X 2 - University of Michigan

Math/Stat 425, Solutions to Quiz #6

Problem # 1. Let X be a continuous random variable with the probability density

function

x

2

f (x) =

0

if 0 < x < 2 otherwise

Let Y = X2. Find the cumulative distribution function of Y . (That is, give FY (t), for t 0.)

Solution: Recall that by definition the cumulative distribution function of Y is

t

FY (t) = P [Y t] = fY (x)dx.

where fY (x) is the density function of Y , which however we don't know. We do know that Y = X2 takes values between 0 and 4, because X takes values between 0 and 2,

so the cumulative distribution function FY (t) will move from 0 to 1 over the interval 0 t 4.

We will derive the cumulative distribution function FY (t) of Y from the cumulative distribution function FX(t) of X. To see why this is possible, we have

FY (t) = P [Y t] = P [X2 t] = P [- t X t].

Since X only takes non-negative values, we deduce

FY (t) = P [- t X t] = P [- < X t] = FX( t).

We now find the cumulative distribution function FX(t) of X, which we can do directly since we know its density fX(x) = f (x). Recall that by definition

t

FX(t) = P [X t] = f (x)dx.

Certainly FX(t) = 0 for t 0 and FX(t) = 1 for t 2, since the density function vanishes outside [0, 1]. For 0 t 2, we obtain

FX(t) =

t

f (x)dx =

-

t 0

1 xdx

2

=

1 4

x2

|xx==0t

=

1 t2. 4

We conclude that , for 0 t 4,

FY (t) := P [Y

t] = FX[ t] =

1

( t)2

4

=

1 t.

4

1

This holds for 0 t 4.

Answer:

0

if t < 0

FY (t) =

1 4

t

if

0t4

1

if t > 4.

Remarks. (1) The probability density function fY (y) of Y can be obtained by differentiation from the cumulative distribution function FY (t). It is

d

1

fY (t) = dt FY (t) = 4 for 0 y 4,

which is the uniform distribution on [0, 4]. (2) Solution Procedure. First find the cumulative distribution function FX(t) of the

original random variable X. Then use this to find the cumulative distribution function FY (t) of the new random variable Y = g(X) that the problem asks for.

2

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