Random Variable - Random Signals and Processes Primer with MATLAB - page 113

ð

1

e joa
ob

e job

e job
2 j

e joa
ob

e joðbyþaÞ 1 f X ðoÞ¼

2 d y ¼

¼ sin ðboÞ: (2.402)

1

Now we will demonstrate how the characteristic function is also useful to
express the density function of the transformed random variable Y ¼ g ( X )in
terms of the PDF of the variable X.
From

y ¼ gðxÞ; (2.403)

we get:

x ¼ g 1 ðyÞ:

(2.404)

Considering ( 2.404 ), we can write:

f X ðxÞj x¼g 1 ðyÞ ¼ f X ðyÞ;
d x ¼ g 1 ðyÞ d y; (2.405)

Placing ( 2.405 ) into ( 2.386 ), we arrive at:

1 1

f Y ðoÞ¼Ef e joY e jogðxÞ f X ðyÞg 1 ðyÞ d y ¼ e joy hðyÞ d y:

g¼ (2.406)

1 1

Comparing the two integrals in ( 2.406 ), we can express density function as:

f Y ðyÞ¼hðyÞ¼f x ðyÞg 1 ðyÞ: (2.407)

The difficulty arises when the transform is not unique, because it is not possible
to simply express the transformed PDF in the form ( 2.407 ).

Example 2.8.10 Find the density function from Fig. 2.52 , using ( 2.401 ).

Solution

ð ð

1 aþb

1
b

y a
b

f Y ðoÞ¼Ef e joY e jogðxÞ f X ðxÞ d x ¼ e joy f X g¼ d y

1 ab

ð ð

aþb aþb

e joy 1
2

1
b

e joy hðyÞ d y:

¼ d y ¼ (2.408)

ab ab