Function of Random Variables

FRV - 1

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Functions of Random Variables

7.1 Introduction

Methods for finding distribution of function of one

or more random variables:

1. Distribution Function Technique

2. Transformation Technique

3. Moment Generating Function Technique

How to find the distribution of a random variable Y

that is a function of several random variables X1, X2,

…, Xn that has a joint probability distribution?

𝑦 = 𝑢(𝑥1, 𝑥2, …, 𝑥𝑛)

7.2 Distribution Function Technique

For finding the probability density function with a

given joint probability density, the probability

density function of 𝑌 = 𝑢(𝑋1, 𝑋2, …, 𝑋𝑛) can be

obtained by first finding the cumulative probability

or distribution function

F(y)= 𝑃 𝑌 ≤ y = 𝑃(𝑢(𝑋1, 𝑋2, …, 𝑋𝑛)

and then differentiate it to get the p.d.f.

𝑓 𝑦 =𝑑𝐹(𝑦)

𝑑𝑦

4

Example: Let X ~ U(0,1), and Y = 𝑋𝑛 , find 𝑝. 𝑑. 𝑓. of 𝑌.

G(y) = P(Y ≤ y)

= P(𝑋𝑛 ≤ y)

= P(X ≤ y1/n )

= F(y1/n )

= y1/n

g 𝑦 = 1

𝑛𝑦

1𝑛−1, 0 ≤ 𝑦 ≤ 1

0, elswhere.

5

Example: Let X has the following p.d.f.,

G(y) = P(Y ≤ y) = P(𝑋3 ≤ y)

= P(X ≤ y1/3 )

= 6𝑥 1 − 𝑥 𝑑𝑥y1/3

0

= 3y2/3 – 2y, for 0< y <1

𝑓 𝑥 = 6𝑥 1 − 𝑥 , 0 < 𝑥 < 1

0, elswhere

g 𝑦 = 2(y −1/3 – 1), 0 < 𝑦 < 10, elswhere.

find the p.d.f. of Y = 𝑋3.

6

Example: Let X have a p.d.f. f(x) and Y = 𝑋2 , find 𝑝. 𝑑. 𝑓. of 𝑌.

G(y) = P(Y ≤ y)

= P(𝑋2 ≤ y)

= P(-y1/2 ≤ X ≤ y1/2)

= F(y1/2) - F(- y1/2)

g(y) = 1

2 𝑦𝑓(y1/2) + f(− y1/2)

Function of Random Variables

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Example: Let X have a p.d.f. f(x) and Y = |X|, find 𝑝. 𝑑. 𝑓. of 𝑌 when 𝑋 ~ 𝑁 0,1 .

G(y) = P(Y ≤ y)

= P(|X|≤ y)

= P(-y ≤ X ≤ y)

= F(y) - F(- y)

g(y) = 𝑓(y) + f(−y), for y > 0

when 𝑋 ~ 𝑁 0,1 , g(y) = 2𝑓(y) , for y > 0

= 21

2𝜋𝑒−

𝑦2

2

-y 0 y 8

Example: Let X have a p.d.f. 𝑓 𝑥 =1

𝜃𝑒−

𝑥

𝜃, for 𝑥 > 0,

find p. d. f. of 𝑌 = ln 𝑋 .

G(y) = P(Y ≤ y)

= P(lnX ≤ y)

= P(X ≤ e y )

= 1

𝜃𝑒−

𝑥

𝜃 𝑑𝑥𝑒𝑦

0

= - 𝑒−𝑥

𝜃 𝑒𝑦

0= 1 − 𝑒−

1

𝜃∙𝑒𝑦

g(y) = G′(y) = 1

𝜃𝑒𝑦𝑒−

1

𝜃∙𝑒𝑦

, - < y <

9

Example: If the joint density of X1 and X2 is given by

Find the probability of Y = X1 + X2

F(y) = P(Y ≤ y) = P(X1 + X2 ≤ y)

𝑓 𝑥1, 𝑥2 = 6𝑒−3𝑥1−2𝑥2 , for 𝑥1 > 0, 𝑥2 > 0

0, elsewhere

= 1 + 2e-3y – 3e-2y

F(y) = f (y) = 6(e-2y – e-3y), for y > 0,

f (y) = 0, elsewhere.

x1 + x2 = y = 6𝑒−3𝑥1−2𝑥2

𝑦−𝑥2

0

𝑦

0𝑑𝑥1𝑑𝑥2

0 y

10

Example: If X1 and X2 are independent random variables

having U(0,1), find the distribution function of Y = X1+X2.

