Normal Random Variable - Random Signals and Processes Primer with MATLAB - page 246

a b

TRANSFORMATION OF NORMAL R.V.

Y=ab s (X)

1

3

0.9

2.5

0.8

0.7

2

OUTPUT PDF

0.6

1.5

0.5

INPUT PDF

0.4

1

0.3

0.5 0.2

0.1

0

-3 -2 -1 0 1 2 3 0

−2 −1 0 1 2 3 −3

X

x, y

Fig. 4.9 Absolute value of a normal random variable. ( a ) Transformation. ( b ) Input and output
PDFs

Solution The random variable Y takes only two discrete values: 0.3 and 0.3.
Consequently, the random variable Y is a discrete random variable and its PDF will
have two delta functions:

f Y ðyÞ¼PfY ¼ 0 : 3 gdðy 0 : 3 ÞþPfY ¼ 0 : 3 gdðy þ 0 : 3 Þ: (4.83)

From ( 4.82 ), the random variable Y has the value 0.3 if the random variable X is
positive. Therefore,

PfY ¼ 0 : 3 g¼PfX> 0 g¼ 0 : 5 : (4.84)

Similarly,

PfY ¼ 0 : 3 g¼PfX< 0 g¼ 0 : 5 : (4.85)

From ( 4.83 )to( 4.85 ), we have:

f Y ðyÞ¼ 0 : 5 dðy 0 : 3 Þþ 0 : 5 dðy þ 0 : 3 Þ: (4.86)

The input and output PDFs along with the transformation are shown in Fig. 4.10 .
A normal random variable can also be transformed into a mixed random
variable, as shown in the following example.

Example 4.3.5 The normal random variable has a mean value equal to 0, and a
variance equal to 1. Find the PDF of the random variable Y ,if

8
< 1 for X> 1

Y ¼ X for 1 <X< 1 ; (4.87)

:

1 X< 1

for

as shown in Fig. 4.11a .