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•3
A
•4 6
2
B
•5
1
C
22% 22%
56% 51%
49%
A beats B with probability 0.56
B beats C with probability 0.62
A beats C with probability 0.51
A > B >C
What if all three spin at once?
Suppose that X denotes the number of robberies in a particular small town in a given month. Suppose that there are on average 4 robberies each month.
What is the probability that there are at least three robberies in a given month? Express the answer in terms of e (no summations).
We will assume that X is Poisson with λ = 4
P X = k( ) = e−4 4 k
k!
The probability of at most two robberies is P X = 0( ) + P X = 1( ) + P X = 2( ) = e−4 + 4e−4 + 8e−4 = 13e−4
The probability of at least three robberies is 1−13e−4
Suppose that X denotes the number of misprints on a page in a given book. Suppose that there are on average 1.2 misprints per page.
(a). What is the probability that there are at least two misprints on a given page? Express the answer as a decimal value.
(b). What is the probability that there are exactly two misprints on a given page? Express the answer as a decimal value.
We will assume that X is Poisson with λ = 1.2
P X ≥ 2( ) = 1− P X ≤1( )
•
•
= 1− 0.663= 0.337λ = 1.2P X ≥ 2( ) = 1− P X ≤1( )
P X = 2( ) = P X ≤ 2( )− P X ≤1( )= 0.879 − 0.663 = 0.216
P X ≤ a( )
Suppose that you toss a biased coin 10 times and that the coin is weighted so that it shows heads 70% of the time.
P X ≤ 3( )
Let X denote the number of heads that appear.
Then X is b(10, 0.7)
What is an exact expression for:What is an approximation to this value?
What is an exact expression for:What is an approximation to this value?
P X ≥ 7( )
What is an exact expression for:What is an approximation to this value?
P X = 8( )
10k
⎛⎝⎜
⎞⎠⎟(0.7)k (0.3)10−k
k=0
3
∑
10k
⎛⎝⎜
⎞⎠⎟(0.7)k (0.3)10−k
k=7
10
∑
108
⎛⎝⎜
⎞⎠⎟(0.7)8 (0.3)2
0.011
0.011
0.650 = 1 – 0.350
0.234 = 0.851 – 0.617
P(X = k) = p(k) =0.3 if k = 80.2 if k = 100.5 if k = 6
⎧⎨⎪
⎩⎪
Suppose that X is a random variable that has the probability function
What is the moment generating function for X ?
MX (t) = ekt p(X = k) = 0.3e8t + 0.2e10t + 0.5e6tk∑
Suppose that Y is a random variable with moment generating function H(t). Suppose further that X is a random variable with moment generating function M(t) given by
�
M(t) =13 (2e
3 t +1)H (t)
Given that the mean of Y is 10 and the variance of Y is 12, then determine the mean and variance of X.
Since the mean of Y is 10, Since the variance of Y is 12,
�
H '(0) =10
�
12 = E(Y 2 ) − E(Y )2 = E(Y 2) −100⇒ E (Y 2) =112
�
H ' '(0) =112
Suppose that Y is a random variable with moment generating function H(t). Suppose further that X is a random variable with moment generating function M(t) given by
�
M(t) =13 (2e
3 t +1)H (t)
Given that the mean of Y is 10 and the variance of Y is 12, then determine the mean and variance of X.
�
H '(0) =10
�
H ' '(0) =112
′M (t) = ddt
132e3t +1( )H (t)⎡
⎣⎢⎤⎦⎥= 2e3tH (t)+ 1
32e3t +1( ) ′H (t)
µ = E X( ) = ′M (0) = 2H (0)+ 133( ) ′H (0) = 2 +10 = 12
E X( ) = 12
Suppose that Y is a random variable with moment generating function H(t). Suppose further that X is a random variable with moment generating function M(t) given by
�
M(t) =13 (2e
3 t +1)H (t)
Given that the mean of Y is 10 and the variance of Y is 12, then determine the mean and variance of X.
