Normal Random Variable - Random Signals and Processes Primer with MATLAB - page 243

From ( 2.138 ) , we have:

x ¼ 2

2

e ð x m X Þ

y b
a

m X

x ¼ ð y b Þ
a

1
2 p

2 s X

p

e

f X ð x Þ
d y
d x

1
2 p

s X

2 s X

p

f Y ðyÞ¼ ¼ ¼

jj

jjs X

yb
a

2

e ð y ð b þ am X ÞÞ

1
2 p

2 a 2 s X

p

¼ :

jjs X

(4.71)
Comparing the result ( 4.71 ) with the expression for the normal PDF, we can
write:

2

e ð y m Y Þ

1
2 p

2 s Y

f Y ðyÞ¼ p ; 1<y<1; (4.72)

s Y

where

m Y ¼ am X þ b;
s Y ¼ðas X Þ

2

¼ a 2 s X :

(4.73)

Observing ( 4.70 ), we can easily conclude that the parameters ( 4.73 ) can be
obtained from ( 4.70 ), using ( 2.244 ) and ( 2.355 ):

m Y ¼ EfYg¼EfaX þ bg¼aEfXgþb ¼ am X þ b;
s aXþb ¼ a 2 s X : (4.74)

The following rule stands:

The linear transformation of the normal random variable results in a normal
variable with the mean value and variance given in ( 4.73 ).

Example 4.3.1 Find the PDF of the random variable Y ¼ 3 X + 2, where X is the
normal random variable with a mean value and variance equal to 1 and 4,
respectively.

Solution The transformed random variable Y is also a normal random variable
with the parameters obtained from ( 4.73 ):

m Y ¼ 3 m X þ 2 ¼ 3 ð 1 Þþ 2 ¼ 5 ;

s 2

2

s 2

3 Xþ 2 ¼ð 3 Þ X ¼ 9 4 ¼ 36

: (4.75)

From here, the PDF of the random variable Y is as follows:

2

2 p 6 e ð y 5 Þ

1

72

f Y ðyÞ¼ for 1<y<1: (4.76)