 .

(5) 𝑋1 … 𝑋72 𝑟. 𝑠. 𝑓𝑟𝑜𝑚 𝑓 𝑥 = 1

𝑥2 𝑥 > 1


𝐷𝑒𝑓𝑖𝑛𝑒 𝑌𝑖 = 1 𝑖𝑓 𝑋𝑖 < 3


𝑃 𝑌𝑖 = 1 = 𝑃 𝑋𝑖 < 3 = 1

𝑥2 𝑑𝑥

3

1

=2

3= 𝜃 𝑠𝑎𝑦

𝑌1 , … , 𝑌72 𝑎𝑟𝑒 𝑖. 𝑖. 𝑑. 𝐵 1, 𝜃

𝑌 = 𝑌𝑖

72

𝑖=1

∼ 𝐵 72, 𝜃 =2

3

𝐶𝐿𝑇 ⟹𝑌 − 72 ×

23

72 ×23 ×

13

⟶ 𝑍 ∼ 𝑁 0, 1

𝑖. 𝑒.𝑌 − 48

4⟶ 𝑍 ∼ 𝑁 0, 1

𝑃 𝑌 > 50 = 1 − 𝑃 𝑌 ≤ 50 = 1 − 𝑃 𝑌 ≤ 50.5 ←

= 1 − 𝑃 𝑌 − 48

4≤

50.5 − 48

4

≈ 1 − 𝛷 2.5

4 = ⋯

(6) 𝑋1 … 𝑋100 𝑖. 𝑖. 𝑑 𝑃 3

𝐸 𝑋1 = 3; 𝑉 𝑋1 = 3 ; 𝑌 = 𝑋1

100

𝑖=1

∼ 𝑃 300 ⟹ 𝐸 𝑌 = 𝑉 𝑦 = 3

𝐶𝐿𝑇 ⟹𝑌 − 300

10 3 =

𝑆𝑛 − 𝐸𝑆𝑛

𝑉𝑆𝑛

→ 𝑁 0, 1

𝑃 100 ≤ 𝑌 ≤ 200 = 𝑃 99.5 ≤ 𝑌 ≤ 200.5 ← 𝑐𝑜𝑛𝑡 𝑐𝑜𝑟𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛

= 𝑃 99.5 − 300

10 3≤

𝑌 − 300

10 3≤

200.5 − 300

10 3

≈ 𝛷 200.5 − 300

10 3 − 𝛷

99.5 − 300

10 3 .

(7) 𝑋 ∼ 𝑏𝑖𝑛 100, 0.6

𝐶𝐿𝑇 ⟹𝑋 − 100 × 0.6

100 × 0.6 × 0.4=

𝑋 − 60

24⟶ 𝑍 ∼ 𝑁(0, 1)

⟹ 𝑃 10 ≤ 𝑋 ≤ 16 = 𝑃 9.5 ≤ 𝑋 ≤ 16.5

= 𝑃 9.5 − 60

24≤

𝑋 − 60

24≤

16.5 − 60

24

≈ 𝛷 16.5 − 60

24 − 𝛷

9.5 − 60

24

= ⋯

(8)𝑋𝑛 has p.d.f.

𝑓𝑛 𝑥 = 1

⎾𝑛𝑒−𝑥𝑥𝑛−1 𝑥 > 0


m. g. f. of 𝑋𝑛

𝑀𝑋𝑛 𝑡 =

1

⎾𝑛 𝑒𝑡𝑥

∞

0

𝑒−𝑥𝑥𝑛−1 𝑑𝑥

= 1

⎾𝑛 𝑒−𝑥 1−𝑡

∞

0

𝑥𝑛−1 𝑑𝑥

=1

(1 − 𝑡)𝑛= (1 − 𝑡)𝑛

m. g. f. of 𝑌𝑛 =𝑋𝑛

𝑛 𝑖𝑠

𝑀𝑌𝑛 𝑡 = 𝐸 𝑒𝑡

𝑋𝑛𝑛 = 1 −

𝑡

𝑛 −𝑛

⟶ 𝑒𝑡 𝑎𝑠 𝑛 → ∞

↑ 𝑚. 𝑔. 𝑓. 𝑟. 𝑣. 𝑑𝑒𝑛𝑔 𝑎𝑡 𝑥 = 0

𝑌𝑛 ⟶ 𝑋 (𝑑𝑒𝑔𝑟𝑒𝑒 𝑎𝑡 1)

(9)

𝑓𝑋𝑛 𝑥 = ⎾𝑝

1

𝛼𝑝 𝑒−

𝑥𝛼 𝑥𝑝−1 𝑥 > 0 𝛼 = 2, 𝑝 = 4


𝐸 𝑋𝑛 = 𝛼𝑝 ; 𝑉 𝑋𝑛 = 𝛼2𝑝 = 16 = 𝜎2

𝐸 𝑋 = 8; 𝑉 𝑋 =16

𝑛=

1

4

By C LT 𝑛 𝑋 −𝛼𝑝

𝛼2𝑝→ 𝑍 ∼ 𝑁 0, 1

𝑖𝑒. 2 𝑋 − 8 ⟶ 𝑍 ∼ 𝑁 0,1

𝑃 7 < 𝑋 < 9 = 𝑃 2 7 − 8 < 2 𝑋 − 8 < 2 9 − 8

≈ 𝑃 −2 < 𝑧 < 2

= 𝛷 2 − 𝛷 −2 = 2𝛷 2 − 1

(10)

𝑋𝑖 ∼ 𝑈 0, 2 𝐸𝑋𝑖 =1

2 𝑥

2

0

𝑑𝑥 = 1

𝐸𝑋𝑖2 =

1

2 𝑥2

2

0

𝑑𝑥 =4

3; 𝑉 𝑋𝑖 =

1

3

𝑋1 , … , 𝑋𝑛 i. i. d. with E𝑋1 = 1 & 𝑉𝑋1 =1

3

𝐵𝑦 𝐶𝐿𝑇 𝑛 𝑋𝑛 − 1 ⟶ 𝑁 0,

1

3 .

𝑖. e. 𝑛 𝑌𝑛 − 1 ⟶ 𝑁 0,1

3 .

Problem Set-10

[1] Let 𝑋1 , 𝑋2 , … , 𝑋𝑛 be a random sample from an exponential distribution with p.d.f.

𝑓𝑥 𝑥 =1

𝛽exp −

𝑥

𝛽 ; 𝑥 > 0

Show that X̅= 𝑋𝑖𝑛𝑖=1 /𝑛 is an unbiased estimator of𝛽.

[2] Let 𝑋1 , 𝑋2 , … , 𝑋𝑛 be a random sample from U (0, ); 𝜃> 0, show that

𝑛+1

𝑛 𝑋(𝑛) 𝑎𝑛𝑑 2𝑋 are both unbiased estimators of 𝜃.

[3] Let 𝑋1 , 𝑋2 , … , 𝑋𝑛 be a random sample from an exponential distribution with p.d.f.

𝑓 𝑥 = 𝛽 exp −𝛽 𝑥 ; 𝑥 > 0

Show that X̅ is an unbiased estimator of 1

𝛽.

[4] Let 𝑋1 , 𝑋2 , … , 𝑋𝑛 be a random sample from 𝑁 0, 𝜃2 , 𝜃 > 0. 𝑆𝑕𝑜𝑤 𝑡𝑕𝑎𝑡 𝑋𝑖

𝑛𝑖=1

2

𝑛(𝑛+1) 𝑎𝑛𝑑

𝑋𝑖2𝑛

𝑖=1

2𝑛 are

both unbiased estimators of 𝜃2.

[5] Let 𝑋1 , 𝑋2 , … , 𝑋𝑛 be a random sample from P( 𝜃); 𝜃 >0. Find an unbiased estimator of 𝜃 𝑒−2𝜃 .

[6] Let 𝑋1 , 𝑋2 , … , 𝑋𝑛 be a random sample from B(1, 𝜃); 0≤ 𝜃 ≤1.

(a) Show that the estimator 𝑇 𝑋 =1

2 𝑛+ 𝑋𝑖

𝑛𝑖=1

𝑛+ 𝑛 is not unbiased 𝜃?

(b) Show that lim𝑛→∞ 𝐸 𝑇 𝑋 = 𝜃.

(An estimator satisfying the condition in (b) is said to be unbiased in the limit)

[7] 𝑋1 , … , 𝑋𝑛 be a random sample from N 𝜇, 𝜎2 , 𝜇 ∈ ℜ, 𝜎 ∈ ℜ+. Find unbiased estimators of 𝜇

𝜎2 𝑎𝑛𝑑𝜇

𝜎.

[8] Let 𝑋1 , 𝑋2 , … , 𝑋𝑛 be a random sample from B(1, 𝜃); 0 ≤ 𝜃 ≤1. Find an unbiased estimator of

𝜃2(1 − 𝜃).

[9] Using Neyman Fisher Factorization Theorem, find a sufficient based on a random sample

𝑋1 , 𝑋2 , … , 𝑋𝑛 from each of the following distributions

(a) 𝑓𝛼 𝑥 = 1

𝛼 exp −

𝑥

𝛼 𝑖𝑓 𝑥 > 0


(b) 𝑓𝛽 𝑥 = exp − 𝑥 − 𝛽 𝑖𝑓 𝑥 > 𝛽


(c) 𝑓𝛼 ,𝛽 𝑥 = 1

𝛼exp −

𝑥−𝛽

𝛼 𝑖𝑓 𝑥 > 𝛽


(d) 𝑓𝜇 ,𝜎 𝑥 = 1

𝑥𝜎 2𝜋 exp −

𝑙𝑜𝑔 𝑥1−𝜇 2

2𝜎2 𝑖𝑓 𝑥 > 0


(e) 𝑓𝜃 𝑥 = 1

𝜃 –

𝜃

2≤ 𝑥 ≤

𝜃

2


[10] Let 𝑋1 𝑎𝑛𝑑 𝑋2 be independent random samples with densities 𝑓1 𝑥1 = 𝜃𝑒−𝜃 𝑥1 𝑎𝑛𝑑 𝑓2 𝑥2 =

2𝜃𝑒−2𝜃 𝑥2 as the respective p.d.f.s where 𝜃 >0 is an unknown parameter and 0<𝑥1 , 𝑥2 < ∞. Using

Neyman Fisher Factorization Theorem find a sufficient stastic for 𝜃.

[11]Let 𝑋1 , … , 𝑋𝑛 be a random sample with densities

𝑓𝑥𝑖 𝑥 =

exp 𝑖𝜃 − 𝑥 𝑖𝑓 𝑥 ≥ 𝑖𝜃0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒.

Using Neyman Fisher Factorization Theorem find a sufficient statistic for 𝜃.

[12] Let 𝑋1 , 𝑋2 , … , 𝑋𝑛 be a random sample from a Beta (𝛼, 𝛽 ) distribution 𝛼 > 0, 𝛽 > 0 𝑤𝑖𝑡𝑕 𝑝. 𝑑. 𝑓.

𝑓 𝑥 =

⎾𝛼 + 𝛽

⎾𝛼⎾𝛽 𝑥𝛼−1 (1 − 𝑥)𝛽−1 0 < 𝑥 < 1


Show that

(a) 𝑋𝑖𝑛𝑖=1 is sufficient for 𝛼 if 𝛽 is known to be a given constant.

(b) (1 − 𝑋𝑖)𝑛𝑖=1 is sufficient for 𝛽 if 𝛼 is known to be given constant.

(c) 𝑋𝑖𝑛𝑖=1 , 1 − 𝑋𝑖

𝑛𝑖=1 is jointly sufficient for (𝛼, 𝛽) if both the parameters are unknown.

[13] Let T and T* be two statistic such that T= 𝛹(𝑇∗), Show that if T is sufficient then 𝑇∗ is also sufficient.

[14] 𝑋1 , … , 𝑋𝑛 be a random sample from U 𝜃 −1

2, 𝜃 +

1

2 , 𝜃 ∊ ℜ. Find a sufficient statistic for 𝜃.

[15] Let 𝑋1 , … , 𝑋𝑛 be independent random variables with 𝑋𝑖(𝑖 = 1, 2, … , 𝑛) having the

𝑓𝑖 𝑥𝑖 = 𝑖𝜃 𝑒−𝑖𝜃𝑥𝑖 𝑥𝑖 > 00 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒

Find a sufficient statistic for 𝜃.

Solution Key

(1) 𝐸 𝑋 = 1

𝛽∫ 𝑥

∞

0 𝑒−

𝛽

𝑥𝑑𝑥 = 𝛽

⟹ 𝐸 𝑋 = 𝐸 1

𝑛 𝑋𝑖 =

1

𝑛 𝐸 𝑋𝑖 = 𝛽

⟹ 𝑋 𝑖𝑠 𝑢. 𝑒. 𝑜𝑓 𝛽

(2) 𝑓𝑋 𝑛 𝑥 =

𝑛

𝜃𝑛 𝑥𝑛−1 0 < 𝑥 < 𝜃

0 𝑜/𝑤

𝐸 𝑋 𝑛 =𝑛

𝜃𝑛 𝑥𝑛

𝜃

0

𝑑𝑥 =𝑛

𝑛 + 1𝜃

⟹ 𝐸 𝑛 + 1

𝑛 𝑋 𝑛 = 𝜃

⟹𝑛 + 1

𝑛𝑋 𝑛 𝑖𝑠 𝑢. 𝑒. 𝑜𝑓 𝜃.

Also 𝑓𝑋 𝑥 = 1

𝜃 0 < 𝑥 < 𝜃

0 𝑜/𝑤

𝐸 𝑋 =𝜃

2

⟹ 𝐸 2𝑋 = 𝐸 2

𝑛 𝑋𝑖

=2

𝑛 𝐸(𝑋𝑖) = 𝜃

⟹ 2 X̅ is u. e. for 𝜃

(3) 𝐸 𝑋 = 𝛽 ∫ 𝑥∞

0 𝑒−𝛽𝑥 𝑑𝑥 =

1

𝛽

𝐸 𝑋 = 𝛽

⟹ 𝑋 𝑖𝑠 𝑢. 𝑒. 𝑜𝑓 𝛽.

