•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