1

STAT 400 UIUC 5.3 – Several Independent Random

Variables

Stepanov Dalpiaz Nguyen

If X 1 , X 2 , … , X n are n random variables and a 0 , a 1 , a 2 , … , a n are n + 1 constants,

then the random variable U = a 0 + a 1 X 1 + a 2 X 2 + … + a n X n has mean

E(U) = a 0 + a 1 E(X 1) + a 2 E(X 2) + … + a n E(X n)

and variance

Var(U) =

=

If X 1 , X 2 , … , X n are n independent random variables and a 0 , a 1 , a 2 , … , a n are n + 1

constants, then the random variable U = a 0 + a 1 X 1 + a 2 X 2 + … + a n X n has variance

Var(U) = a 1

2 Var(X 1) + a 22 Var(X 2) + … + a n

2 Var(X n) If X 1 , X 2 , … , X n are n independent random variables and a 0 , a 1 , a 2 , … , a n are n + 1

constants, and U = a 0 + a 1 X 1 + a 2 X 2 + … + a n X n , then

M U ( t ) = e a 0 t × M X 1

( a 1 t ) × M X 2 ( a 2 t ) × … × M X n ( a n t ).

If X 1 , X 2 , … , X n are normally distributed random variables,

then U is also normally distributed.

( ) ( )ååå=

¹==

+n

i

n

jij

jijin

iii

1 11

2

X,XCov XVar aaa

( ) ( )ååå<=

+ji

jijin

iii

X,XCov 2 XVar aaa1

2

does not have to be independent

yEl!

+ ELY)

-→ Cov ( Xi , Xj ) = O for it j

-

-

O

T

U = ao t a , X, t- - - t an Xn

2

Example 1: Models of the pricing of stock options often make the assumption of a normal distribution. An investor believes that the price of an Burger Queen stock option is a normally distributed random variable with mean $18 and standard deviation $3. He also believes that the price of an Dairy King stock option is a normally distributed random variable with mean $14 and standard deviation $2. Assume the stock options of these two companies are independent. The investor buys 8 shares of Burger Queen stock option and 9 shares of Dairy King stock option. What is the probability that the value of this portfolio will exceed $300? Example 2: A machine fastens plastic screw-on caps onto containers of motor oil. If the machine applies more torque than the cap can withstand, the cap will break. Both the torque applied and the strength of the caps vary. The capping machine torque has the normal distribution with mean 7.9 inch-pounds and standard deviation 0.9 inch-pounds. The cap strength (the torque that would break the cap) has the normal distribution with mean 10 inch-pounds and standard deviation 1.2 inch-pounds. The cap strength and the torque applied by the machine are independent. What is the probability that a cap will break while being fastened by the capping machine? That is, find the probability P ( Strength < Torque ).

X = price of an Burger Ameen stock option, Xu N ( µ= 18 , 8--3)Y =

price of a Dairy King stock option, Yu N ( µ = 14, 8=2)

X and Y are indeep .-

Let U = 8 X t g y.

So , U~N(µ=270,6=3#E ( U ) = E ( 8X t 9 Y) = 8 E (x) t 9 E ( Y) = 8 ( 18) t 9 ( 147=270 .

Var ( U) = 82 Var (X) t 9 -Var ( y) = 82 ( 5) t 9422) = 900 .

⇒ SD (u) =a = 30

I ( u > 300) = I ( z > 300-32070-1 = I (Z > 1) = I - I CZ Ll)= I - 0.8413 = 0.15-870

-

=

3

Example 3: In Neverland, the weights of adult men are normally distributed with mean of 170 pounds and standard deviation of 10 pounds, and the weights of adult women are normally distributed with mean of 125 pounds and standard deviation of 8 pounds. Six women and four men got on an elevator. Assume that all their weights are independent. What is the probability that their total weight exceeds 1500 pounds?

S ~ N ( µ = 10,6 = I

. 2)T N N ( µ = 7.9, 6 = 0.9)

S and T are indep .