F(y) = 0, if y ≤ 0.

𝑓1 𝑥1 = 1 = 𝑓2 𝑥2 𝑓 𝑥1, 𝑥2 ,

for 0 < 𝑥1< 1, 0 < 𝑥2< 1.

y = x1 + x2

0 1

1

x1

x2

F(y) = 1

2𝑦2, if 0 < y ≤ 1.

F(y) = 1 −(2−𝑦)2

2, if 1 < y ≤ 2.

F(y) = 1, if y > 2.

11

7.3 Transformation Technique: One Variable

For discrete random variable, whether X and Y = u(X)

is one-to-one or not, finding the distribution of Y is

straight forward substitution.

Example: Let X be the number of heads in tossing a

balanced coin three times, find the probability

distribution of Y = 1/(1+X) . (One-to-one function)

x 0 1 2 3

f(x) 1/8 3/8 3/8 1/8

y 1 1/2 1/3 1/4

g(y) 1/8 3/8 3/8 1/8 12

Example: Let X be the number of heads in tossing a

balanced coin three times, find the probability

distribution of Y = (1 - X)2 . (Not one-to-one function)

x 0 1 2 3

f(x) 1/8 3/8 3/8 1/8

y* 1 0 1 4

g(y*) 1/8 3/8 3/8 1/8

y 0 1 4

g(y) 3/8 4/8 1/8

Function of Random Variables

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Inverse Function Theorem :

For functions of a single variable, if u is a continuously

differentiable function with nonzero derivative at the

point x, then u is invertible in a neighborhood of x, the

inverse is continuously differentiable, and

where y = u(x).

𝐷𝑦𝑢−1(𝑦) =1

𝐷𝑥𝑢(𝑥)

14

Theorem 7.1: (Univariate Transformation Theorem)

Let f(x) be the probability density of the continuous

random variable X at x. If the function given by y =

u(x) is differentiable and either (monotone) increasing

and decreasing for all values within the range of X that

has density, then the equation y = u(x) is one-to-one and

x = w(y), and the probability density of Y = u(X) is

given by

g 𝑦 = 𝑓 𝑤 𝑦 ∙ 𝑤′ 𝑦 , for 𝑢′(𝑥) ≠ 0

0, elsewhere.

(𝑢−1 𝑦 )′ = 𝑤′(𝑦) =1

𝑢′ 𝑥=

𝑑𝑥

𝑑𝑦

15

g 𝑦 = 𝑓 𝑢−1 𝑦 ∙

𝑑

𝑑𝑦𝑢−1 𝑦 , for 𝑢′(𝑥) ≠ 0

0, elsewhere.

Another version of the formula: u-1 (y) = w(y)

16

y

x

y = u(x)

b

a

w(a) w(b)

P(u(X) ≤ y) = P(X ≤ w(y))

u(x) is increasing function

x

y

Y = 2X

0 .5 1

0 1 2

f(x)

g(y)

Example: X ~ U (0, 1)

17

y

x

y = u(x) b

a

w(b) w(a)

P(u(X) ≤ y) = P(X w(y))

u(x) is decreasing function

x

y

Y = -2X

0 .5 1

- 2 - 1 0

f(x)

g(y)

Example: X ~ U (0, 1)

18

Let u(x) be a strictly decreasing function in the range

of X, and w be the inverse function of u, i.e., u-1 .

G(y) = P(Y ≤ y) = P(u(X) ≤ y)

= P(X w(y)) = 1 - P(X < w(y)) =1 - F(w(y))

g(y) = G (y) = - F (w(y)) = -f(w(y))w′(y)

Since u(x) is a decreasing function then w′(y) < 0. If

u(x) is a increasing function in the range of X, then

G(y) = P(Y ≤ y) = P(u(X) ≤ y)

= P(X ≤ w(y)) = P(X < w(y)) = F(w(y))

g(y) = G (y) = F (w(y)) = f(w(y))w′(y), with w′(y) > 0.

g 𝑦 = 𝑓 𝑤 𝑦 ∙ 𝑤′ 𝑦 , for 𝑢′(𝑥) ≠ 0

0, elsewhere.