𝐸 𝑋 = 𝐸 1

𝑛 𝑋𝑖 =

1

𝑛 𝐸 𝑋𝑖

(4) 𝑇1 = 𝑋𝑖 , 𝑇2 = 𝑋𝑖2

𝐸 𝑇12 = 𝑉 𝑇1 + 𝐸2 𝑇1

= 𝑛𝜃2 + 𝑛2𝜃2 = 𝜃2𝑛 𝑛 + 1

⟹ 𝐸 𝑇1

2

𝑛 𝑛 + 1 = 𝜃2

⟹𝑇1

2

𝑛 𝑛 + 1 𝑖𝑠 𝑢. 𝑒. 𝑜𝑓 𝜃2

𝐸 𝑇2 = 𝐸 𝑋𝑖2 = 𝐸 𝑋𝑖

2

= 𝑉 𝑋𝑖 + 𝐸2 𝑋𝑖

= 𝜃2 + 𝜃2 = 2𝑛𝜃2

⟹𝑇2

2𝑛 𝑖𝑠 𝑢. 𝑒. 𝑜𝑓 𝜃2

(5) 𝑔 𝜃 = 𝜃𝑒−2𝜃

𝛿0 𝑋 = 1 𝑖𝑓 𝑋1 = 0, 𝑋2 = 0

0 𝑜/𝑤

𝐸 𝛿0 𝑋 = 1. 𝑃 𝑋1 = 0, 𝑋2 = 1

= 𝑃 𝑋1 = 0 𝑃 𝑋2 = 1

= 𝑒−𝜃 .𝑒−𝜃𝜃1

1!= 𝜃𝑒−2𝜃

⟹ 𝛿0 𝑋 is u. e. of𝜃𝑒−2𝜃 .

(6) 𝑋1 , … , 𝑋𝑛 i. i. d. B(1, 𝜃)

𝑋𝑖 ∼ 𝐵 𝑛, 𝜃

𝐸 𝑇 𝑋 =

12 𝑛 + 𝐸( 𝑋𝑖)

𝑛 + 𝑛=

12 𝑛 + 𝑛𝜃

𝑛 + 𝑛≠ 𝜃

⟹ T(X̲) is not u. e. of 𝜃.

lim𝑛→∞

𝐸(𝑇 𝑋 ) = lim𝑛→∞

12 + 𝑛𝜃

𝑛 + 𝑛= 𝜃

⟹ 𝑇 𝑋 is unbiased in the limit for 𝜃

(7) 𝑋1 , … , 𝑋𝑛r. s. from 𝜇, 𝜎2

⟹𝑋 ∼ 𝑁 𝜇,

𝜎2

𝑛

𝑌 =(𝑛 − 1)𝑆2

𝜎2 ∼ 𝜒𝑛−12

> 𝑖𝑛𝑑𝑒𝑝.

If Z∼ 𝜒𝑚2 , 𝑡𝑕𝑒𝑛

𝐸 1

𝑧 =

1

2𝑚2 𝑚/2

𝑧−1𝑒−𝑧2𝑧

𝑚2

−1∞

0

𝑑𝑧

= 1

2𝑚2

𝑚2

𝑒−𝑧2𝑧

𝑚2

−1−1∞

0

𝑑𝑧

=

𝑚2 − 1 2

𝑚 2 − 1

2𝑚2

𝑚2

=1

𝑚 − 2

& 𝐸 1

𝑧 =

1

2𝑚 2

𝑚2

𝑒−𝑧2

∞

0

𝑍𝑚2

−12−1 𝑑𝑧

= 𝑚 − 1

2 2𝑚 2

−12

2𝑚 2

𝑚2

= 𝑚 − 1

2

2 𝑚2

⟹ 𝐸 1

𝑌 = 𝐸

𝜎2

𝑛 − 1 𝑠2 =1

𝑛 − 1 − 2=

1

𝑛 − 3

⟹ 𝐸 1

𝑠2 =

𝑛 − 1

𝑛 − 3.

1

𝜎2.

& E 1

𝑌 = 𝐸

𝜎

𝑛−1 𝑆 =

𝑛−2

2

2⎾𝑛−1

2

⟹ 𝐸 1

𝑆 =

𝑛 − 1⎾𝑛 − 1

2

2 ⎾𝑛 − 1

2

=1

𝜎

𝑠𝑖𝑛𝑐𝑒 X̅ & 𝑠2 are indep.

𝐸 X

𝑠2 = 𝐸 𝑋 . 𝐸 1

𝑠2

= 𝜇.𝑛 − 1

𝑛 − 3.

1

𝜎2.

⟹ 𝐸 𝑛−3

𝑛−1.𝑋

𝑠2 =𝜇

𝜎2

⟹ 𝑛−3

𝑛−1.𝑋

𝑠2 is an unbiased estimator of 𝜇

𝜎2.

Further

𝐸 X

𝑠 = 𝐸 𝑋 . 𝐸

1

𝑆

= 𝜇. 𝑛 − 1⎾

𝑛 − 12

2 ⎾𝑛 − 1

2

.1

𝜎

⟹ 𝐸 2

𝑛 − 1

⎾𝑛 − 1

2

⎾𝑛 − 1

2

X

𝑠 =

𝜇

𝜎

⟹ 2

𝑛−1.⎾

𝑛−1

2

⎾𝑛−1

2

.X

𝑠 is an unbiased estimator of

𝜇

𝜎.

(8) 𝑋1 , … , 𝑋𝑛 are i. i. d. B(1, 𝜃)

𝑔 𝜃 = 𝜃2(1 − 𝜃)

𝐷𝑒𝑓𝑖𝑛𝑒 (X̲)= 1 𝑖𝑓 𝑋1 = 1, 𝑋2 = 1, 𝑋3 = 0

0 𝑜/𝑤

𝐸𝜃 δ X = P 𝑋1 = 1, 𝑋2 = 1, 𝑋3 = 0

= P 𝑋1 = 1 𝑃 𝑋2 = 1 𝑃 𝑋3 = 0

= 𝜃2(1 − 𝜃)

⟹ (X̲) is an u. e. of g(𝜃)= 𝜃2(1 − 𝜃).

(9)

(a) f(x│𝛼)= 1

𝛼 𝑒−

𝑥

𝛼 ; 𝑥 > 0

jt p. d. f. 𝑓 𝑥 𝛼 =1

𝛼𝑛 𝑒−1

𝛼 𝑥𝑖 𝑋1 , … , 𝑋𝑛 > 0

= 1

𝛼𝑛 𝑒−

1𝛼

𝑥𝑖 . 1

𝑖. 𝑒. 𝑓 𝑥 𝛼 = 𝑔 𝛼, 𝑥𝑖

𝑛

1

. 𝑕 𝑥 𝑕 𝑥 = 1 .

By NFFT, T (X̲) = 𝑥𝑖𝑛1 is suff for 𝛼

(9)(b) 𝑓 𝑥 𝛽 = 𝑒− 𝑥−𝛽 𝑥 > 𝛽

𝑓 𝑥 𝛽 = 𝑒− (𝑥𝑖−𝛽), 𝑋1 , … , 𝑋𝑛 > 𝛽

0 𝑜/𝑤

𝑖. 𝑒. 𝑓 𝑥 𝛽 = 𝑒− 𝑥𝑖+𝑛𝛽 , 𝑥 1 > 𝛽

0 𝑜/𝑤

𝑖. 𝑒. 𝑓 𝑥 𝛽 = 𝑒𝑛𝛽 − 𝑥𝑖 𝐼 𝛽 ,𝑥 1 𝐼 𝑎 ,𝑏 =

1 𝑎 < 10 𝑜/𝑤

= 𝑒− 𝑥𝑖 𝑒𝑛𝛽 𝐼 𝛽 ,𝑥 1

= 𝑕 𝑥 𝑔 𝛽, 𝑥 1

By NFFT, T(X̲)= 𝑋 1 is a suff statistic.

(9) (c) 𝑓 𝑥 𝛼, 𝛽 =1

𝛼𝑛 exp − 𝑥𝑖

𝛼+

𝑛𝛽

𝛼 𝐼 𝛽 ,𝑥 1

= 1

𝛼𝑛exp

− 𝑥𝑖

𝛼+

𝑛𝛽

𝛼 . 𝐼 𝛽 ,𝑥 1

. 1

↓

= 𝑔 𝛼, 𝛽 ; 𝑥𝑖 𝑥 1 . 𝑕(𝑥 )

By, NFFT, T(X̲)= 𝑥𝑖 𝑥 1 is jointly sufficient for (𝛼, 𝛽).

(9)(d) 𝑓 𝑥 𝜇, 𝜎 = 1

𝜎 2𝜋 𝑛

1

𝑥𝑖 𝑛

𝑖=1 . exp −1

2𝜎2 (𝑙𝑜𝑔𝑥𝑖 − 𝜇)2

= 1

𝜎𝑛exp −

𝑛𝜇2

2𝜎2−

1

2𝜎2 (𝑙𝑜𝑔𝑥𝑖)

2 +𝜇

𝜎2 𝑙𝑜𝑔𝑥𝑖 ×

1

2𝜋 𝑛

𝑥𝑖−1

𝑛

𝑖=1

= 𝑔 𝜇, 𝜎 ; 𝑙𝑜𝑔𝑥𝑖 , (𝑙𝑜𝑔𝑥𝑖)2 . 𝑕(𝑥 )

1

2𝜋 𝑛

𝑥𝑖−1

𝑛

𝑖=1

By NFFT 𝑙𝑜𝑔𝑥𝑖 , (𝑙𝑜𝑔𝑥𝑖)2 is jointly sufficient for (𝜇, 𝜎)

(9)(e)f(x│̰𝜃)= 1

𝜃𝑛 ; −𝜃

2< 𝑥1 …𝑥𝑛 < 𝜃/2

0 𝑜/𝑤

=

1

𝜃𝑛; │𝑥𝑖│ <

𝜃

2; 𝑖 = 1, 2, …𝑛

0 𝑜/𝑤

i.e. f(x│̰𝜃)= 1

𝜃𝑛 ; Max𝑖 │𝑥𝑖│ <𝜃

2; 𝑖 = 1, 2, …𝑛

0 𝑜/𝑤

⟹ f (x│̰𝜃)= 1

𝜃𝑛 𝐼 Max𝑖 │𝑥𝑖│ ,𝜃

2

⟹By NFFT T(X̲)= Max𝑖

│𝑥𝑖│ is suff for 𝜃.

(10) Joint p. d. f. of 𝑋1 & 𝑋2

𝑓 𝑥1 , 𝑥2 = 𝜃𝑒−𝜃𝑥1 . 2𝜃 𝑒−2𝜃𝑥2

𝑓 𝑥1 , 𝑥2 = 𝜃2𝑒−𝜃 𝑥1+2𝑥2

𝑓 𝑥1 , 𝑥2 = 𝜃2𝑒−𝜃 𝑥1+2𝑥2 2

BY NFFT T 𝑋1 , 𝑋2 = 𝑋1 + 2𝑋2 is suff for 𝜃.

(11) jt p. d. f.

f (x│̰𝜃)= exp 𝑖𝜃 − 𝑥𝑖 𝑖𝑓

𝑥1

1,𝑥2

2, … ,

𝑥𝑛

𝑛≥ 𝜃 𝑛

𝑖=1

0 𝑜/𝑤

= 𝑒− 𝑥𝑖

𝑛1 𝑒𝜃

𝑛(𝑛+1)2 𝑖𝑓

𝑥𝑖

𝑖≥ 𝜃

0 𝑜/𝑤

𝑖. 𝑒. f x θ = 𝑒𝜃𝑛 𝑛+1

2 I θ, Mini

𝑥𝑖

𝑖 𝑒− 𝑥𝑖

𝑛𝑖=1

↙ ↙

= g θ, Mini

𝑥𝑖

𝑖 × h(x )

⟹ T (X̰) = Mini𝑋𝑖

𝑖 is suff.

(12) 𝑋1 , … , 𝑋𝑛 𝑖. 𝑖. 𝑑 𝐵𝑒𝑡𝑎 𝛼, 𝛽

(a) 𝛽 is known – 𝛼 is the unknown parameter

f (x│̰𝛼) =

⎾𝛼+𝛽

⎾𝛼 𝑛

𝜋𝑥𝑖 𝛼−1

𝜋 1 − 𝑥𝑖 𝛽−1

1

⎾𝛽 𝑛

𝑕 𝑥 0 < 𝑥1 , … , 𝑥𝑛 < 1

0 𝑜/𝑤

By NFFT 𝑋𝑖𝑛𝑖=1 is suff for 𝛼.

(b) 𝛼 is known – 𝛽 is the unknown parameter

f (x│̰𝛼) =

⎾𝛼+𝛽

⎾𝛽 𝑛

𝜋 1 − 𝑥𝑖 𝛽−1

𝜋𝑥𝑖

𝛼−1 1

⎾𝛼 𝑛

𝑕 𝑥 0 < 𝑥1 , … , 𝑥𝑛 < 1

0 𝑜/𝑤

By NFFT (1 − 𝑋𝑖)𝑛𝑖=1 is suff for 𝛽.

(c) 𝛼, 𝛽 both unknown

f (x│̰𝛼, 𝛽) =

⎾𝛼+𝛽

⎾𝛼⎾𝛽 𝑛 𝜋𝑥𝑖

𝛼−1 𝜋 1 − 𝑥𝑖 𝛽−1

. 1 0 < 𝑥1 , … , 𝑥𝑛 < 1

0 𝑜/𝑤

By NFFT (𝜋𝑥𝑖 , 𝜋 1 − 𝑥𝑖 ) is jointly sufficient for (𝛼, 𝛽).

(13) T is suff for 𝜃∊𝛩 & T= 𝛹(𝑇∗)

By NFFT T is suff for 𝜃 iff.

f (x│̰𝜃) = g(𝜃, t(x )) h(x )

i.e. f (x│̰𝜃) = g(𝜃, 𝛹(𝑡∗(x )) . h(x )

f (x│̰𝜃 = g’ 𝜃, 𝑡∗(x )) . h(x )

⟹ 𝑇∗(X ) is suff for 𝜃.

