Chapter 3:Random Variables and Distributions

3.9 Functions of Two or More Random Variables

Random Variables with a Discrete Joint Distribution

Theorem: Binomial and Bernoulli Distributions. As-

sume that X1, . . . , Xn are i.i.d. random variables

having the Bernoulli distribution with parameter p. Let

Y = X1 + . . . + Xn. Then Y has the binomial distri-

bution with parameters n and p.

Example

The joint p.f. f (x, y) of X and Y is as specified in the

following table:

y = 1 y = 2 y = 3 y = 4

x = 1 0.1 0 0.1 0

x = 2 0.3 0 0.1 0.2

x = 3 0 0.2 0 0

Table 1: Joint p.f.

Find the p.f. of Z = X + Y .

1

Solution: Z can take 2, 3,...,7.

P (Z = 2) = P (X + Y = 2) = P (X = 1, Y = 1) = 0.1

P (Z = 3) = P (X + Y = 3)

= P [X = 1, Y = 2 or X = 2, Y = 1]

= P (X = 1, Y = 2) + P (X = 2, Y = 1)

= 0.3

... ...

Z 2 3 4 5 6 7

f (z) 0.1 0.3 0.1 0.3 0.2 0

Random Variables with a Continuous Joint Distribu-tion

(i) Y = X1 + X2

Theorem: Let X1 and X2 be independent contin-

uous random variables and let Y = X1 + X2. The

distribution of Y is called the convolution of the

distributions of X1 and X2. The p.d.f. of Y is

g(y) =

f1(y z)f2(z)dz

2

Proof. The c.d.f. F (y) of Y is

F (y) = P (Y y)= P (X1 + X2 y)

=

x1+x2y

f (x1, x2)dx1dx2

=

x1+x2y

f1(x1)f2(x2)dx1dx2

=

[ yx2

f1(x1)f2(x2)dx1

]dx2

x1

x2

x1 + x2 = y

0

Take derivative with respect to y on both sides,

g(y) = F (y)

=

[ yx2

f1(x1)f2(x2)dx1

]dx2

=

f1(y x2)f2(x2)dx2

=

f1(y z)f2(z)dz

3

Example: Suppose that X1 and X2 are independent

random variables with common distribution having p.d.f.

f (x) =

{2e2x x > 0

0 otherwise

find the p.d.f. of Y = X1 + X2.

Consider the support: {(z, y) : y z > 0and z > 0}.

z

y

y = z

0

If y 0, g(y) = 0.If y > 0,

g(y) =

f1(y z)f2(z)dz

=

y0

2e2(yz)2e2zdz

= 4ye2y

g(y) =

{4ye2y y > 0

0 otherwise

4

(ii) Maximum and Minimum of a Random Sample.

Suppose that X1, ..., Xn form a random sample of size

n from a distribution for which the p.d.f. is f and the

c.d.f. is F. The largest value Yn and the smallest value Y1

in the random sample are defined as follows:

Yn = max{X1, ..., Xn}, Y1 = min{X1, ..., Xn}

Consider Yn first. Let Gn stand for its c.d.f., and let gn

be its p.d.f.

Gn(y) = P (Yn y) = P (X1 y, ..., Xn y)= P (X1 y)... P (Xn y)= [F (y)]n

gn(y) =dGn(y)

dy= n[F (y)]n1f (y), y R

consider Y1 with c.d.f. G1 and p.d.f. g1.

G1(y) = P (Y1 y) = 1 P (Y1 > y)= 1 P (X1 > y, ..., Xn > y)= 1 P (X1 > y)... P (Xn > y)= 1 [1 F (y)]n

g1(y) =dG1(y)

dy= n[1 F (y)]n1f (y), y R

5

(iii) Direct Transformation of a Multivariate p.d.f.

Theorem: Let X1, X2 have a continuous joint distribu-

tion for which the joint p.d.f. is f (x1, x2) . Assume that

there is a subset S of R2 such that P [(X1, X2) S] = 1.Define two new random variables Y1, Y2 as follows:

Y1 = r1(X1, X2)

Y2 = r2(X1, X2)

where we assume that the functions r1, r2 define a one-

to-one differentiable transformation of S onto a subset T

of R2. Let the inverse of this transformation be given as

follows:

x1 = s1(y1, y2)

x2 = s2(y1, y2)

Then the joint p.d.f. g(y1, y2) of Y1, Y2 is

g(y1, y2) =

{f (s1, s2) |J | (y1, y2) T

0 otherwise

where J is the determinant

J = det

[s1y1

s1y2

s2y1

s2y2

]6

Example: Suppose that two random variables X1 and

X2 have a continuous joint distribution for which the joint

p.d.f. is as follows:

f (x1, x2) =

{4x1x2 0 < x1 < 1, 0 < x2 < 1

0 otherwise

determine the joint p.d.f. of two new random variables

Y1 =X1X2

and Y2 = X1X2.

{Y1 = r1(X1, X2) =

X1X2

Y2 = r2(X1, X2) = X1X2

S = {(x1, x2)|0 < x1 < 1, 0 < x2 < 1}r1, r2 define a one-to-one differentiable transformation of

S onto a subset T of R2. The inverse functions can be

found as X1 = s1(Y1, Y2) =Y1Y2

X2 = s2(Y1, Y2) =

Y2Y1

Y1 > 0, Y2 > 0, 0 0, 0 < y1y2 < 1, 0