Function of Random Variables

FRV - 4

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Example: Let X have the exponential distribution with

p.d.f. f(x) given by

find the p.d.f. of the random variable Y = 𝑋. Sol: For y > 0, y = 𝑥 x = y2 .

w(y) = y2 , w(y) = 2y

g(y) = 𝑓(𝑦2 ) ∙ 2𝑦 = 𝑒−𝑦2|2𝑦|

𝑓 𝑥 = 𝑒−𝑥, for 𝑥 > 0,0, elsewhere

g 𝑥 = 2𝑦𝑒−𝑦2, for 𝑦 > 0,

0, elsewhere. (Weibull Distribution)

for y > 0

20

Example: Let X be the a random variable takes the

distance for 0 to a point on the x-axis where the double

arrow will point to, when it is spun. The random

variable Q is the angle that has uniform density

find the p.d.f. of the random variable X.

𝑓 𝜃 = 1

𝜋, for −

𝜋

2< 𝜃 <

𝜋

2,

0, elsewhere

a q

0 x

x = a tanq

21

a q

0 x

x = a tanq

Sol: x = a tan q 𝑑𝜃

𝑑𝑥=

𝑎

𝑎2 + 𝑥2

1

𝜋∙

𝑎

𝑎2 + 𝑥2 g(x) =

=1

𝜋∙

𝑎

𝑎2 + 𝑥2 for - < x < .

22

Example: If F(x) is the distribution function of the

continuous random variable X, find the p.d.f. of Y = F(x).

Sol: Let y = F(x), )()( xfxFdx

dy

.0)(for ,)(

11 xf

xf

dx

dydy

dx

g(y) = f(x) = 1, for 0 < y <1. )(

1

xf

* Distribution function technique for random number

generation using U(0,1) random number generator.

23

Distribution Function Method for Random Numbers:

1. Generate a U(0, 1) random number

2. set this random numbers equal to F(x) and solve

for x.

3. The value x would be a random number from the

distribution that has a distribution function F(x).

Example: Generate a random number from

exponential distribution with parameter q using

U(0,1) random number.

F(x) = 1 – e-x/q = u u is a random # from U(0, 1)

The random # from the exponential distribution

would be: x = -q ln (1 – u) 24

Example: If X has the standard normal distribution find

the probability density of Z = X 2.

Sol: z = x 2 is not one-to-one.

First let Y = |X|, then Z = Y 2 = X 2

p.d.f. of Y g(y)= 2n(y; 0, 1) = 2

2𝜋𝑒−

12𝑦2

for y > 0

z = y2 , w(z) = y = 𝑧

p.d.f. of Z h(z)= g( 𝑧 ) |w(y)|

=2

2𝜋𝑒−

12𝑧

1

2𝑧−

12 =

1

2𝜋𝑧−

12𝑒−

12𝑧

for z > 0

(Chi-square distribution with degrees of freedom = 1.)

Function of Random Variables

FRV - 5

25

Example: Let X ~ U(0,1), and Y = 𝑋𝑛 , find 𝑝. 𝑑. 𝑓. of 𝑌.

g 𝑦 = 1

𝑛𝑦

1𝑛−1, 0 ≤ 𝑦 ≤ 1

0, elswhere. Answer:

26

7.4 Transformation Method: Several Variables

For random variable Y = u(X1, X2) where the joint

distribution or density of X1 and X2 is given, and one

can find the joint distribution or density for Y and X2 or

X1 and Y by holding the other variable fixed, if possible,

and then find the marginal distribution or density

function for Y.

In continuous case, one can first use the transformation

technique with the formula, by holding x1 or x2 fixed,

g 𝑦, 𝑥2 = 𝑓(𝑥1, 𝑥2) ∙𝜕𝑥1

𝜕𝑦

g 𝑥1, 𝑦 = 𝑓(𝑥1, 𝑥2) ∙𝜕𝑥2

𝜕𝑦

or

then find the marginal density of Y.

27

Example: If X1 and X2 are independent random having

Poisson distribution with the parameters l1 and l2 , find

the probability distribution of the random variable Y =

X1 + X2.

Sol: Since X1 and X2 are independent, the joint density is

𝑓 𝑥1, 𝑥2 =𝑒−𝜆1𝜆1

x1

𝑥1!∙𝑒−𝜆2𝜆2

x2

𝑥2!=

𝑒−(𝜆1+𝜆2)𝜆1x1𝜆2

x2

𝑥1! 𝑥2!

for x1 = 0, 1, 2, …, and x2 = 0, 1, 2, … .