(14) f (x│̰𝜃) = 1, 𝜃 −

1

2< 𝑥(1), … , < 𝑥(𝑛) < 𝜃 +

1

2

0 𝑜/𝑤

i.e. f (x│̰𝜃) = 𝐼 𝜃−

1

2,𝑥 1

𝐼 𝑥 𝑛 ,𝜃+

1

2

= 𝑔 𝜃, 𝑥 1 , 𝑥 𝑛 𝑕(𝑥 )

By NFFT, T(X )= 𝑋 1 , 𝑋 𝑛 is jointly suff for 𝜃.

(15) f (x│̰𝜃)= 𝜃 − 𝑒−𝜃𝑥1 2𝜃 − 𝑒−2𝜃𝑥2 … 𝑛𝜃 − 𝑒−𝑛𝜃𝑥𝑛

i.e. f (x│̰𝜃)= 𝜃𝑛 𝑖𝑛𝑖=1 𝑒−𝜃 𝑖𝑥𝑖

𝑛𝑖=1

= 𝑖

𝑛

𝑖=1

𝜃𝑛 𝑒−𝜃 𝑖𝑥𝑖𝑛𝑖=1

= 𝑕 𝑥 𝑔 𝜃, 𝑖𝑥𝑖

𝑛

𝑖=1

By NFFT, T(X )= 𝑖𝑋𝑖𝑛𝑖=1 is sufficient for 𝜃.

Problem Set-11

[1] Let 𝑋1 , … , 𝑋𝑛 be a random sample from P(𝜃), 𝜃∊ (0, ∞). Show that T= 𝑋𝑖𝑛𝑖=1 is complete

sufficient statistic. Find the Uniformly Minimum Variance Unbiased Estimator (UMVUE) of the

following parametric functions: (a) g(𝜃) = 𝜃, (b) g(𝜃) = 𝑒−𝜃 (c) g(𝜃) = 𝑒−𝜃(1 + 𝜃).

[2] Suppose 𝑋1 , … , 𝑋𝑛 be a random sample from B(1, 𝜃), 𝜃∊ (0, 1). Show that T= 𝑋𝑖𝑛𝑖=1 is

complete sufficient statistic and hence find the UMVUE for each of the following parametric

functions : (a) g(𝜃) = 𝜃, (b) g(𝜃)= 𝜃4 and (c) g(𝜃) = 𝜃 1 − 𝜃 2 .

[3] Let 𝑋1 , … , 𝑋𝑛 be a random sample from Exp (𝜃, 1), i. e.

f (x│𝜃)= 𝑒− 𝑥−𝜃 𝑖𝑓 𝑥 > 𝜃0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒

Show that T= 𝑋 1 = min{𝑋1 , … , 𝑋𝑛} is a complete sufficient statistic and hence find the UMVUE of g

(𝜃) =𝜃2.

[4] 𝑋1 , … , 𝑋𝑛 is a random sample from U(0, 𝜃), 𝜃 > 0. Show that T=𝑋 𝑛 = max{𝑋1, … , 𝑋𝑛} is a

complete sufficient statistic and find the UMVUE of g(𝜃) =𝜃𝑘 ; 𝑘 > −𝑛.

[5] 𝑋1 , … , 𝑋𝑛 is a random sample from Gamma (2, 𝜃), 𝜃> 0, i.e.

f x│𝜃) = 1

⎾2𝜃2 𝑒−𝑥/𝜃𝑥 𝑖𝑓 𝑥 > 0


Show that T= 𝑋𝑖𝑛𝑖=1 is complete sufficient statistic and find the UMVUE of 𝜃.

[6] Let 𝑋1 , … , 𝑋𝑛 be a random sample from U(𝜃 – ½ , 𝜃 + ½). Show that sufficient statistic is not

complete.

[7] Suppose the statistic T is UMVUE of 𝜃 such that V(T) ≠ 0. Show that 𝑇2 cannot be UMVUE of 𝜃2.

[8] Let 𝑋1 , … , 𝑋𝑛 be a random sample from N(0, 𝜃). Find the UMVUE of𝜃2.

[9] Let 𝑋1 , … , 𝑋𝑛 be a random sample from N(𝜇, 𝜃). Find the UMVUE of (a) 𝜃 when 𝜇 is known and

(b) 𝜃 when 𝜇 is not known.

[10] Let 𝑋1 , … , 𝑋𝑛 be a random sample from N (𝜇, 𝜎2). Find the UMVUE of (a) 𝜎𝑟 when 𝜇 is known,

(b) 𝜎𝑟 when 𝜇 is not known and (c) 𝛿, where 𝛿 is given by P X≥ 𝛿)= p for a given p.

[11] 𝑋1 , … , 𝑋𝑛 is a random sample from U (0, 𝜃), 𝜃 > 0. Consider the following 3 estimators for 𝜃;

𝑇1 𝑋 =𝑛 + 1

𝑛𝑋(𝑛), 𝑇2 𝑋 = 2𝑋 𝑎𝑛𝑑 𝑇3 𝑋 = 𝑋(1) + 𝑋(𝑛).

Show that all the estimators are unbiased for 𝜃. Among the three estimators, which are would you

prefer and why?

[12] 𝑋1 , … , 𝑋𝑛 be a random sample from N (𝜇, 𝜎2), 𝜇∊ ℜ, 𝜎∊ℜ+. Assuming completeness of the

associated sufficient statistic find the UMVUE of 𝜇2 and 𝜇 +.

[13] Let 𝑋1 , … , 𝑋𝑛 be a random sample from Exp (a, b). Assuming completeness of the associated

sufficient statistic, find the (a) UMVUE of a when b is known and (b) UMVUE of b when a is known.

Solution Key

(1) (a) g(𝜃)= 𝜃

E(T) = n𝜃 ⟹ E(T/n) = 𝜃

T/n u. e. based on CSS.

⟹ 𝑇

𝑛 is UMVUE of 𝜃

(b) g(𝜃) = 𝑒−𝜃

Consider 𝛿0 𝑋 = 1, 𝑖𝑓 𝑋1 = 0

0, 𝑜/𝑤

E𝛿0 𝑋 = 𝑃 𝑋1 = 0 = 𝑒−𝜃 ⟹ 𝛿0 𝑋 𝑖𝑠 𝑢. 𝑒. 𝑜𝑓𝑒−𝜃

Rao-blackwellization of 𝛿0 𝑋

(t) = E(𝛿0 𝑋 │𝑇 = 𝑡)

= P(𝑋1 = 0│𝑇 = 𝑡) = 𝑃 𝑋1=0,𝑇=𝑡

𝑃 𝑇=𝑡 =

𝑃 𝑋1=0, 𝑋𝑖=𝑡𝑛𝑖=2

𝑃 𝑇=𝑡

=𝑃 𝑋1 = 0 𝑃 𝑋𝑖 = 𝑡𝑛

𝑖=2

𝑃 𝑇 = 𝑡 =

𝑒−𝜃 𝑒− 𝑛−1 𝜃 𝑛 − 1 𝜃

𝑡

𝑡!

𝑒−𝑛𝜃 𝑛𝜃 𝑡

𝑡!

= 𝑛 − 1

𝑛 𝑡

(T)= 𝑛−1

𝑛 𝑇

is u. e. based on CSS

⟹ 𝑛−1

𝑛 𝑇

is UMVUE of 𝑒−𝜃 .

(c)g(𝜃) = 𝑒−𝜃(1 + 𝜃)

𝛿0 𝑋 = 1, 𝑖𝑓 𝑋1 = 0

0, 𝑜/𝑤

E𝛿0 𝑋 = 𝑃 𝑋1 ≤ 1 = 𝑃 𝑋1 = 0 + 𝑃(𝑋1 = 1)

= 𝑒−𝜃 + 𝜃𝑒−𝜃 = 𝑒−𝜃(𝜃 + 1)

⟹ 𝛿0 𝑋 is u.e. of 𝑒−𝜃(𝜃 + 1).


(t) = E(𝛿0 𝑋 │𝑇 = 𝑡)

= P(𝑋1 ≤ 1│𝑇 = 𝑡)

=P 𝑋1 ≤ 1 𝑇 = 𝑡

𝑃 𝑇 = 𝑡

=𝑃 𝑋1 = 0 ∪ 𝑋1 = 1

𝑃 𝑇 = 𝑡

=𝑃 𝑋1 = 0, 𝑇 = 𝑡) + 𝑃( 𝑋1 = 1, 𝑇 = 𝑡

𝑃 𝑇 = 𝑡

=𝑃 𝑋1 = 0, 𝑋𝑖

𝑛2 = 𝑡 + 𝑃 𝑋1 = 1, 𝑋𝑖

𝑛2 = 𝑡 − 1

𝑃 𝑇 = 𝑡

=𝑃 𝑋1 = 0 𝑃 𝑋𝑖

𝑛2 = 𝑡 + 𝑃 𝑋1 = 1 𝑃 𝑋𝑖

𝑛2 = 𝑡 − 1

𝑃 𝑇 = 𝑡

=

𝑒−𝜃 𝑒− 𝑛−1 𝜃 𝑛 − 1 𝜃

𝑡

𝑡! + 𝜃𝑒−𝜃 𝑒− 𝑛−1 𝜃 𝑛 − 1 𝜃

𝑡−1

𝑡 − 1 !

𝑒−𝑛𝜃 𝑛𝜃 𝑡

𝑡!

= 𝑛 − 1

𝑛 𝑡

1 +𝑡

𝑛 − 1

(T)= 𝑛−1

𝑛 𝑇 1 +

𝑇

𝑛−1 is u.e. of g(𝜃) based on C.S.S T

⟹ 𝑛−1

𝑛 𝑇 1 +

𝑇

𝑛−1 is UMVUE of g(𝜃) = 𝑒−𝜃(1 + 𝜃).

(2) 𝑋1 , … , 𝑋𝑛 r. s. from B(1, 𝜃), 𝜃∊ (0, 1)= 𝛩

By NFFT T= 𝑋𝑖𝑛𝑖=1 is sufficient

T ∼B n, 𝜃)

Note that p. m. f. of 𝑋𝑖 ∼ 𝐵(1, 𝜃) is

f(x)= 𝜃𝑥(1 − 𝜃)1−𝑥

= 𝜃

1−𝜃

𝑥(1 − 𝜃)

= exp(x log 𝜃

1−𝜃 + log(1 − 𝜃))

With h(x) =1, (𝜃)= log 𝜃

1−𝜃 , T(x)= x & 𝛽(𝜃)= - log (1-𝜃)

This is 1-parameter exponential family

Natural parameter space {: 𝜂∊(0, ∞)} ({𝜂(𝜃) : 𝜃∊ (0, 1)}).

Natural parameter space contains open intervals.

⟹ The 1-parameter expo family is of full rank (complete)

⟹T(X ) = 𝑋𝑖𝑛𝑖=1 is C.S.S.

Alternate proof using convergent power series argument done class.

(a) g(𝜃)= 𝜃

E(T) = n𝜃 ⟹ E 𝑇

𝑛 = 𝜃

𝑇

𝑛u. e. based on CSS T

⟹ 𝑇

𝑛 is UMVUE of 𝜃.

(b) g(𝜃)= 𝜃4

𝛿0 𝑋 = 1, 𝑖𝑓 𝑋1 = 1, 𝑋2 = 1, 𝑋3 = 1, 𝑋4 = 1

0, 𝑜/𝑤

E𝛿0 𝑋 = 𝑃 𝑋1 = 1, 𝑋2 = 1, 𝑋3 = 1, 𝑋4 = 1 = 𝑃 𝑋𝑖 = 1 = 𝜃44𝑖=1

⟹ 𝛿0 𝑋 is u.e. of 𝜃4.

Rao-blackwellization of 𝜃4

(t) = E(𝛿0 𝑋 │𝑇 = 𝑡)

= 𝑋1 = 1, 𝑋2 = 1, 𝑋3 = 1, 𝑋4 = 1│𝑇 = 𝑡

=𝑃 𝑋1 = 1, 𝑋2 = 1, 𝑋3 = 1, 𝑋4 = 1, 𝑋𝑡

𝑛𝑖=5 = 𝑡 − 4

𝑃 𝑇 = 𝑡

=𝑃 𝑋1 = 1 𝑃 𝑋2 = 1 𝑃 𝑋3 = 1 𝑃 𝑋4 = 1 𝑃 𝑋𝑡

𝑛𝑖=5 = 𝑡 − 4

𝑃 𝑇 = 𝑡

=𝜃. 𝜃. 𝜃. 𝜃. 𝑛−4

𝑡−4 𝜃𝑡−4 1 − 𝜃 𝑛−𝑡

𝑛𝑡 𝜃𝑡 1 − 𝜃 𝑛−𝑡

= 𝑛−4

𝑡−4

𝑛𝑡

=𝑡 𝑡 − 1 𝑡 − 2 (𝑡 − 3)

𝑛 𝑛 − 1 𝑛 − 2 (𝑛 − 3)

(T)= 𝑇 𝑇−1 𝑇−2 (𝑇−3)

𝑛 𝑛−1 𝑛−2 (𝑛−3) is u.e. based on C.S.S.

⟹(T) is UMVUE of 𝜃4.

(c) g(𝜃) = 𝜃 1 − 𝜃 2

𝛿0 𝑋 = 1, 𝑖𝑓 𝑋1 = 1, 𝑋2 = 0, 𝑋3 = 0

0, 𝑜/𝑤

E𝛿0 𝑋 = 𝑃 𝑋1 = 1, 𝑋2 = 0, 𝑋3 = 0 = θ 1 − 𝜃 2

⟹ 𝛿0 𝑋 is u.e. of θ 1 − 𝜃 2


(t) = E(𝛿0 𝑋 │𝑇 = 𝑡)

= 𝑃 𝑋1 = 1, 𝑋2 = 0, 𝑋3 = 0 𝑇 = 𝑡

=𝑃 𝑋1 = 1, 𝑋2 = 0, 𝑋3 = 0, 𝑋𝑖

𝑛4 = 𝑡 − 1

𝑃 𝑇 = 𝑡

=𝑃 𝑋1 = 1 𝑃 𝑋2 = 0 𝑃 𝑋3 = 0 𝑃 𝑋𝑖

𝑛4 = 𝑡 − 1

𝑃 𝑇 = 𝑡

=𝜃. 1 − 𝜃 . 1 − 𝜃 𝑛−3

𝑡−1 𝜃𝑡−1 1 − 𝜃 𝑛−3−𝑡+1

𝑛𝑡 𝜃𝑡 1 − 𝜃 𝑛−𝑡

= 𝑛−3

𝑡−1

𝑛𝑡

=

𝑛 − 3 ! 𝑡 − 1 ! 𝑛 − 𝑡 − 2 !

