Transformations - Stony Brook

Transformations

Dear students,

Since we have covered the mgf technique extensively already, here we only review the cdf and the pdf techniques, first for univariate (one-to-one and more-to-one) and then for bivariate (one-to-one and more-to-one) transformations.

1. The cumulative distribution function (cdf) technique

Suppose Y is a continuous random variable with cumulative distribution function (cdf) () ( ). Let = () be a function of Y, and our goal is to find the distribution of U. The cdf technique is especially convenient when the cdf () has closed form analytical expression. This method can be used for both univariate and bivariate transformations.

Steps of the cdf technique:

1. Identify the domain of Y and U. 2. Write() = ( ), the cdf of U, in terms of (),

the cdf of Y . 3. Differentiate () to obtain the pdf of U, ().

Example 1. Suppose that ~ (0,1). Find the distribution of = () = - ln . Solution. The cdf of ~ (0,1) is given by

0, 0 () = {, 0 < 1

1, 1

The domain (*domain is the region where the pdf is non-zero) for ~ (0,1) is = {: 0 < < 1}; thus, because = - ln > 0, it follows that the domain for U is = {: > 0}. The cdf of U is:

() = ( ) = (- ln )

= (ln > -)

= ( > -) = 1 - ( -) = 1 - (-)

1

Because () = for 0 < y < 1; i.e., for u > 0, we have () = 1 - (-) = 1 - -

Taking derivatives, we get, for u > 0,

()

=

()

=

(1

-

-)

=

-

Summarizing,

() = {0-, ,

> 0 otherwise

This is an exponential pdf with mean 1 = 1; that is, U ~

exponential( = 1).

Example 2. Suppose that ~ (- /2 , /2) . Find the distribution of the random variable defined by U = g(Y ) = tan(Y ). Solution. The cdf of ~ (- /2 , /2) is given by

0,

() =

+ /2 ,

{

1,

- /2 - /2 < /2

/2

The domain for Y is = {: - /2 < < /2}. Sketching a graph of the tangent function from - /2 to /2, we see that

- < < .

Thus, = { : - < < } , the set of all reals. The cdf of U is:

() = ( ) = [tan() ]

= [ tan-1()] = [tan-1()]

Because

()

=

+/2

for

-

/2

<

<

/2

;

i.

e.

,

for

,

we have

()

=

[tan-1()]

=

tan-1()

+

/2

The pdf of U, for , is given by

tan-1() + /2

1

() = () = [

] = (1 + 2).

Summarizing,

2

1 () = {(1 + 2) , - < <

0,

otherwise.

A random variable with this pdf is said to have a (standard) Cauchy distribution. One interesting fact about a Cauchy random variable is that none of its moments are finite. Thus, if U has a Cauchy distribution, E(U), and all higher order moments, do not exist. Exercise: If U is standard Cauchy, show that(||) = +.

2. The probability density function (pdf) technique, univariate

Suppose that Y is a continuous random variable with cdf () and domain , and let = (), where : is a continuous, one-to-one function defined over . Examples of such functions include continuous (strictly)

increasing/decreasing functions. Recall from calculus that if is one-to-one, it has an unique inverse -1. Also recall that if is increasing (decreasing), then so is -1.

Derivation of the pdf technique formula using the cdf

method:

Suppose that () is a strictly increasing function of y defined over . Then, it follows that = () -1() = and

() = ( ) = [() ] = [ -1()] = [-1()]

Differentiating () with respect to u, we get

()

=

()

=

[-1()]

=

[-1()]

-1()

(by chain rule)

Now as is increasing, so is -1; thus, -1() > 0. If () is

strictly decreasing, then

()

=

1

-

[-1()]

and

-1()

<

0,

which

gives

3

 ()

=

()

=

{1

-

[-1()]}

=

-

[-1

()]

-1()

Combining both cases, we have shown that the pdf of U, where nonzero, is given by

()

=

[-1()]

|

-1()

|.

It is again important to keep track of the domain for U. If denotes the domain of Y, then , the domain for U, is given by = {: = (); }.

Steps of the pdf technique:

1. Verify that the transformation u = g(y) is continuous and

one-to-one over . 2. Find the domains of Y and U. 3. Find the inverse transformation = -1() and its

derivative (with respect to u). 4. Use the formula above for ().

Example 3. Suppose that Y ~ exponential(); i.e., the pdf of Y is

()

=

{

1

-/ ,

> 0

0,

otherwise.

Let = () = , . Use the method of transformations to find the pdf of U.

Solution. First, we note that the transformation () = is a continuous strictly increasing function of y over = {: > 0}, and, thus, () is one-to-one. Next, we need to find the

domain of U. This is easy since y > 0 implies = > 0 as well. Thus, = {: > 0}. Now, we find the inverse transformation:

() = = = -1() = 2 (by inverse transformation)

and its derivative:

4

-1()

=

(2)

=

2.

Thus, for u > 0,

()

=

[-1()]

|

-1()

|

=

1

-2

?

|2|

=

2

-2 .

Summarizing,

()

=

2 {

-2 ,

> 0

0,

otherwise.

This is a Weibull distribution. The Weibull family of distributions is common in life science (survival analysis), engineering and actuarial science applications.

Example 4. Suppose that Y ~ beta( = 6; = 2); i.e., the pdf of Y is given by

() = {420, 5(1 - ),

0 < < 1 otherwise.

What is the distribution of U = g(Y ) = 1 -Y ? Solution. First, we note that the transformation g(y) = 1 -Y is a continuous decreasing function of y over = {: 0 < < 1}, and, thus, g(y) is one-to-one. Next, we need to find the domain of U. This is easy since 0 < y < 1 clearly implies 0 < u < 1. Thus, = {: 0 < < 1}. Now, we find the inverse transformation:

() = = 1 - = -1()

= 1 - (by inverse transformation)

and its derivative:

-1()

=

(1

-

)

=

-1.

Thus, for 0 < u < 1,

()

=

[-1()]

|

-1()

|

5

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