Since y = x1 + x2 then x1 = y - x2

𝑔 𝑦, 𝑥2 =𝑒−(𝜆1+𝜆2)𝜆1

𝑦−x2𝜆2x2

(𝑦 − 𝑥2)! x2!

for y = 0, 1, 2, …, and x2 = 0, 1, 2, …, y .

Joint Distribution

of Y and X2

28

ℎ(𝑦) = 𝑒−(𝜆1+𝜆2)𝜆1

𝑦−x2𝜆2x2

(𝑦 − 𝑥2)! x2!

𝑦

𝑥2=0

for y = 0, 1, 2, … .

=𝑒−(𝜆1+𝜆2)

𝑦!

𝑦!

(𝑦 − 𝑥2)! x2! 𝜆1

𝑦−x2𝜆2x2

𝑦

𝑥2=0

=𝑒− 𝜆1+𝜆2 (𝜆1 + 𝜆2)𝑦

𝑦!

The sum of two independent Poisson random

variables with parameters l1 and l2 is a Poisson

random variable with parameter l1 + l2 .

29

Example: Let random variables X1 and X2 have the

joint p.d.f. as

find the p.d.f. of the random variable Y =

Sol: Since y decreases as x2 increases and x1 hold

constant, we can find a joint density of X1 and Y and

then use transformation technique to find density of Y.

𝑓 𝑥1, 𝑥2 = 𝑒−(𝑥1+𝑥2), for 𝑥1 > 0, 𝑥2 > 0,

0, elsewhere

𝑋1

𝑋1+𝑋2

𝑦 =𝑥1

𝑥1 + 𝑥2⟹ 𝑥2 = 𝑥1 ∙

1 − 𝑦

𝑦 for 0 < y < 1

⟹ 𝜕𝑥2

𝜕𝑦= −

𝑥1

𝑦2 𝑔 𝑥1, 𝑦 = 𝑒−𝑥1/𝑦 −𝑥1

𝑦2 30

𝑔 𝑥1, 𝑦 = 𝑒−𝑥1/𝑦 −𝑥1

𝑦2

=𝑥1

𝑦2 𝑒−𝑥1/𝑦 for x1 > 0, 0 < y < 1.

ℎ 𝑦 = 𝑥1

𝑦2

∞

0

∙ 𝑒−𝑥1/𝑦𝑑𝑥1 Let u = x1 / y

du = -1/y2 dx1

= 𝑢∞

0

∙ 𝑒−𝑢𝑑𝑥1

= 1 for 0 < y < 1.

It is a U(0, 1)!!!

Function of Random Variables

FRV - 6

31

Example: Let random variables X and Y have the joint

p.d.f. as

find the joint p.d.f. of X and Z = X + Y & marginal p.d.f.

of Z.

𝑓 𝑥, 𝑦 = 2, for 𝑥 > 0, 𝑦 > 0, 𝑥 + 𝑦 < 10, elsewhere

z = x + y y = z – x , and 0 < z – x and 0 < z < 1

x

z

0 1

1

z = x 𝑔 𝑥, 𝑧 = 𝑓(𝑥, 𝑦)𝑑𝑦

𝑑𝑧

= 2 ∙ 1 = 2

ℎ 𝑧 = 2𝑑𝑥 = 2𝑥 𝑧0

= 2𝑧𝑧

0

for x < z, 0 < z < 1

for 0 < z < 1 32

Theorem 7.2: (Generalization of Theorem 7.1)

Let f(x1, x2) be the joint probability density of the continuous

random variable X1 and X2. If the function given by y1 = u1(x1,

x2) and y2 = u2(x1, x2) are partially differentiable w.r.t. both x1

and x2 and are one-to-one for which f(x1, x2) ≠ 0, ∀ x1, x2 in

the space of X1 and X2 , and the inverse functions x1 = w1(y1,

y2) and x2 = w2(y1, y2) can be uniquely determined (by solving

for x1 and x2), the joint p.d.f. of Y1 = u1(X1, X2) and Y2 = u2(X1,

X2) is

where J is the Jacobian of the transformation, is the

determinant

g 𝑦1, 𝑦2 = 𝑓[w1(y1, y2),w2(y1, y2)] ∙ 𝐽

𝐽 =

𝜕𝑥1

𝜕𝑦1

𝜕𝑥1

𝜕𝑦2

𝜕𝑥2

𝜕𝑦1

𝜕𝑥2

𝜕𝑦2

33

Example: Let random variables X1 and X2 have the

joint p.d.f. as

a) find the joint p.d.f. of Y1= X1 + X2 , and Y2 =

𝑓 𝑥1, 𝑥2 = 𝑒−(𝑥1+𝑥2), for 𝑥1 > 0, 𝑥2 > 0,

0, elsewhere

𝑋1

𝑋1+𝑋2.