𝑛!𝑡! 𝑛 − 𝑡 !

=𝑡. 𝑛 − 𝑡 . (𝑛 − 𝑡 − 1)

𝑛 𝑛 − 1 (𝑛 − 2)

(T)= 𝑇. 𝑛−𝑇 .(𝑛−𝑇−1)

𝑛 𝑛−1 (𝑛−2) is u.e. based on C.S.S. T

⟹(T) is UMVUE of θ 1 − 𝜃 2.

(3) 𝑋1 , … , 𝑋𝑛 r.s. from f(x)= 𝑒− 𝑥−𝜃 𝑖𝑓 𝑥 > 𝜃0 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒

By NFFT T= 𝑋(1) is sufficient.

𝑓𝑇 𝑡 =𝑛𝑒−𝑛 𝑡−𝜃 𝑖𝑓 𝑡 > 𝜃

0 𝑜/𝑤

Now E g(T)= 0 ∀ 𝜃∊ 𝛩

⟹ ∫ 𝑔(𝑡)∞

𝜃𝑛𝑒−𝑛 𝑡−𝜃 𝑑𝑡 = 0 ∀ 𝜃∊ 𝛩

i.e. ∫ 𝑔(𝑡)∞

𝜃𝑒−𝑛𝑡 𝑑𝑡 = 0 ∀ θ ∊ Θ

Differentiating w.r.t. 𝜃, we get

g(𝜃) 𝑒−𝑛𝜃 = 0 ∀ θ ∊ Θ

⟹ g(𝜃)= 0 ∀ 0<𝜃 <t ; 0< t < ∞

⟹ g(t) = 0 a.e. (𝜃<t <∞)

↔

↓

Range of T=𝑋(1)

⟹T= 𝑋(1) is complete.

g(𝜃)= 𝜃2

E 𝑋 1 =∫ 𝑡∞

𝜃𝑛𝑒−𝑛 𝑡−𝜃 𝑑𝑡

= 𝑛 𝑦 + 𝜃 ∞

0

𝑒−𝑛𝑦 𝑑𝑦 = 𝑛 ⎾2

𝑛2+ 𝜃.

1

𝑛 = 𝜃 +

1

𝑛

E(𝑋 1 −1

𝑛)= 𝜃

Sly E 𝑋 1 2 = 𝑛 ∫ 𝑡2∞

𝜃𝑒−𝑛 𝑡−𝜃 𝑑𝑡

y= t- 𝜃

= n∫ 𝑦 + 𝜃 2∞

0𝑒−𝑛𝑦 𝑑𝑦

= 𝑛 (𝑦2 + 𝜃2 + 2𝜃𝑦∞

0

𝑒−𝑛𝑦 𝑑𝑦

= 𝑛 ⎾3

𝑛3+ 𝜃2

1

𝑛+ 2𝜃

⎾2

𝑛2

=2

𝑛2+ 𝜃2 +

2𝜃

𝑛

⟹ E 𝑋 1 2 =

2

𝑛2 + 𝜃2 +2

𝑛 𝐸 𝑋 1 −

1

𝑛

⟹ 𝐸 𝑋 1 2 −

2

𝑛2 −2

𝑛 𝑋 1 −

1

𝑛 = 𝜃2

𝑖. 𝑒. 𝐸(𝑋 1 2 −

2

𝑛2−

2

𝑛𝑋 1 +

2

𝑛2= 𝜃2

⟹ E 𝑋 1 2 −

2

𝑛𝑋 1 = 𝜃2

𝑋 1 2 −

2

𝑛𝑋 1 is u.e. of 𝜃2 based on C.S.S. 𝑋 1

⟹ 𝑋 1 2 −

2

𝑛𝑋 1 is UMVUE of 𝜃2

(4) 𝑋1 , … , 𝑋𝑛 r.s. from U(0, 𝜃)

T= 𝑋(𝑛) is C.S.S. (proved in class)

g(𝜃) = 𝜃𝑘

p.d.f. of T; 𝑓𝑇 𝑡

𝑛

𝜃𝑛 𝑡𝑛−1, 0 < 𝑡 < 𝜃

0 , 𝑜/𝑤

E 𝑘𝑋 𝑛

=𝑛

𝜃𝑛 ∫ 𝑡𝑘𝜃

0𝑡𝑛−1 𝑑𝑡 =

𝑛

𝑛+𝑘𝜃𝑘 .

⟹𝐸 𝑛+𝑘

𝑛𝑋 𝑛

𝑘 = 𝜃𝑘 .

𝑛+𝑘

𝑛 𝑋 𝑛

𝑘 is u.e. based of C.S.S. 𝑋 𝑛

⟹ 𝑛+𝑘

𝑛 𝑋 𝑛

𝑘 is UMVUE of 𝜃𝑘 .

(5) 𝑋1 , … , 𝑋𝑛 r.s. from G(2, 𝜃); 𝜃> 0

p.d.f.

f(x)= 1

⎾2𝜃2 𝑒−𝑥/𝜃𝑥 2−1 𝑖𝑓 𝑥 > 0

0 𝑜/𝑤

Note that the p.d.f. can be written as

f(x)= x exp (−𝑥

𝜃− 2 𝑙𝑜𝑔𝜃)

with h(x)= x ; 𝜂(𝜃)= −1

𝜃 𝑇 𝑥 = 𝑥 ; 𝛽 𝜃 = 2 𝑙𝑜𝑔𝜃

the above is 1-parameter expo family distn.

Natural parameter space

{:𝜂 < 0} ({(𝜃) : 𝜃∊ (0, ∞)}).

The above contains open intervals

⟹ 1-parameter expo family is full rank (complete)

⟹ T(X ) = 𝑋𝑖𝑛𝑖=1 is C.S.S.

E (T) = E 𝑋𝑖𝑛𝑖=1 = 𝐸 𝑋𝑖

𝑛𝑖=1 = 2𝑛𝜃

𝐸 𝑋𝑖 =1

𝜃2 𝑥2𝑒

−𝑥𝜃

∞

0

𝑑𝑥 =⎾3 𝜃3

𝜃2= 2𝜃

⟹𝐸 𝑇

2𝑛 = 𝜃

𝑇

2𝑛 is u.e. based on C.S.S. T= 𝑋𝑖

𝑛𝑖=1

⟹ 𝑇

2𝑛 is UMVUE of 𝜃.

(6) 𝑋1 , … , 𝑋𝑛 be a r.s. from U(𝜃 – ½ , 𝜃 + ½).

jt. P.d.f f(x )= 1, 𝜃 −

1

2< 𝑥(1) < ⋯ < 𝑥(𝑛) < 𝜃 +

1

2

0 𝑜/𝑤

i.e. f(x )= 1. 𝐼 𝜃−

1

2,𝑥 1

𝐼 𝑥 𝑛 ,𝜃+

1

2

𝐼 𝑎 ,𝑏 = 1, 𝑎 < 𝑏

0, 𝑜/𝑤

By NFFT, T(X )= 𝑋 1 , 𝑋 𝑛 is jointly suff for 𝜃

𝑓𝑋 1 𝑥 = 𝑛 1 − 𝐹𝑋 𝑥

𝑛−1𝑓𝑋 𝑥 ; 𝜃 −

1

2< 𝑥 < 𝜃 +

1

2

= 𝑛 𝜃 − 𝑥 +1

2 𝑛−1

, 𝜃 −1

2< 𝑥 < 𝜃 +

1

20 𝑜/𝑤

𝐸 𝑋 1 = 𝑛 𝑥 𝜃 − 𝑥 +1

2 𝑛−1𝜃+

12

𝜃−12

𝑑𝑥 = 𝜃 +1

2−

𝑛

𝑛 + 1

𝑓𝑋 𝑛 𝑥 = 𝑛 𝐹𝑋 𝑥

𝑛−1𝑓𝑋 𝑥 ; 𝜃 −

1

2< 𝑥 < 𝜃 +

1

2

= 𝑛 𝑥 − 𝜃 +1

2 𝑛−1

, 𝜃 −1

2< 𝑥 < 𝜃 +

1

20 𝑜/𝑤

𝐸 𝑋 𝑛 = 𝑛 𝑥 𝑥 − 𝜃 +1

2 𝑛−1𝜃+

12

𝜃−12

𝑑𝑥 =𝑛

𝑛 + 1−

1

2+ 𝜃

⟹𝐸 𝑋 𝑛 − 𝑋 1 −𝑛−1

𝑛+1 = 0 ∀ 𝜃

⇏ 𝑋 𝑛 − 𝑋 1 =𝑛−1

𝑛+1 a.e.

i.e. with g(T)= 𝑋 𝑛 − 𝑋 1 −𝑛−1

𝑛+1

E g(T )= 0 ∀ 𝜃 ⇏ g t = 0 a.e.

⟹ T = 𝑋 𝑛 − 𝑋 1 is not complete.

(7) T is UMVUE of 𝜃

⟹ E(T)= 𝜃

Suppose 𝑇2 is UMVUE of 𝜃2⟹𝑇2 is u.e. of 𝜃2

⟹𝐸 𝑇2 = 𝜃2

⟹ V(T)= 𝐸 𝑇2 − 𝐸2(𝑇)

i.e. V(T)= 𝜃2 − 𝜃2 = 0

whis is a cantvadiction

⟹𝑇2 cannot be u.e. of 𝜃2

⟹ 𝑇2 cannot be UMVUE of 𝜃2

(8) 𝑋1 , … , 𝑋𝑛 r.s. from N(0, 𝜃)

T= 𝑋𝑖2𝑛

𝑖=1 𝑖𝑠 𝐶. 𝑆. 𝑆.

𝑇

𝜃 ∼ 𝜒𝑛

2; 𝐸 𝑇

𝜃 = 𝑛; 𝑉

𝑇

𝜃 = 2𝑛

𝐸 𝑇 = 𝑛𝜃; 𝑉 𝑇 = 2𝑛𝜃2

𝐸𝑇2 = 𝑉 𝑇 + 𝐸 𝑇 2

= 2𝑛𝜃2 + 𝑛2𝜃2 = 𝑛 𝑛 + 2 𝜃2

⟹𝐸 𝑇2

𝑛 𝑛+2 = 𝜃2

𝑇2

𝑛 𝑛+2 is u.e. based on C.S.S. 𝑋𝑖

2 = 𝑇

⟹ 𝑇2

𝑛 𝑛+2 is UMVUE of 𝜃2.

(9) (a) 𝑋1 , … , 𝑋𝑛 r.s. from N(𝜇, 𝜃); 𝜇 is known

T= (𝑋𝑖 − 𝜇)2𝑛𝑖=1 is CSS (𝜇 is known)

𝐸 𝑇

𝜃 = 𝑛

(𝑋𝑖 − 𝜇)2

𝜃∼ 𝜒𝑛

2

⟹ (𝑋𝑖−𝜇)2

𝑛 is UMVUE for 𝜃 when 𝜇 is known.

(9) (b) 𝑋1 , … , 𝑋𝑛 r.s. from N(𝜇, 𝜃); 𝜇, 𝜃 both known

𝑋𝑖 , 𝑋𝑖2𝑛

𝑖=1 ⇔ 𝑋 , 𝑠2 =1

𝑛−1 (𝑋𝑖 − 𝑋 )2 𝑖𝑠 𝐶. 𝑆. 𝑆. (2-parameter full rank expo family)

𝑛 − 1 𝑠2

𝜃=

(𝑋𝑖 − 𝑋 )2

𝜃 ∼ 𝜒𝑛−1

2

⟹ 𝐸 (𝑋𝑖 − 𝑋 )2

𝑛 − 1 = 𝜃

(𝑋𝑖−𝑋 )2

𝑛−1 is u.e. of 𝜃 based on C.S.S.

⟹ (𝑋𝑖−𝑋 )2

𝑛−1 is UMVUE of 𝜃.

(10) (a) 𝑋1 , … , 𝑋𝑛 r.s. from N(𝜇, 𝜎2) 𝜇 is known

T= (𝑋𝑖 − 𝜇)2 is C.S.S.

g(𝜎)= 𝜎𝑟

Y= 𝑇

𝜎2 ∼ 𝜒𝑛2

𝐸 𝑌𝑟2 =

1

2𝑛2⎾

𝑛2

𝑦𝑟2 𝑒−

𝑦2 𝑦−

𝑛2−1

∞

0

𝑑𝑦

= 1

2𝑛2⎾

𝑛2

𝑒−𝑦2 𝑦−

𝑛+𝑟2

−1∞

0

𝑑𝑦

=2

𝑟2 ⎾

𝑛 + 𝑟2

⎾𝑛2

⟹ 𝐸 𝑌𝑟2 = 𝐸

𝑇𝑟2

𝜎𝑟 =

2𝑟2 ⎾

𝑛 + 𝑟2

⎾𝑛2

⟹ 𝐸 ⎾

𝑛2

2𝑟2 ⎾

𝑛 + 𝑟2

= 𝜎𝑟

⎾𝑛

2

2𝑟2 ⎾

𝑛+𝑟

2

𝑇𝑟

2 is u.e. of 𝜎𝑟 based on C.S.S.

⟹⎾

𝑛

2

2𝑟2 ⎾

𝑛+𝑟

2

(𝑋𝑖 − 𝜇)2 𝑟

2 is UMVUE of 𝜎𝑟 (when 𝜇 is known).

(b) 𝑋1 , … , 𝑋𝑛 r.s. from N(𝜇, 𝜎2) ; 𝜇 & 𝜎 both unknown

𝑋 , (𝑋𝑖 − 𝜇)2 is C.S.S.

