Random Variable - Random Signals and Processes Primer with MATLAB - page 64

f X ðxÞ d x ¼ f Y ðyÞ d y: (2.136)

From here, we can write:

f X ðxÞ
d y=

f Y ðyÞ ¼

d x :

(2.137)

The derivation d y /d x can be positive as well as negative. However, the density
function can be only positive. To solve this problem, the absolute value of the
derivative d y /d x must be used in ( 2.137 ), as demonstrated in the following relation:

f X ðxÞ
d y=

f Y ðyÞ ¼

j :

(2.138)

j d x

Note that the left side of ( 2.138 ) is a function of y , while the right side is a
function of x . To overcome this problem, we solve ( 2.132 ) to obtain:

x ¼ g 1

ðyÞ; (2.139)

in the final expression for the desired density function. This gives us:

x¼g 1 ðyÞ :

f X ðxÞ
d y=

f Y ðyÞ ¼ (2.140)

j d x j

The desired density f Y ( y ) is equal to zero for values of y where for y there is no
solution ( 2.132 ).

Example 2.6.1 Show that the linear transformation of the uniform random variable
results in a uniform random variable.

Solution Let the input variable X be in the interval [ x 1 , x 2 ] with the corresponding
density, as shown in Fig. 2.31 .

8
<

1
x 2 x 1

x 1 xx 2 ;

for

f X ðxÞ ¼ (2.141)

:

0 otherwise

A linear transformation is given as:

y ¼ ax þ b; (2.142)

where a and b are constants.
From ( 2.142 ), we obtain:

x ¼ ðybÞ=a (2.143)