𝑦1 = 𝑥1 + 𝑥2 and 𝑦2 =𝑥1

𝑥1 + 𝑥2

⟹ 𝑥1 = 𝑦1𝑦2 and 𝑥2 = 𝑦1(1 − 𝑦2)

𝐽 =

𝜕𝑥1

𝜕𝑦1

𝜕𝑥1

𝜕𝑦2

𝜕𝑥2

𝜕𝑦1

𝜕𝑥2

𝜕𝑦2

=𝑦2 𝑦1

1 − 𝑦2 −𝑦1= −𝑦1

0 < y1 and 0 < y2 <1

34

for 0 < y1 and 0 < y2 <1, and 0 elsewhere.

g 𝑦1, 𝑦2 = 𝑒−𝑦1 −𝑦1 = 𝑦1𝑒−𝑦1

b) Find the marginal density of Y2 .

ℎ 𝑦2 = 𝑔 𝑦1, 𝑦2 𝑑𝑦1

∞

0

= 𝑦1𝑒−𝑦1 𝑑𝑦1

∞

0

= (2)

= 1

for 0 < y2 <1, and 0 elsewhere.

35

Example: Let random variables X1 and X2 have the

joint p.d.f. as

a) find the joint p.d.f. of Y = X1 + X2 , and Z = X2 .

𝑓 𝑥1, 𝑥2 = 1, for 0 < 𝑥1 < 1,0 < 𝑥2 < 1,0, elsewhere

𝐽 =

𝜕𝑥1

𝜕𝑦

𝜕𝑥1

𝜕𝑧𝜕𝑥2

𝜕𝑦

𝜕𝑥2

𝜕𝑧

=1 −10 1

= 1

y = x1 + x2 , and z = x2

x1= y - x2 , and x2 = z, z < y < z + 1, 0 < z < 1

y

z

0 1 2

1

y = z + x1 , z = x2

36

𝑓 𝑦, 𝑧 = 1 ∙ 1 = 1,

for z < y < z + 1, 0 < z < 1, and 0, else where.

b) Find the marginal density of Y.

ℎ 𝑦 =

0, 𝑦 ≤ 0

1𝑑𝑧 = 𝑦,𝑦

0

for 0 < 𝑦 < 1

1𝑑𝑧 = 2 − 𝑦, 1

𝑦−1

for 1 < 𝑦 < 2

0, 𝑦 ≥ 2

y

z

0 1 2

1

Function of Random Variables

FRV - 7

37

Example: Let random variables X1 and X2 have the

joint p.d.f. as

a) find the joint p.d.f. of Z = X + Y , and W = X - Y.

𝑓 𝑥, 𝑦 = 2, for 𝑥 > 0, 𝑦 > 0, 𝑥 + 𝑦 < 1,0, elsewhere

𝐽 =

1

2

1

21

2−

1

2

= −1/2

z = x + y and w = x - y x=(z + w)/2; y=(z-w)/2,

z + w > 0, z - w > 0, and 0< z < 1.

x

z

0 1

1

-1

z – w = 0

z + w = 0

𝑔 𝑧,𝑤 = 2 ∙ −1

2= 1

for 0 < z < 1, z > -w, and z > w. 38

b) find the marginal p.d.f. of Z = X + Y .

z

w

0 1

1

-1

z – w = 0

z + w = 0

𝑔 𝑧 = 𝑔(𝑧, 𝑤)𝑧

−𝑧

𝑑𝑤

for 0 < z < 1.

= 1𝑧

−𝑧

𝑑𝑤 = 𝑤 𝑧

−𝑧= 2𝑧

𝑔 𝑧, 𝑤 = 1, for 0 < z < 1, z > -w, and z > w.

39

Example: Let random variables X1 and X2 have the

joint p.d.f. as

find the marginal p.d.f. of Z = X + Y .

𝑓 𝑥, 𝑦 = 2, for 𝑥 > 0, 𝑦 > 0, 𝑥 + 𝑦 < 1,0, elsewhere

𝑔 𝑧 = 𝑔(𝑥, 𝑧)𝑧

0

𝑑𝑥

for 0 < z < 1.