= (𝑋 , 𝑆𝑋2)

Y= 𝑆𝑋

2

𝜎2 ∼ 𝜒𝑛−12

𝐸 𝑌𝑟

2 = 𝐸 𝑆𝑋

𝑟

𝜎𝑟 =2

𝑟2 ⎾

𝑛−1+𝑟

2

⎾𝑛−1

2

(following the derivation in part (a))

⟹ 𝐸 ⎾

𝑛−1

2

2𝑟2 ⎾

𝑛−1+𝑟

2

𝑆𝑋𝑟 = 𝜎𝑟

⎾𝑛−1

2

2𝑟2 ⎾

𝑛−1+𝑟

2

𝑆𝑋𝑟 is u.e. of 𝜎𝑟 based on C.S.S.

⟹⎾

𝑛−1

2

2𝑟2 ⎾

𝑛−1+𝑟

2

𝑆𝑋𝑟 is UMVUE of 𝜎𝑟 .

(c)p= P X ≤𝛿) = P 𝑋−𝜇

𝜎≤

𝛿−𝜇

𝜎 = 𝛷

𝛿−𝜇

𝜎

↑

Given value ; To find the UMVUE of 𝛿

p= 𝛷 𝛿−𝜇

𝜎 ⟹

𝛿−𝜇

𝜎= 𝛷−1(𝑝)

i.e. 𝛿= 𝜇 +𝜎 𝛷−1 𝑝 ⎵ = 𝑔(𝜃 )

↓

Known 𝜃 = (𝜇, 𝜎 ’.

Note that E(X̅ )= 𝜇 & E ⎾

𝑛−1

2

212 ⎾

𝑛

2

𝑆𝑋 = 𝜎 ↖(from part (b) with r= 1)

⟹ [X̅ +⎾

𝑛−1

2

212 ⎾

𝑛

2

𝑆𝑋 𝛷−1(𝑝) ] is u.e. of 𝜇 + 𝜎 𝛷−1(𝑝) based on C.S.S.

⟹ X̅ +⎾

𝑛−1

2

212 ⎾

𝑛

2

𝑆𝑋 𝛷−1(𝑝) is UMVUE of 𝛿= 𝜇 + 𝛷−1(𝑝) .

(11) 𝑋1 , … , 𝑋𝑛 r.s. from U(0, 𝜃)

It’s easy to check that

𝐸𝑋𝑖 =𝜃

2 ∀ 𝑖

𝐸𝑋 𝑛 =𝑛

𝑛 + 1𝜃 & 𝐸𝑋 1 =

𝜃

𝑛 + 1

⟹ 𝐸 𝑇1 𝑋 =𝑛 + 1

𝑛 𝐸𝑋 𝑛 = 𝜃

𝐸 𝑇2 𝑋 = 𝜃 & 𝐸 𝑇3 𝑋 =𝜃

𝑛 + 1+

𝑛

𝑛 + 1𝜃 = 𝜃

𝑇1 , 𝑇2 , 𝑇3 are all u.e. of

Among the 3 estimators 𝑇1 is UMVUE and hence is the preferred estimator.

(12) 𝑋1 , … , 𝑋𝑛 r.s. from N(𝜇, 𝜎2)

𝑋 , 𝑆2 =1

𝑛 − 1 (𝑋𝑖 − 𝑋 )2 𝑖𝑠 𝐶. 𝑆. 𝑆

X̅∼ N(𝜇, 𝜎2/𝑛) > indep.

𝑛−1 𝑠2

𝜎2 ∼ 𝜒𝑛−12

𝐸 𝑋 2 = 𝑉 𝑋 + 𝐸 𝑋 2

i.e. 𝐸 𝑋 2 =𝜎2

𝑛+ 𝜇2

𝐸 𝑋 2 =1

𝑛𝐸 𝑆2 + 𝜇2

i.e. 𝐸 𝑋 2 −1

𝑛𝑆2 = 𝜇2

⟹ 𝑋 2 −1

𝑛𝑆2 is u.e. of 𝜇2 based only on the C. S. S.

⟹𝑋 2 −1

𝑛𝑆2 is UMVUE of 𝜇2 .

If g(𝜇, 𝜎) = 𝜇 + 𝜎

𝐸 𝑋 = 𝜇

And E ⎾

𝑛−1

2

2 ⎾𝑛

2

(𝑋𝑖 − 𝑋 )2 1

2 = 𝜎 (from problem # 10 (c)/(b))

⟹ E 𝑋 + ⎾

𝑛−1

2

2 ⎾𝑛

2

(𝑋𝑖 − 𝑋 )2 1

2 = 𝜇 + 𝜎

⟹ 𝑋 + ⎾

𝑛−1

2

2 ⎾𝑛

2

(𝑋𝑖 − 𝑋 )2 1

2 is u.e. of 𝜇 + 𝜎 based on C.S.S.

⟹𝑋 + ⎾

𝑛−1

2

2 ⎾𝑛

2

(𝑋𝑖 − 𝑋 )2 1

2 is UMVUE of 𝜇+ 𝜎.

(13) 𝑋1 , … , 𝑋𝑛 r.s. from

f(x)= 1

𝑏 𝑒

− 𝑥−𝑎

𝑏

, 𝑥 > 𝑎

0 𝑜/𝑤

This is exponential distn with location parameter ‘a’ and scale parameter ‘b’

(a) If b is known then 𝑋(1) is C.S.S., in such a situation

𝐸𝑋(1) = 𝑎 +𝑏

𝑛

i.e. 𝐸 𝑋 1 −𝑏

𝑛 = 𝑎

𝑋 1 −𝑏

𝑛 is u.e. based on C.S.S.

⟹𝑋 1 −𝑏

𝑛 is UMVUE of a, when b is known

(b) If a is known, then 𝑋𝑖 is C.S.S.

𝐸 𝑋𝑖 = 𝐸𝑋𝑖 = 𝑛 𝑎 + 𝑏

⟹ 𝐸 𝑋𝑖

𝑛− 𝑎 = 𝑏

⟹ X̅ - a is u.e. based on C.S.S. and hence is UMVUE of b.

Remark:

[when both a & b are unknown, then

T(X )= (𝑋 1 , 𝑋𝑖) or (𝑋 1 , (𝑋𝑖 − 𝑋 1 )) is C.S.S

We may note that under such a setup

𝑋 1 ∼ 𝐸𝑥𝑝 𝑎,𝑏

𝑛 &

2

𝑏 𝑋𝑖 − 𝑋 1 ∼ 𝜒2 𝑛−1

2

And the 2 are independent.]

i.e. 𝑋 1 & 2

𝑏 𝑋𝑖 − 𝑋 1 are indep.

Problem Set-12

[1] 𝑋1 , … , 𝑋𝑛 be a random sample from N(𝜇, 𝜎2) distribution. Find the Cramer-Rao Lower Bounds

(CRLB) on the variances of unbiased estimators of 𝜇 and𝜎2. Can you find unbiased estimators 𝜇 and

𝜎2 whose variance attains the respective CRLB?

[2] 𝑋1 , … , 𝑋𝑛 is a random sample from Gamma(𝛼, 𝛽)

f x∣𝛼, 𝛽) = 1

⎾𝛼 𝛽𝛼 𝑒−𝑥/𝛽 𝑥𝛼−1 𝑖𝑓 𝑥 > 0


𝛼is assumed to be known. Find the Fisher Information I(𝛽) and CRLB on the variances of unbiased

estimators of 𝛽.

[3] 𝑋1 , … , 𝑋𝑛 be a random sample from P(𝜃), 𝜃∊(0, ∞). Find the CRLB on the variances of unbiased

estimators of the following estimands: (a) g(𝜃)= 𝜃 , (b) g(𝜃)= 𝜃2 and (c) g(𝜃)= 𝑒−𝜃 .

[4] Suppose 𝑋1 , … , 𝑋𝑛 be a random sample from B(1, 𝜃), 𝜃∊ (0, 1). Find the CRLB on the variances of

unbiased estimators of the following estimands: (a) g(𝜃)= 𝜃4 (b) g(𝜃)= 𝜃(1- 𝜃).

[5] 𝑋1 , … , 𝑋𝑛 be a random sample from U(0, 𝜃), 𝜃 > 0. Show that (a) 𝑛

𝑛+1𝑋(𝑛) is a consistent

estimator of 𝜃 and (b) 𝑒𝑋(𝑛 ) is consistent for 𝑒𝜃 , where 𝑋(𝑛) = max 𝑋1 , … , 𝑋𝑛 .

[6] 𝑋1 , … , 𝑋𝑛 be a random sample from U(𝜃- ½ , 𝜃 + ½ ), 𝜃∊ℜ. Show that

𝑋 1 +1

2, 𝑋 𝑛 −

1

2 𝑎𝑛𝑑

𝑋 1 + 𝑋 𝑛

2 are all consistent estimators of 𝜃, 𝑋 𝑛 = max 𝑋1 , … , 𝑋𝑛 and

𝑋 1 = min 𝑋1 , … , 𝑋𝑛 .

[7] 𝑋1 , … , 𝑋𝑛 be a random sample from

f(x)= 1

2 1 + 𝜃𝑥 − 1 < 𝑥, 1


Where, 𝜃∊(-1, 1). Fin a consistent estimator for 𝜃.

[8] 𝑋1 , … , 𝑋𝑛 be a random sample from P(𝜃). Find a consistent estimator of 𝜃3 3 𝜃 + 𝜃 + 12 .

[9] Let 𝑋1 , … , 𝑋𝑛 be a random sample from Gamma (𝛼, 𝛽) with density

f(x)= 1

⎾𝛼𝛽𝛼 𝑒−𝑥/𝛽𝑥𝛼−1 𝑥 > 0


Where, 𝛼 is a known constant and 𝛽 is an unknown parameter, show that 𝑋𝑖

𝑛𝑖=1

𝑛𝛼 is a consistent

estimator of 𝛽.

[10] Let 𝑋1 , … , 𝑋𝑛 be a random sample from each of the following distributions having the following

density or mass functions. Find the maximum likelihood estimator (MLE) of 𝜃 in each case.

(a) f(x; 𝜃)= 𝑒−𝜃𝜃𝑥

𝑥 ! 𝑥 = 0, 1, 2, …


(b) f(x; 𝜃)= 𝜃 𝑥𝜃−1 0 < 𝑥 < 10 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒.

(c) f(x; 𝜃)= 1

𝜃 𝑒−𝑥/𝜃 𝑥 > 0


(d) f(x; 𝜃)= 1

2 𝑒−∣𝑥−𝜃∣ − ∞ < 𝑥, ∞


e X∼ U −𝜃

2,𝜃

2 .

[11] Let 𝑋1 , … , 𝑋𝑛 be a random sample from the distribution having p.d.f.

f(x ; 𝜃1, 𝜃2)= 1

𝜃2𝑒−(𝑥−𝜃1)/𝜃2 𝑥 ≥ 𝜃1


Find the MLEs of 𝜃1 𝑎𝑛𝑑 𝜃2.

[12] Let 𝑋1 , … , 𝑋𝑛 be a random sample from the distribution having p.d.f.

f(x ; 𝛼, 𝜆)= 𝜆𝛼

⎾𝛼𝑒−𝜆𝑥𝑥𝛼−1 𝑥 ≥ 0


Find the MLEs of 𝛼 and 𝜆.

[13] Let 𝑋1 , … , 𝑋𝑛 be a random sample from the function having p.d.f.

f(x ; 𝜇, 𝜎)= 1

2 3𝜎 𝜇 − 3 𝜎 , 𝑥 < 𝜇 + 3 𝜎


Find the MLEs of 𝜇 and 𝜎.

[14] Let 𝑋1 , … , 𝑋𝑛 be a random sample from U (𝜃- ½ , 𝜃 + ½ ), 𝜃∊ℜ. Show that any statistic

u(𝑋1 , … , 𝑋𝑛) such that it satisfies

𝑋(𝑛) −1

2≤ 𝑢 𝑋1 , … , 𝑋𝑛 ≤ 𝑋(1) +

1

2

Is a maximum likelihood estimator of 𝜃. in particular 𝑋 1 + 𝑋 𝑛

2 𝑎𝑛𝑑

3

4 𝑋 1 +

1

2 +

1

4 𝑋 𝑛 −

1

2 are

MLEs of 𝜃.

[15] The lifetimes of a component are assumed to be exponential with parameter 𝜆. Ten of these

components were placed on a test independently. The only data recorded were the number of

components that had failed (out of 10 put to test) in less than 100 hours, which was recorded to be

3. Find the maximum likelihood estimate of 𝜆.

[16] A salesman of used cars is willing to assume that the number of sales he makes per day is a

Poisson random variable with parameter 𝜇. Over the past 30 days he made no sales on 20 days and

one or mare sales on each of the remaining 10 days. Find the maximum likelihood estimate of 𝜇.

[17] Let 𝑋1 , … , 𝑋𝑛 be a random sample from each of the following distributions. Find the method of

moments estimator (MOME) of the corresponding unknown parameters in each of the situations.

a X∼P 𝜃 ; b X∼ -𝜃/2 , 𝜃/2);

c X∼ Exp 0, 𝜃 ; d X∼ Exp 𝛼, 𝛽);

e X∼ G 𝛼, 𝛽) with p.d.f. f(x ; 𝛼, 𝛽)= 1

𝛽𝛼⎾𝛼 𝑒−𝑥/𝛽 𝑥𝛼−1 𝑖𝑓 𝑥 ≥ 0


Solution Key

(1) 𝑋1 , … , 𝑋𝑛 i.i.d. N(𝜇, 𝜎2)

f x∣ 𝜇, 𝜎2)= 1

2𝜋 𝜎2 𝑒

−1

2 𝜎2 𝑥−𝜇 2

log 𝑓 = 𝑘 −1

2log 𝜎2 −

1

2 𝜎2 𝑥 − 𝜇 2

𝜕 log 𝑓

𝜕𝜇=

𝑥 − 𝜇

𝜎2;𝜕2 log 𝑓

𝜕𝜇2= −

1

𝜎2.