I ( S C T) = I ( s - T so )linear combination of S and T

Let U = S - T. U N N ( µ = 2. I , 6 = 1.5 )

E (u) = E ( s -T) = E (s) - E ( T) = 10 - 7.9=2 . IVar (U) = Var ( S - T) = Var ( s) t C-D

-

Var (T) = 2.25

⇒ SD (U) = V2.LT = 1.5

I ( s - T so) = I ( z s 01.3€) = I (Z C - l. 40) e 0.08080

-independent andidentically distributed

Wi , We , Ws , Wy , Ws , Woo"du N ( µ = 125

,6=8)

Mi , Mz, Ms , My"id N ( µ = 170

,6 = 10)

Total = W , t Wat W gt W,t Wo t W g

t M,t Mz t Ms t My

E ( Total) = II,

E (Walt II,

E (Mi) = G E ( Wi) t 4 E (Mi)I

= f ( 125) t 4 ( 170) = 1430 ,

identically distributedVar ( Total) = II.Var (Wa) t.FI

,

var ( Mi) = 6 ( 82) t 4 ( 107=784

⇒ Total ~ N ( µ = 1430 , 6--1787--28)I ( Total 71500) = I ( z y

'50%1-430) = I ( Z > 2.50)⇐0.00620

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Example 4: Let X and Y be two independent Poisson random variables with mean l 1 and l 2 , respectively.

Let W = X + Y. a) What is the probability distribution of W? b) What is the conditional distribution of X given W = n?

X : ⑤④ 2,3, .

- inY : O, I , 2 , 3 , . . . , n

→discrete RVs

w -- X t Y -

- n

- discrete

X - Poisson ( di ) , Y- Poisson ( Az)n

Method : I (W -- n) ⇐E I ( x -- k) Icy -- n - k) =¥.oA"ej . dF"

=

e- (dit de)K -- O ( na )

( n- K) !

T.EE?diiIn---e-IitEo.Yin.m.aixi- kk ! (n - k) !

=

( dit Az )"

e- lditdz) -

#← Poisson ( D= Kit de) Binomial Theorem

S" W ~ Ia'sson ( d, t da ) .= ( dit da)

"

Method : use mgfMw ( t) = Mx ( t) My Ct) = e

'll let - 1) edz let - D= ecditdz) ( et - I?⇒ W N Poisson (dit de)

I ( X = KI w = n) =I(X=khW=n#

=ICK-k.nl/=n-k)ICW--n) I ( W -

- n)independence1 I ( X -- K) . I ( Y -- n - K)=-

I ( W -- n)

dike- d'

.

air-ke""- -

k ! C n - K) !=.

( dit da)"

e- (ditdatTn !

= I.¥a.)"

(EET.

)" ! osksn .

⇒ X I W --

n - Binomial ( n, p = !¥d

.

) .

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If random variables X and Y are independent, then

M X + Y ( t ) = M X ( t ) × M Y ( t ). If X and Y are independent, X is Bernoulli ( p ), Y is Bernoulli ( p ) Þ X + Y is Binomial ( n = 2, p ); X is Binomial ( n 1 , p ), Y is Binomial ( n 2 , p ) Þ X + Y is Binomial ( n 1 + n 2 , p ); X is Geometric ( p ), Y is Geometric ( p ) Þ X + Y is Neg. Binomial ( r = 2, p ); X is Neg. Binomial ( r 1 , p ), Y is Neg. Binomial ( r 2 , p )

Þ X + Y is Neg. Binomial ( r 1 + r 2 , p ); X is Poisson ( l 1 ), Y is Poisson ( l 2 ) Þ X + Y is Poisson ( l 1 + l 2 ); X is Exponential ( q ), Y is Exponential ( q ) Þ X + Y is Gamma ( a = 2, q ); X is Gamma ( a 1 , q ), Y is Gamma ( a 2 , q ) Þ X + Y is Gamma ( a 1 + a 2 , q ); X is Normal ( µ 1 , s 1

2 ), Y is Normal ( µ 2 , s 22 )

Þ X + Y is Normal ( µ 1 + µ 2 , s 12 + s 2

2 ).