= 2𝑧

0

𝑑𝑥 = 2𝑥 𝑧0

= 2𝑧

Previous Example: g 𝑥1, 𝑦 = 𝑓(𝑥1, 𝑥2) ∙𝜕𝑥2

𝜕𝑦

z = x + y y = z – x

𝑔 𝑥, 𝑧 = 𝑓(𝑥, 𝑦) ∙𝑑𝑦

𝑑𝑧 = 2∙ 1 = 2

for 0 < z < 1 and 0 < z – x .

x

z

0 1

1 z – x = 0

40

Example: Let random variables X1 and X2 have the

joint p.d.f. as

a) find the joint p.d.f. of Y1 = X12 and Y2 = X1 X2 .

𝑓 𝑥1, 𝑥2 = 4𝑥1𝑥2 , for 0 < 𝑥1 < 1,0 < 𝑥2 < 1

0, elsewhere

𝐽 =

1

2 𝑦10

−1

2𝑦2𝑦1

−3/21

𝑦1

=1

2𝑦1

y1 = x12 and y2 = x1x2 x1= 𝑦1 , x2 =

𝑦2

𝑦1,0 <y2< 𝑦1

𝑔 𝑦1, 𝑦2 = 4 𝑦1 ∙𝑦2

𝑦1𝐽

= 4y2 /2y1 =2y2 /y1

y1

y2

0 1

1 y2

2 = y1

for 0 < y2< 𝑦1

y22 < y1

41

X1, X2 , … , Xn ~ f(x1, x2 , … , xn)

Y1 ~ u1(x1, x2 , … , xn)

Y2 ~ u2(x1, x2 , … , xn)

Yn ~ un(x1, x2 , … , xn) …

g(x1, x2 , … , xn) = f(x1, x2 , … , xn)|J|

𝐽 =

𝜕𝑥1

𝜕𝑦1…

𝜕𝑥1

𝜕𝑦𝑛

⋮ ⋱ ⋮𝜕𝑥𝑛

𝜕𝑦1⋯

𝜕𝑥𝑛

𝜕𝑦𝑛

42

7.5 Moment-Generating Function Technique

Theorem 7.3 (Generalized Version): If X1, X2, …, Xn

are independent random variables with m.g.f.’s is

i = 1, 2,…, n, then the m.g.f. of is

)(tMiX

n

i

iXY taMtMi

1

)()(

n

i ii XaY1

Function of Random Variables

FRV - 8

43

Example: Find the probability distribution of the

sum of n independent random variables X1, X2, …, Xn

that have Poisson distribution with parameters l1, l2,

…, ln , respectively.

)1()(

-

ti

i

e

X etMl

The m.g.f. of Poisson distribution is

So, for Y = X1+ X2 + …+ Xn , the m.g.f. is

)1)((

1

)1( 21)(-+++

-

tn

ti e

n

i

e

Y eetMllll

which is the m.g.f. of a Poisson distribution with

parameter l = l1+ l2 + …+ ln , therefore, Y has a

Poisson distribution with l = l1+ l2 + …+ ln . 44

Example: Find the probability distribution of the

sum of n independent random variables X1, X2, …, Xn

that have Poisson distribution with parameters l1, l2,

…, ln , respectively.

)1()(

-

ti

i

e

X etMl

The m.g.f. of Poisson distribution is

So, for Y = X1+ X2 + …+ Xn , the m.g.f. is

)1)((

1

)1( 21)(-+++

-

tn

ti e

n

i

e

Y eetMllll

which is the m.g.f. of a Poisson distribution with

parameter l = l1+ l2 + …+ ln , therefore, Y has a

Poisson distribution with l = l1+ l2 + …+ ln .

45

Example: If X1, X2, …, Xn are mutually independent

random variables from normal distributions with

means m1, m2, m3, …, mn, and variances s12, s2

2, s32,

…, sn2, then the linear function

has the normal distribution N(Scimi , Sci2si

2).

n

i

ii XcY1

.1

if :Mean Samplen

cXY i

46

Sol:

+

+

2

1 1

2/

2

1

22

1

222

)()(

tctc

n

i

n

i

tctc

iXY

n

iii

n

i

ii

iiii

i

e

etcMtM

sm

sm

m.g.f. of

n

i

n

i

ii iiccN

1

22

1

, sm

47

Distribution of

If X1, X2, …, Xn are observations of a random

sample of size n from the normal distribution

N(m, s 2), then the distribution of the sample

mean is N(m, s 2/n)

n

i

iXn

X1

1

X

nX

X

ss

mm