−𝐸 𝜕2 log 𝑓

𝜕𝜇2 =1

𝜎2= 𝐼(𝜇)

CRLB for an u.e. for 𝜇= 𝜎2

𝑛.

Since V(X̅)= 𝜎2

𝑛; X̅ attains CRLB.

𝜕 log 𝑓

𝜕𝜎2 = − 1

2 𝜎2 + 1

2 𝜎4 𝑥 − 𝜇 2

𝜕2 log 𝑓

𝜕 𝜎2 2=

1

2 𝜎4−

𝑥 − 𝜇 2

𝜎6.

𝐼 𝜎2 = −𝐸 𝜕2 log 𝑓

𝜕 𝜎2 2 = −1

2 𝜎4+

1

𝜎4=

1

2 𝜎4

CRLB for an u.e. for𝜎2 =2𝜎4

𝑛.

Now 𝑆2 =1

𝑛−1 (𝑋𝑖 − 𝑋 )2𝑛

1 is UMVUE for 𝜎2 with

V(𝑆2) =2𝜎4

𝑛−1 > CRLB

Since UMVUE is the unbiased with lowest variance in the class of all unbiased estimators, CRLB

can’t be attained by any unbiased estimator of 𝜎2.

(2) f x∣𝛼, 𝛽) = 1

⎾𝛼 𝛽𝛼 𝑒−

𝑥

𝛽 𝑥𝛼−1 𝑖𝑓 𝑥 > 0

log 𝑓 = −𝑙𝑜𝑔⎾𝛼 − 𝛼 log 𝛽 −𝑥

𝛽+ 𝛼 − 1 log 𝑥

𝜕 log 𝑓

𝜕𝛽= −

𝛼

𝛽+

𝑥

𝛽2

𝜕2 log 𝑓

𝜕𝛽2=

𝛼

𝛽2− 2

𝑥

𝛽3.

𝐼 𝛽 = −𝐸 𝜕2 log 𝑓

𝜕𝛽2 = −𝛼

𝛽2+ 2

𝛼𝛽

𝛽3=

𝛼

𝛽2.

⟹ CRLB for u.e. of 𝛽 : 1

𝑛 .𝛼

𝛽2

=𝛽2

𝑛𝛼.

(3) 𝑋1 , … , 𝑋𝑛 i.i.d. P (𝜃)

f x∣𝜃)=𝑒−𝜃𝜃𝑥

𝑥!

log f x ∣ θ = −𝜃 + 𝑥𝑙𝑜𝑔𝜃 − log 𝑥!

𝜕 log 𝑓

𝜕𝜃= −1 +

𝑥

𝜃;𝜕2 log 𝑓

𝜕𝜃2= −

𝑥

𝜃2 .

𝐼 𝜃 = −𝐸 𝜕2 log 𝑓

𝜕𝜃2 =1

𝜃

CRLB for any u.e. of 𝜃= 1

𝑛 .1

𝜃

=𝜃

𝑛

CRLB for any u.e. of g (𝜃) = 𝜃2:(2𝜃)2

𝑛

𝜃

=4𝜃3

𝑛

CRLB for any u.e. of g (𝜃) = 𝑒−𝜃 ∶ −𝑒−𝜃

2

𝑛

𝜃

=𝜃 𝑒−2𝜃

𝑛.

(4) 𝑋1 , … , 𝑋𝑛 i.i.d. B (1, 𝜃)

f x∣𝜃)= 𝜃𝑥 1 − 𝜃 1−𝑥

log f x ∣ θ = 𝑥𝑙𝑜𝑔 𝜃 + 1 − 𝑥 log 1 − 𝜃

𝜕 log 𝑓

𝜕𝜃=

𝑥

𝜃+

1 − 𝑥

1 − 𝜃 −1 =

𝑥

𝜃 1 − 𝜃 −

1

1 − 𝜃.

𝐸 𝜃 = 𝐸 𝜕 log 𝑓

𝜕𝜃 2 = 𝑉

𝜕 log 𝑓

𝜕𝜃 =

𝜃(1 − 𝜃)

𝜃 1 − 𝜃 2 =

1

𝜃(1 − 𝜃).

CRLB for u.e. of 𝜃4: 4𝜃3

2

𝑛 .1

𝜃(1−𝜃 )

=16𝜃7(1−𝜃)

𝑛

CRLB for u.e. of 𝜃(1-𝜃): (1−2𝜃)2

𝑛 .1

𝜃 (1−𝜃 )

=(1−2𝜃)2𝜃(1−𝜃)

𝑛.

(5) 𝑋1 , … , 𝑋𝑛 i.i.d. U(0, 𝜃)

P[∣𝑋 𝑛 − 𝜃∣ ≥𝜖] ≤ 𝐸 𝑋 𝑛 −𝜃

2

𝜖2 =𝐸𝑋 𝑛

2+𝜃2−2𝜃 𝐸𝑋 𝑛

𝜖2

𝑓𝑋 𝑛 𝑥 =

𝑛

𝜃𝑛𝑥𝑛−1 , 0 < 𝑥 < 𝜃

0, 𝑜/𝑤

𝐸𝑋 𝑛 =𝑛

𝜃𝑛 𝑥𝑛

𝜃

0

𝑑𝑥 =𝑛

𝑛 + 1. 𝜃

𝐸𝑋 𝑛 2 =

𝑛

𝜃𝑛 𝑥𝑛+1

𝜃

0

𝑑𝑥 =𝑛

𝑛 + 2 𝜃2

⟹ P[∣ 𝑋 𝑛 − 𝜃 ∣ ≥ ϵ] ≤1

𝜖2 𝑛

𝑛 + 2𝜃2 + 𝜃2 − 2𝜃

𝑛

𝑛 + 1. 𝜃

→ 0 as n→∞

⟹ 𝑋 𝑛 → 𝜃.

⟹ 𝑛

𝑛+1𝑋 𝑛 → 𝜃

⟹ 𝑛

𝑛+1𝑋 𝑛 is a consistent estimator for 𝜃

Further since 𝑋 𝑛 → 𝜃

𝑒𝑋 𝑛 = 𝑔 𝑋 𝑛 → 𝑔 𝜃 = 𝑒𝜃 .

⟹ 𝑒𝑋 𝑛 is a consistent estimator for 𝑒𝜃 .

(6) 𝑋1 , … , 𝑋𝑛 i.i.d. U(𝜃- ½ , 𝜃 + ½ )

𝐹𝑋 𝑥 = 𝑑𝑥𝑥

𝜃−12

= 𝑥 − 𝜃 +1

2

𝑓𝑋 1 𝑥 = 𝑛 1 − 𝐹𝑋 𝑥

𝑛−1𝑓 𝑥

𝑖. 𝑒. 𝑓𝑋 1 𝑥 = 𝑛 𝜃 − 𝑥 +

1

2 𝑛−1

, 𝜃 −1

2≤ 𝑥 ≤ 𝜃 +

1

20 , 𝑜/𝑤

𝐸𝑋 1 = 𝑛 𝑥 𝜃 − 𝑥 +1

2 𝑛−1𝜃+

12

𝜃−12

𝑑𝑥 = 𝜃 +1

2−

𝑛

𝑛 + 1.

𝐸𝑋 1 2 = 𝑥2 𝜃 − 𝑥 +

1

2 𝑛−1𝜃+

12

𝜃−12

𝑑𝑥

= 𝜃 +1

2

2

+𝑛

𝑛 + 2−

𝑛

𝑛 + 1 2𝜃 + 1

𝑃 𝑋 𝑛 − 𝜃 −1

2 ≥ ϵ ≤

𝐸 𝑋 1 − 𝜃 −12

2

𝜖2

r.h.s. = 1

𝜖2 𝐸 𝑋 1 2 + 𝜃 −

1

2

2− 2 𝜃 −

1

2 𝐸 𝑋 1

= 1

𝜖2 𝜃 +

1

2

2

+𝑛

𝑛 + 2−

𝑛

𝑛 + 1 2𝜃 + 1 + 𝜃 −

1

2

2

− 2 𝜃 −1

2 𝜃 +

1

2−

𝑛

𝑛 + 1

⟶1

𝜖2 𝜃 +

1

2

2

+ 1 − 2𝜃 + 1 + 𝜃 −1

2

2

− 2 𝜃 −1

2 𝜃 −

1

2 𝑎𝑠 𝑛 → ∞

= 0

⟹ 𝑃 𝑋 𝑛 − 𝜃 −1

2 ≥ ϵ ⟶ 0 𝑎𝑠 𝑛 → ∞

⟹𝑋 1 ⟶ 𝜃 −1

2. ___________(1)

We can similarly prove that

𝑋 𝑛 ⟶ 𝜃 +1

2. _________________(2)

Combining (1) & (2), we get.

𝑋 1 + 𝑋 𝑛

2⟶ 𝜃

⟹𝑋 1 +𝑋 𝑛

2 is a consistent estimator for 𝜃

So,

𝑋 1 +1

2 is a consistent estimator for 𝜃 (from (1))

& 𝑋 𝑛 −1

2 is a consistent estimator for 𝜃 (from (2)).

(7) 𝑋1 , … , 𝑋𝑛 i. i. d. 𝑓𝑋 𝑥 = 1

2 1 + 𝜃𝑥 − 1 < 𝑥, 1


𝐸 𝑋 =1

2 (1 + 𝜃𝑥)

1

−1

𝑑𝑥 =𝜃

3

⟹ 𝑋1 , … , 𝑋𝑛are i. i. d. with 𝐸 𝑋1 =𝜃

3

By khintchime’s WLLN

1

𝑛 𝑋𝑖

𝑛

1

⟶ 𝐸 𝑋1

𝑖. 𝑒. 𝑋 ⟶ 𝜃

3⟹ 3𝑋 ⟶ 𝜃

⟹3𝑋 ⟶ 𝜃 is a consistent estimator for 𝜃.

(8) 𝑋1 , … , 𝑋𝑛 i. i. d. P (𝜃)

𝐸 𝑋𝑖 = 𝜃 ∀ 𝑖 = 1(1)𝑛

By WLLN 𝑋𝑛 ⟶ 𝜃

⟹ g(𝑋𝑛 )⟶g(𝜃)

⟹𝑋𝑛 3

3 𝑋𝑛 + 𝑋𝑛

+ 12 ⟶ 𝜃3 3 𝜃 + 𝜃 + 12

⟹ 𝑋𝑛 3

3 𝑋𝑛 + 𝑋𝑛

+ 12 is a consistent estimator for 𝜃3 3 𝜃 + 𝜃 + 12 .

(9) 𝑋1 , … , 𝑋𝑛 i. i. d. Gamma (𝛼, 𝛽)

𝛼 is known

𝐸 𝑋 = 𝛼𝛽 𝑓𝑜𝑟 𝑋 ∼ 𝐺𝑎𝑚𝑚𝑎 𝛼, 𝛽

By WLLN

1

𝑛 𝑋𝑖

𝑛

1

⟶ 𝐸 𝑋1

𝑖. 𝑒.1

𝑛 𝑋𝑖

𝑛

1

⟶ 𝛼𝛽

⟹1

𝑛𝛼 𝑋𝑖

𝑛1 ⟶ 𝛽.

⟹ 1

𝑛𝛼 𝑋𝑖

𝑛1 is a consistent estimator for 𝛽.

[Note: T= 𝑋𝑖𝑛1 ∼ 𝐺𝑎𝑚𝑚𝑎 𝑛𝛼, 𝛽 can be proved using m. g. f. approach]

(10) (a) 𝑋1 , … , 𝑋𝑛 i.i.d. P (𝜃)

Likelihood function L(𝜃|x )= 𝜃 𝑥𝑖𝑒−𝑛𝜃 𝑥𝑖! −1

𝑙 θ x = log L θ x = 𝑥𝑖 𝑙𝑜𝑔𝜃 − 𝑛𝜃 − log 𝑥𝑖 !

𝜕𝑙

𝜕𝜃=

𝑥𝑖

𝜃− 𝑛;

𝜕𝑙

𝜕𝜃= 0 ⟹ 𝜃ˆ = �̅�

𝜕2𝑙

𝜕𝜃2 𝜃 = 𝜃ˆ = −

𝑥𝑖

𝜃ˆ2< 0

⟹ 𝜃ˆ𝑀𝐿𝐸 = 𝑋

(10) (b) 𝑋1 , … , 𝑋𝑛 i.i.d. with p. d. f.

𝑓𝑋 𝑥 = 𝜃 𝑥𝜃−1 0 < 𝑥 < 10 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒.

L θ x = 𝜃𝑛 𝑥𝑖

𝑛

1

𝜃−1

𝑙 θ x = log L θ x = n log𝜃 + 𝜃 − 1 log 𝑥𝑖

𝑛

1

𝜕𝑙

𝜕𝜃=

𝑛

𝜃+ 𝑙𝑜𝑔𝑥𝑖

𝜕𝑙

𝜕𝜃= 0 ⟹ 𝜃ˆ = −

𝑛

log 𝑥𝑖𝑛1

𝜕2𝑙

𝜕𝜃2 𝜃 = 𝜃ˆ = −

𝑛

𝜃ˆ2< 0 ⟹ 𝜃ˆ𝑀𝐿𝐸 = −

𝑛

log 𝑥𝑖𝑛1

.

(c) 𝑋1 , … , 𝑋𝑛 i.i.d. with p. d. f.

𝑓𝑋 𝑥 = 1

𝜃 𝑒−𝑥/𝜃 𝑥 > 0


L θ x =1

𝜃𝑛𝑒− 𝑥𝑖/𝜃

𝑙 θ x = log L θ x = −n log𝜃 +1

𝜃 𝑥𝑖

𝜕𝑙

𝜕𝜃= −

𝑛

𝜃+

1

𝜃2 𝑥𝑖

𝜕𝑙

𝜕𝜃= 0 ⟹ 𝜃ˆ = �̅�

𝜕2𝑙

𝜕𝜃2 𝜃 = 𝜃ˆ =

𝑛

𝜃2−

2𝑛�̅�

𝜃3 𝜃 = 𝜃ˆ =

𝑛

�̅�2−

2𝑛�̅�

�̅�3= −

𝑛

�̅�2< 0

⟹ 𝜃ˆ𝑀𝐿𝐸 = 𝑋 .

