Digital Signal Processing Reference
In-Depth Information
Y
X
1
X
2
Fig. 3.15
Transformation of two input random variables
Now we have the transformation:
Y ¼ gðX
1
; X
2
Þ;
r ¼ X
1
;
(3.169)
or
Y ¼ gðX
1
; X
2
Þ:
r ¼ X
2
:
(3.170)
The joint PDF
f
Yr
(
y
,
z
) is obtained from (
3.159
). Finally, the required PDF
f
Y
(
y
)
is obtained from
f
Yr
(
y
,
z
), using (
3.42a
,
3.42b
):
1
f
Y
ðyÞ¼
f
Yr
ðy; zÞ
d
z:
(3.171)
1
Example 3.4.2
Find the PDF of the variable
Y
Y ¼ X
1
X
2
;
(3.172)
if it is known then the joint PDF is
f
X
1
X
2
ðx
1
; x
2
Þ
.
Solution
The auxiliary variable
r ¼ X
1
(3.173)
is introduced, resulting in the following transformation:
y ¼ x
1
x
2
:
z ¼ x
1
:
(3.174)
From here,
x
1
¼ z;
x
2
¼ y=z:
(3.175)
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