Digital Signal Processing Reference
In-Depth Information
Exercise 4.18
The normal random variables
X
1
and
X
2
have the mean values
EfX
1
g¼
4
;
EfX
2
g¼
2
;
(4.221)
and the covariance matrix
:
43
=
4
C ¼
(4.222)
=
3
49
Find the joint density function.
Answer
From (
4.222
), we have:
s
1
¼
4
s
2
¼
9
;
;
C
1
;
2
¼ C
2
;
1
¼
3
=
4
:
(4.223)
The correlation coefficient is:
C
1
;
2
s
1
s
2
¼
4
2
3
¼
1
3
=
r ¼
=
8
¼
0
:
125
:
(4.224)
From (
4.124
), the joint density is:
"
#
2
2
5079
ð
x
1
4
Þ
þ
ð
x
2
2
Þ
25
ð
x
1
4
Þð
x
2
2
Þ
6
e
0
:
0
:
1
4
9
f
X
1
;X
2
ðx
1
; x
2
Þ¼
:
:
906
p
11
(4.225)
Exercise 4.19
Consider a Gaussian random variable with
m ¼
0 and
s
2
¼
1,
given in Fig.
4.26
.
x
2
2
1
2
p
e
f
X
ðxÞ¼
p
:
(4.226)
The condition event
A
is defined such that the random variable
X
is positive,
A ¼fX>
0
g:
(4.227)
Find
E
{
X
|
A
}.
Fig. 4.26
Gaussian probability density in Example 4.19
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