(a) (10) (d) 𝑋1 , … , 𝑋𝑛 i.i.d. with p. d. f.

𝑓𝑋 𝑥 = 1

2 𝑒−∣𝑥−𝜃∣ − ∞ < 𝑥, ∞


L θ x is maximized if ∣ 𝑥𝑖 − 𝜃 ∣ is minimized

Realize that ∣ 𝑥𝑖 − 𝜃 ∣ is minimized w.r.t. 𝜃 at

𝜃ˆ = 𝑚𝑒𝑑𝑖𝑎𝑛(𝑥1 , … , 𝑥𝑛)

⟹ 𝜃ˆ𝑀𝐿𝐸 = 𝑚𝑒𝑑𝑖𝑎𝑛 𝑋1 , … , 𝑋𝑛

(10) (e) 𝑋1 , … , 𝑋𝑛 i.i.d. with U −𝜃

2,𝜃

2

Likelihood function L θ x = 1

θn , −θ

2≤ 𝑥1 , … , 𝑥𝑛 ≤

𝜃

2

0, otherwise

L θ x =

1

θn, if 𝑥𝑖 ≤

θ

2, i = 1(1)n

0, otherwise

i.e.

L θ x = 1

θn, if max

i 𝑥𝑖 ≤

θ

20, otherwise

L θ x is maximized at minimum value of 𝜃 given x

⟹𝜃ˆ𝑀𝐿𝐸 = 2 maxi 𝑥𝑖

(11) 𝑋1 , … , 𝑋𝑛 i.i.d. Exp(𝜃1 , 𝜃2)

𝜃1ˆ𝑀𝐿𝐸 = 𝑋(1)

𝜃2ˆ𝑀𝐿𝐸 =1

𝑛 𝑋𝑖 − 𝑋 1

𝑛

1

Done in class.

(12) 𝑋1 , … , 𝑋𝑛 i.i.d. 𝑓𝑋 𝑥 = 𝜆𝛼

⎾𝛼𝑒−𝜆𝑥𝑥𝛼−1 𝑥 ≥ 0


L θ x =𝜆𝑛𝛼

⎾𝛼 𝑛𝑒−𝜆 𝑥𝑖

𝑛1 𝑥𝑖

𝑛

1

𝛼−1

𝑙 θ x = log L θ x = n𝛼 log 𝜆 − 𝑛 log ⎾𝛼 + 𝛼 − 1 𝑙𝑜𝑔𝑥𝑖 − 𝜆 𝑥𝑖

𝑛

1

Likelihood equations

𝜕 𝑙𝑜𝑔𝑙

𝜕𝜆=

𝑛𝛼

𝜆− 𝑥𝑖 = 0__________(1)

𝜕 𝑙𝑜𝑔𝑙

𝜕𝛼= 𝑛𝑙𝑜𝑔𝜆 − 𝑛

⎾𝛼′

⎾𝛼+ 𝑙𝑜𝑔𝑥𝑖 = 0 ________(2)

(1) ⟹ 𝜆= 𝛼

𝑥̅

From (2), we get

N log 𝛼

𝑥̅ − 𝑛

⎾𝛼′

⎾𝛼+ 𝑙𝑜𝑔𝑥𝑖= 0 (*)

Solving (*) by numerical method gives 𝛼ˆ𝑀𝐿𝐸

& 𝜆ˆ𝑀𝐿𝐸 =𝛼ˆ𝑀𝐿𝐸

𝑋 .

(13) 𝑋1 , … , 𝑋𝑛 i.i.d. with p.d.f.

f(x ; 𝜇, 𝜎)= 1

2 3𝜎, 𝜇 − 3 𝜎 , 𝑥 < 𝜇 + 3 𝜎


Likelihood function

L (μ, θ) x =

1

2 3𝜎 𝑛

, 𝜇 − 3 𝜎 ≤ 𝑥 1 ≤ ⋯ ≤ 𝑥 𝑛 ≤ 𝜇 + 3 𝜎

∗

0 , 𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒.

Using condition (*),

𝜇 − 3 𝜎 ≤ 𝑥 1 & 𝑥 𝑛 ≤ 𝜇 + 3 𝜎

⟹𝜇≤ 𝑥 1 + 3 𝜎 &𝑥 𝑛 − 3 𝜎 ≤ 𝜇

⟹ 𝑥 𝑛 − 3 𝜎 ≤ 𝜇 ≤ 𝑥 1 + 3 𝜎

For a given 𝜎, L (μ, θ) x is maximized w.r.t. 𝜇 if

𝜇∊ 𝑥 𝑛 − 3 𝜎, 𝑥 1 + 3 𝜎 (o/w L (μ, θ) x = 0)

⟹ Any value of 𝜇 in the above internal is an MLE of 𝜇

In particular

𝑋 𝑛 − 3 𝜎 + 𝑋 1 + 3 𝜎

2=

𝑋 𝑛 + 𝑋 1

2= 𝜇ˆ(𝜎)𝑀𝐿𝐸

Since the above MLE is indep of 𝜎, it is MLE of 𝜇 ∀ 𝜎

⟹ 𝜇ˆ𝑀𝐿𝐸 =𝑋 𝑛 +𝑋 1

2

Further, L(𝜇ˆ, 𝜎) is maximized w.r.t. 𝜎 if 𝜎 is minimum.

Observe that

3 𝜎 ≥ μ − 𝑥 1 & 3 𝜎 ≥ 𝑥 𝑛 − 𝜇

At the MLE of 𝜇;

3 𝜎 ≥𝑥 𝑛 − 𝑥 1

2

⟹𝜎ˆ𝑀𝐿𝐸 =𝑋 𝑛 −𝑋 1

2 3

(14) 𝑋1 , … , 𝑋𝑛 i.i.d. U(𝜃- ½ , 𝜃 + ½ ), 𝜃∊ℜ

L θ x = 1, θ −

1

2≤ 𝑥 1 ≤ ⋯ ≤ 𝑥 𝑛 ≤ 𝜃 +

1

2

0, o/w

L is maximized w.r.t. 𝜃 of

θ −1

2≤ 𝑥 1 &𝑥 𝑛 ≤ 𝜃 +

1

2 Max𝜃 𝐿 = 1

𝑖. 𝑒. 𝑥 𝑛 −1

2≤ 𝜃 ≤ 𝑥 1 +

1

2.

⟹Any statistic U(X̠ )∋

𝑥 𝑛 −1

2≤ 𝑢(𝑥1 , … , 𝑥𝑛) ≤ 𝑥 1 +

1

2 is an MLE of 𝜃

In particular, 𝑋 1 +𝑋 𝑛

2 is an MLE of 𝜃

In general, 𝛼(𝑋 1 +1

2) +(1- 𝛼)( 𝑋 𝑛 −

1

2); ∀ 0 < 𝛼 < 1 is an MLE of 𝜃

With 𝛼= 3

4, we have the above estimator as

3

4 𝑋 1 +

1

2 +

1

4 𝑋 𝑛 −

1

2 is an MLE of 𝜃.

(15)X : r.v. denoting lifetime of the component

X∼ Exp distn with mean 𝜆

𝑓𝑋 𝑥 = 1

𝜆 𝑒−𝑥/𝜆 𝑥 > 0


𝐷𝑒𝑓𝑖𝑛𝑒 𝑌𝑖 = 1, 𝑖𝑓 𝑖𝑡𝑕 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑕𝑎𝑠 𝑙𝑖𝑓𝑒 < 100𝑕𝑟𝑠

0, 𝑜/𝑤

𝑃 𝑌𝑖 = 1 = 𝑃 𝑋 < 100 =1

𝜆 𝑒−𝑥/𝜆

100

0

𝑑𝑥 = 1 − 𝑒−

100𝜆

𝑌1 … 𝑌𝑛 i.i.d. 𝐵 1, 1 − 𝑒−100

𝜆 ≡ 𝐵(1, 𝜃)

(n= 10) with 𝜃= 1 − 𝑒−100

𝜆 .

⟹𝜃ˆ𝑀𝐿𝐸 = 𝑌 (done in class)

Further, 𝜆= −100

log (1−𝜃)= 𝑔(𝜃)

⟹ MLE of 𝑔(𝜃) is g 𝜃ˆ𝑀𝐿𝐸 .

⟹ 𝜆ˆ𝑀𝐿𝐸 = −100

log 1−𝜃ˆ𝑀𝐿𝐸 = −

100

log (1−𝑋 )

From the given data X̅ = 3

10

⟹ The maximum likelihood estimate of 𝜆 computed

From the given data is −100

log 7

10

(16) X: r.v. denoting the no. of sales per day

X∼ P(𝜇) (from the assumptions)

Define 𝑌𝑖= 1, 𝑖𝑓 0 𝑠𝑎𝑙𝑒𝑠 𝑜𝑛 𝑑𝑎𝑦 𝑖

0, 𝑜/𝑤

P(𝑌𝑖 = 1) = 𝑃 𝑋 = 0 = 𝑒−𝜇 ;

𝑌1 … 𝑌30 i.i.d. B(1, 𝑒−𝜇 )≡B(1, 𝜃)(𝜃= 𝑒−𝜇 )

𝜃ˆ𝑀𝐿𝐸 = 𝑌

Further, 𝜇= -log 𝜃

⟹ 𝜇ˆ𝑀𝐿𝐸 = − log 𝜃ˆ𝑀𝐿𝐸

⟹ MLE estimate of 𝜇 from the given data is

Given by -log 20

30 .

(17) (a)

𝑋1 , … , 𝑋𝑛 i.i.d. P(𝜃)

𝜇11 = 𝐸 𝑋 = 𝜃

MOME of 𝜃, 𝜃ˆ𝑀𝑂𝑀𝐸 = 𝑚𝑖 = 𝑋

(b) 𝑋1 , … , 𝑋𝑛 i.i.d. 𝑈 −𝜃

2,𝜃

2

𝑑𝑜𝑛𝑒 in class.

(c) 𝑋1 , … , 𝑋𝑛 i.i.d. Exp (0, 𝜃)

𝜇𝑖 = 𝐸 𝑋 =1

𝜃∫ 𝑥 𝑒−

𝑥

𝜃∞

0 𝑑𝑥 = 𝜃

⟹𝜃ˆ𝑀𝑂𝑀𝐸 = 𝑚𝑖 = 𝑋

(d) 𝑋1 , … , 𝑋𝑛 i.i.d. Exp(𝛼, 𝛽)

Done in class.

(e) 𝑋1 , … , 𝑋𝑛 i.i.d. G(𝛼, 𝛽) with p.d.f.

𝑓 𝑥; 𝛼, 𝛽 =

1

𝛽𝛼⎾𝛼 𝑒−𝑥/𝛽 𝑥𝛼−1 𝑖𝑓 𝑥 ≥ 0


𝜇11 = 𝐸 𝑋 =

1

𝛽𝛼⎾𝛼 𝑥𝛼+1−1 𝑒−𝑥/𝛽

∞

0

𝑑𝑥 =⎾𝛼 + 1 𝛽𝛼+1

⎾𝛼𝛽𝛼= 𝛼𝛽

𝜇21 = 𝐸 𝑋2 =

1

𝛽𝛼⎾𝛼 𝑥𝛼+2−1 𝑒

−𝑥𝛽

∞

0

𝑑𝑥

= ⎾𝛼 + 2 𝛽𝛼+2

⎾𝛼𝛽𝛼= (𝛼 + 1)𝛼𝛽2

𝐸𝑞𝑢𝑎𝑡𝑒 𝑋 =𝑚 𝑖= 𝛼𝛽

1

𝑛 𝑋𝑖

2= 𝑚21= 𝛼(𝛼+1)𝛽2

}

⟹ 𝑚2

1

𝑚11 = 𝛼𝛽 + 𝛽 = 𝑚𝑖 + 𝛽

⟹ 𝛽ˆ𝑀𝑂𝑀𝐸 =𝑚2

1 − (𝑚11)2

𝑚𝑖=

1𝑛

𝑋𝑖2 − 𝑋 2

𝑋

𝑖. 𝑒. 𝛽ˆ𝑀𝑂𝑀𝐸 =

1𝑛

(𝑋𝑖 − 𝑋 )2

𝑋

& 𝛼ˆ𝑀𝑂𝑀𝐸 =𝑋

1𝑛

𝑋𝑖 − 𝑋 2

𝑋

=𝑋 2

1𝑛

(𝑋𝑖 − 𝑋 )2 .

Assignment-13

[1] The observed value of the mean of a random sample of size 20 from N (𝜇, 80) be 81.2. Find the equal

tail 95% and the equal tail 99% confidence interval of 𝜇. Which one is shorter?

[2] Let X̅ be the mean of a random sample of size n from N (𝜇, 9). Find n such that, approximately, P (X̅ -1

<𝜇 < X̅ +1) = 0.90

[3] Let a random sample of size 25 from a normal distribution N (𝜇, 𝜎2) yield x ̅= 4.7 and 𝑠2 =1

𝑛−1 (𝑥𝑖 − �̅�)2𝑛

𝑖=1 = 5.76. Determine a 90% confidence interval for 𝜇.

[4] Let 𝑋1 , … , 𝑋𝑛 be random sample of size be random sample of size 9 from N (𝜇, 𝜎2). Find the

expected length of 95% confidence interval for 𝜇 when (a) 𝜎 is known and (b) 𝜎 is unknown.

[5] Let 𝑋1 , … , 𝑋𝑛 be random sample of size n from a distribution with p.d.f.

f(x|𝜃)= 3𝑥2

𝜃3 0 < 𝑥 < 𝜃


(a) Find the distribution of 𝑋 𝑛

𝜃, where 𝑋(𝑛) = max 𝑋1 , … , 𝑋𝑛 .

(b) Show that 𝑋 𝑛 , 𝛼−1

3𝑛𝑋 𝑛 gives a 100(1- 𝛼)% confidence interval for 𝜃.

[6] Let 𝑋1 , … , 𝑋𝑛 be random sample of size n from U(0, 𝜃), 𝜃∊ℜ. Show that 𝑋 𝑛 , 𝛼−

1

𝑛𝑋 𝑛 𝑎𝑛𝑑 ( 1 −

𝛼)−1

𝑛𝑋 𝑛 , ∞ are both 100(1- 𝛼)% confidence intervals for 𝜃.