�
H '(0) =10
�
H ' '(0) =112 E X( ) = 12
′′M t( ) = ddt
2e3tH t( ) + 132e3t +1( ) ′H t( )⎡
⎣⎢⎤⎦⎥
= 6e3tH t( ) + 4e3t ′H t( ) + 132e3t +1( ) ′′H t( )
E X 2( ) = ′′M 0( ) = 6H 0( ) + 4 ′H 0( ) + 133( ) ′′H 0( ) = 6 + 40 +112 = 158
E X 2( ) = 158
Suppose that Y is a random variable with moment generating function H(t). Suppose further that X is a random variable with moment generating function M(t) given by
�
M(t) =13 (2e
3 t +1)H (t)
Given that the mean of Y is 10 and the variance of Y is 12, then determine the mean and variance of X.
�
H '(0) =10
�
H ' '(0) =112 E X( ) = 12 E X 2( ) = 158
Var X( ) = E X 2( )− E X( )2 = 158 −122 = 158 −144 = 14
Suppose that the Moment generating function for X is M (t) = et
3− 2et
Then determine
�
µ and σ 2 for X.
M (t) = et
3− 2et=
13et
1− 2 3 et
And so X must be a geometric random variable
with probability of success.p = 13
µ = 1 p =113= 3 σ 2 = q
p2 =1− 13
19= 6
M (t) = 1+ 2et
3⎛⎝⎜
⎞⎠⎟1+ 3et
4⎛⎝⎜
⎞⎠⎟
Suppose that the moment generating function of the random variable X is
What is the probability P(X = 1)?
MX (t) = ektP X = k( )k∑
If X is a discrete random variable, then
The probability X equals k is the coefficient of ekt
M (t) = 1+ 2et
3⎛⎝⎜
⎞⎠⎟1+ 3et
4⎛⎝⎜
⎞⎠⎟
Suppose that the moment generating function of the random variable X is
What is the probability P(X = 1)?
M t( ) = 112
1+ 5et + 6e2t( ) = 112
+ 512et + 1
2e6t
P X = 1( ) = 512
, P X = 0( ) = 112
, P X = 2( ) = 12
Suppose that the moment generating function of the random variable X is
What is the mean and variance of X ?
This is the moment generating function for a binomial random variable with n = 10 and p = 0.75.
µ = 10 × 0.75 = 7.5
σ 2 = 10 × 0.75 × 0.25 = 158
M (t) = 1+ 3et
4⎛⎝⎜
⎞⎠⎟
10
= 14+ 34et⎛
⎝⎜⎞⎠⎟10
M (t) = 0.7et + 0.3( )10The moment generating function of X is
E X( ) = 10 × 0.7 = 7, Var(X) = 10 × 0.7 × 0.3= 2.1
X is a binomial random variable with n = 10, p = 0.7
M (t) = et + 23
⎛⎝⎜
⎞⎠⎟
5
The moment generating function of X is
X is a binomial random variable with n = 5, p = 1/3
E(X) = 53
Var(X) = 109
M (t) = et
2 − et
The moment generating function of X is
X is a geometric random variable with p = 1/2
E(X) = 2 Var(X) =1214= 2
M (t) =
12et
1− 12et
The moment generating function of X is MX (t)
Y = aX + b
MY (t) = E etY( ) = et ak+b( )P(X = k) =k∑ et ak+b( )P(X = k)
k∑
e ak+b( )tP(X = k)k∑ = ebt ⋅eatkP(X = k)
k∑ = ebt e(at )kP(X = k)
k∑ = ebtE(eatX )
= ebtMX (at)
= ebtMX (at)MY (t)
The moment generating function of X is 3.158
MX (t)Y = aX + b = ebtMX (at)MY (t)
′MY (t) = aebt ′MX (at)+ be
btMX at( )E Y( ) = ′MY (0) = a ′MX (0)+ bMX 0( ) = aE X( ) + b
′′MY (t) = a2ebt ′′MX (at)+ 2abe
bt ′MX at( )+ b2ebtMX (at)′′MY (0) = a
2 ′′MX (0)+ 2ab ′MX 0( )+ b2MX (0)
Var Y( ) = E Y 2( )− E Y( )2 = a2E X 2( )+ 2abE X( )+ b2⎡⎣ ⎤⎦ − aE(X)+ b[ ]2
= a2E X 2( )− a2E(X)2 = a2 E X 2( )− E X( )2⎡⎣ ⎤⎦ = a2Var X( )
So, E Y 2( ) = ′′MY (0) = a2E X 2( )+ 2abE X( )+ b2