[7] Let two independent random samples, each of size 5, from two normal distributions N

𝜇1 , 𝜎12 𝑎𝑛𝑑 (𝜇2 , 𝜎2

2) are; 1.5, 2.8 , 3.3 , 3.9 , 7.2 and 2.8 , 1.8 , 3.1 , 6.5 , 6.9 respectively.

(a) If it is known𝜎12 = 𝜎2

2 = 3.5, find a 95% confidence interval for 𝜇1 − 𝜇2.

(b) If it is known that 𝜇1 = 𝜇2 = 4, find a 95% confidence interval for 𝜎12/ 𝜎2

2 .

Solution Key

(1) 𝑋1 , … , 𝑋20 r.s. from N (𝜇, 80)

Confidence int for 𝜇

X∼ N 𝜇,𝜎2

𝑛 ), i.e. N (𝜇, 4)

⟹ Y= 𝑋 −𝜇

2 ∼ 𝑁 0, 1

⟹ 1 − 𝛼 = 𝑃 −𝑧𝛼2

≤𝑋 − 𝜇

2≤ 𝑧𝛼

2 𝑧𝛼

2 𝑖𝑠 ∋ 𝑓𝑜𝑟 𝑧 ∼ 𝑁 0, 1 𝑃 𝑧 > 𝑧𝛼

2 =

𝛼

2

⟹ 1 − 𝛼 = 𝑃 𝑋 − 2𝑧𝛼2

≤ 𝜇 ≤ 𝑋 + 2𝑧𝛼2

For 100(1- 𝛼)Y= 95%, 𝛼= 0.05; 𝛼

2= 0.025

C I ⟶ 𝑋 − 2𝑧0.025 , 𝑋 + 2𝑧0.025

Observed value of X̅ is 81.2 & 𝑧0.025 = 1.96

81.2 − 2 × 1.96, 81.2 + 2 × 1.96 _________(1)

For 100(1- 𝛼)% = 99% , 𝛼= 0.01, 𝛼

2= 0.005

C I ⟶ 𝑋 − 2𝑧0.025 , 𝑋 + 2𝑧0.025

x=̅ 81.2, 𝑧0.025 = 2.575

81.2 − 2 × 2.575, 81.2 + 2 × 2.575 _________(2)

(2) 𝑋1 …𝑋𝑛 r.s. N (𝜇, 9)

𝑋 ∼ 𝑁 𝜇,9

𝑛 ⟹

𝑋 − 𝜇

3

𝑛

∼ 𝑁 0, 1

𝑃[𝑋 − 1 < 𝜇 < 𝑋 + 1] = 𝑃 −1 ≤ 𝑋 − 𝜇 ≤ 1

= 𝑃 −1

3

𝑛

≤ 𝑋 − 𝜇

3

𝑛

≤1

3

𝑛

= 𝛷 𝑛

3 − 𝛷 −

𝑛

3 = 2𝛷

𝑛

3 − 1 = 0.90 𝑔𝑖𝑣𝑒𝑛 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛

⟹ 𝛷 𝑛

3 = 0.95 = 𝛷 1.96

⟹ 𝑛 = 3 × 1.96 ⟹ 𝑛 = ⋯

(3) 𝑋1 …𝑋𝑛 r.s. N (𝜇, 𝜎2)

𝑋 ∼ 𝑁 μ,

𝜎2

𝑛

𝑛 − 1 𝑠2

𝜎2∼ 𝜒𝑛−1

2

𝑖𝑛𝑑𝑒𝑝

⟹ 𝑋 − 𝜇

𝑠

𝑛

∼ 𝑡𝑛−1

𝑃 −𝑡𝛼2

;𝑛−1≤

𝑋 − 𝜇 𝑠

𝑛

≤ 𝑡𝛼2

;𝑛−1 = 1 − 𝛼

𝐹𝑜𝑟 1 − 𝛼 = 0.90; 𝛼 = 0.1, 𝑡𝛼2

;𝑛−1= 𝑡0.05,24 = 1.711

⟹ 𝑃 𝑋 −𝑠

𝑛𝑡0.05,24 ≤ 𝜇 ≤ 𝑋 +

𝑠

𝑛𝑡0.05,24 = 0.90

100 1 − 𝛼 % 𝐶𝐼 𝑤𝑖𝑡𝑕 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 �̅� = 4.7, 𝑠2 = 5.76

4.7 − 5.76

25 × 1.711, 4.7 +

5.76

25 × 1.711

= ⋯

(4) 𝑋1 …𝑋9 r.s. N (𝜇, 𝜎2)

(a) CI for 𝜇 when 𝜎2 is known at 95% level

𝑃 −𝑧𝛼

2≤

𝑋 −𝜇

2≤ 𝑧𝛼

2 = 0.95

𝛼 = 0.05 𝑃 −𝑧0.025 ≤𝑋 − 𝜇

𝜎3

≤ 𝑧0.025 = 0.95

= 𝑃 𝑋 −𝜎

3× 1.96 ≤ 𝜇 ≤ 𝑋 +

𝜎

3× 1.96 = 0.95[𝑧0.025 = 1.96]

𝐶𝐼 𝑋 −𝜎

3× 1.96, 𝑋 +

𝜎

3× 1.96

Length 2𝜎

3× 1.96 = 𝐿

𝐸 𝐿 = 2 ×𝜎

3× 1.96 = ⋯

(b) 𝜎2 unknown case

𝐶𝐼 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛

𝑃 −𝑡𝛼2

;𝑛−1≤

𝑋 − 𝜇 𝑠

𝑛

≤ 𝑡𝛼2

;𝑛−1 = 0.95

𝑡0.025,8 = 2.306,

𝐶𝐼 𝑋 −𝑠

𝑛𝑡0.025,8 , 𝑋 +

𝑠

𝑛𝑡0.025,8

𝐿𝑒𝑛𝑔𝑡𝑕 𝐿 = 2 ×𝑠

𝑛× 2.306

𝐸 𝐿 =2 × 2.306

𝑛× 𝐸 𝑆

𝑢𝑠𝑖𝑛𝑔 𝑛 − 1 𝑠2

𝜎2∼ 𝜒𝑛−1

2 𝑙𝑒𝑡 𝑌 = 𝑛 − 1 𝑠2

𝜎2

𝐸 𝑌 = 𝐸 𝑛 − 1

𝜎𝑠 =

1

2𝑛−1

2 ⎾𝑛 − 1

2

𝑦 12 𝑒−

𝑦2

∞

0

𝑦𝑛−1

2 − 1 𝑑𝑦

= 1

2𝑛−1

2 ⎾𝑛 − 1

2

𝑒−𝑦2

∞

0

𝑦𝑛2−1𝑑𝑦

=⎾

𝑛2 2

𝑛2

2𝑛−1

2 ⎾𝑛 − 1

2

⟹ 𝐸 𝑆 =𝜎

𝑛 − 1.

⎾𝑛2 2

𝑛2

2𝑛−1

2 ⎾𝑛 − 1

2

𝐸 𝐿 =2 × 2.306

𝑛.

𝜎

𝑛 − 1.

⎾𝑛2

2𝑛2

2𝑛−1

2 ⎾𝑛 − 1

2

= ⋯

(5) 𝑋1 …𝑋𝑛 r.s. from f(x|𝜃)= 3𝑥2

𝜃3 0 < 𝑥 < 𝜃

(a) 𝑋 𝑛 = 𝑀𝑎𝑥 𝑋1 …𝑋𝑛

𝑓𝑋 𝑛 𝑥 = 𝑛 𝐹𝑋 𝑥

𝑛−1𝑓𝑋 𝑥 . 𝐹𝑋 𝑥 =

3𝑦2

𝜃3

𝑥

0

𝑑𝑦 = 3

𝜃3.𝑥3

3=

𝑥

𝜃

3

𝑓𝑋 𝑛 𝑥 = 𝑛

𝑥3

𝜃3

𝑛−1

.3𝑥2

𝜃3. 0 < 𝑥 < 𝜃

𝑖. 𝑒. 𝑓𝑋 𝑛 𝑥 =

3𝑛

𝜃3𝑛𝑥3𝑛−1 0 < 𝑥 < 𝜃

𝑌 =𝑋 𝑛

𝜃 ; 𝑓𝑌 𝑦 =

3𝑛

𝜃3𝑛(𝑦𝜃)3𝑛−1𝜃; 0 < 𝑦 < 1

𝑓𝑌 𝑦 = 3𝑛 𝑦3𝑛−1 0 < 𝑦 < 1

(b) 𝑃 𝑋 𝑛 ≤ 𝜃 ≤ 𝛼−1

3𝑛𝑋 𝑛 = 𝑃 𝛼1

3𝑛 ≤ 𝑋 𝑛 ≤ 1

= 𝑃 𝛼1

3𝑛 ≤ 𝑌 ≤ 1

= 3𝑛 𝑦3𝑛−11

𝛼1

3𝑛

𝑑𝑦 =3𝑛

3𝑛 1 − 𝛼 = 1 − 𝛼

⟹ 𝑋 𝑛 , 𝛼−

1

3𝑛𝑋 𝑛 provides at 100(1- 𝛼) % CI for 𝜃.

(6) 𝑋1 …𝑋𝑛 r.s. U (0, 𝜃)

𝑓𝑋 𝑛 𝑥 =

𝑛

𝜃𝑛𝑥𝑛−1 0 < 𝑥 < 𝜃

𝑌 =𝑋 𝑛

𝜃; 𝑓𝑌 𝑦 = 𝑛 𝑦𝑛−1; 0 < 𝑦 < 1

𝑃 𝑋 𝑛 ≤ 𝜃 ≤ 𝛼−1𝑛𝑋 𝑛

= 𝑃 1 ≤𝜃

𝑋 𝑛 ≤ 𝛼−

1𝑛

= 𝑃 𝛼1𝑛 ≤

𝑋 𝑛

𝜃≤ 1

= 𝑛 𝑦𝑛−11

𝛼1𝑛

𝑑𝑦 =𝑛

𝑛 1 − 𝛼 = 1 − 𝛼

⟹ 𝑋 𝑛 , 𝛼−

1𝑛𝑋 𝑛 𝑖𝑠 100 1 − 𝛼 % 𝐶𝐼 𝑓𝑜𝑟 𝜃

𝐴𝑙𝑠𝑜 𝑃 1 − 𝛼 −1𝑛𝑋 𝑛 ≤ 𝜃 < ∞

= 𝑃 1 − 𝛼 −1𝑛 ≤

𝜃

𝑋 𝑛 < ∞

= 𝑃 0 <𝑋 𝑛

𝜃< 1 − 𝛼 −

1𝑛

⟹ 1 − 𝛼 −1

𝑛𝜃

𝑋 𝑛 , ∞ is also 100(1- 𝛼)% CI for 𝜃.

(7) 𝑋1 , … , 𝑋5 i.i.d. N ( 𝜇1 , 𝜎12 - value(1.5, 2.8 , 3.3 , 3.9 , 7.2)

𝑌1 , … , 𝑌5 i.i.d. N ( 𝜇2 , 𝜎22 - value(2.8 , 1.8 , 3.1 , 6.5 , 6.9)

(a) 𝜎12 = 𝜎2

2 = 3.5 = 𝜎2 𝑠𝑎𝑦

𝑋 ∼ 𝑁 𝜇1 ,𝜎2

5

𝑌 ∼ 𝑁 𝜇2 ,𝜎2

5


⟹𝑋 − 𝑌 − 𝜇1 − 𝜇2

𝜎 25

∼ 𝑁 0, 1

100 1 − 𝛼 % 𝐶𝐼 𝑓𝑜𝑟 ( 𝜇1 − 𝜇2 𝑎𝑡 𝛼 = 0.05

𝑋 − 𝑌 − 𝑧0.025 3.5 2

5, 𝑋 − 𝑌 + 𝑧0.025 3.5

2

5

𝑢𝑠𝑒 𝑧0.025 = 1.96 & 𝑐𝑜𝑚𝑝𝑢𝑡𝑒𝑑 �̅�& 𝑦 to get the computed CI

𝜇1 − 𝜇2 = 𝜇 (say); 𝜇 known

(b)

𝑛 𝑋 − 𝜇 2

𝜎12 ∼𝜒1

2 𝜇1= 𝜇

𝑌 ∼𝑁 𝜇 ,𝜎2

2

𝑛

𝑛 𝑌 − 𝜇 2

𝜎22 ∼𝜒1

2


⟹

𝑛 𝑋 − 𝜇 2

𝜎12 /1

𝑛 𝑌 − 𝜇 2

𝜎22 /1

=𝜎2

2

𝜎12

. 𝑋 − 𝜇

𝑌 − 𝜇

2

∼ 𝐹1 ,1

⟹ 𝑃 𝐹1 ,1;1−

𝛼2

≤𝜎2

2

𝜎12

𝑋 − 𝜇

𝑌 − 𝜇

2

≤ 𝐹1 ,1;

𝛼2 = 1 − 𝛼

⟹ 𝑃 𝑋 − 𝜇

𝑌 − 𝜇

2

1

𝐹1 ,1;

𝛼2

≤𝜎1

2

𝜎22

≤ 𝑋 − 𝜇

𝑌 − 𝜇

2

1

𝐹1 ,1;1−

𝛼2

= 1 − 𝛼

100 1 − 𝛼 % 𝐶𝐼 𝑓𝑜𝑟 𝜎1

2

𝜎22

𝑖𝑠

𝑋 − 𝜇

𝑌 − 𝜇

2

1

𝐹1 ,1;

𝛼2

, 𝑋 − 𝜇

𝑌 − 𝜇

2

1

𝐹1 ,1;1−

𝛼2

.

Using computed x ̅, y ̅and 𝐹1 ,1;𝛼

2, 𝐹1 ,1;1−

𝛼

2

At 𝛼 = 0.05 gives the computed CI.

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Problem & Solutions on Probability & Statistics · Problem & Solutions on Probability & Statistics Problem Set-1 [1] A coin is tossed until for the first time the same result appear