Digital Signal Processing Reference
In-Depth Information
Using (
3.270
) in the middle, the covariance is:
C
YZ
¼ ðY YÞðZ ZÞ¼YZ ¼
0
:
(3.271)
From here, it follows that the random variables are uncorrelated.
Exercise 3.10
The discrete random variables
X
1
and
X
2
have the possible values:
1, 0, 1. Find the mean values for the variables
X
1
and
X
2
, if it is known:
PfX
1
¼
1
;X
2
¼
1
g¼
0
;
PfX
1
¼
1
;X
2
¼
0
g¼
1
=
15
;
PfX
1
¼
1
;X
2
¼
1
g¼
4
=
15
;
PfX
1
¼
0
;X
2
¼
1
g¼
2
=
15
;
PfX
1
¼
0
;X
2
¼
0
g¼
2
=
15
;
PfX
1
¼
0
;X
2
¼
1
g¼
1
=
15
;
PfX
1
¼
1
;X
2
¼
1
g¼
0
;
PfX
1
¼
1
;X
2
¼
0
g¼
1
=
15
;
PfX
1
¼
1
;X
2
¼
1
g¼
4
=
15
:
(3.272)
Answer
In order to find the mean values for
X
1
and
X
2
, we need the corresponding
probabilities
P
(
X
1
i
) and
P
(
X
2
j
),
i ¼
1,
,3;
j ¼
1,
,3.
...
...
From (
1.67
), we have:
PðX
1
¼
1
Þ¼PfX
1
¼
1
; X
2
¼
1
gþPfX
1
¼
1
; X
2
¼
0
g
þ PfX
1
¼
1
; X
2
¼
1
g¼
1
=
15
þ
4
=
15
¼
5
=
15
¼
1
=
3
:
(3.273)
PðX
1
¼
0
Þ¼PfX
1
¼
0
; X
2
¼
1
gþPfX
1
¼
0
; X
2
¼
0
gþPfX
1
¼
0
; X
2
¼
1
g
¼
2
=
15
þ
2
=
15
þ
1
=
15
¼
5
=
15
¼
1
=
3
:
(3.274)
PðX
1
¼
1
Þ¼PfX
1
¼
1
;X
2
¼
1
gþPfX
1
¼
1
;X
2
¼
0
þPfX
1
¼
1
;X
2
¼
1
gg
¼
1
=
15
þ
4
=
15
¼
5
=
15
¼
1
=
3
:
(3.275)
PðX
2
¼
1
Þ¼PfX
2
¼
1
; X
1
¼
1
gþPfX
2
¼
1
; X
1
¼
0
g
(3.276)
þPfX
2
¼
1
; X
1
¼
1
g¼
2
=
15
:
PðX
2
¼
0
Þ¼PfX
2
¼
0
; X
1
¼
1
gþPfX
2
¼
0
; X
1
¼
0
gþPfX
2
¼
0
; X
1
¼
1
g
¼
1
=
15
þ
2
=
15
þ
1
=
15
¼
4
=
15
:
(3.277)
PðX
2
¼
1
Þ¼PfX
2
¼
1
; X
1
¼
1
gþPfX
2
¼
1
; X
1
¼
0
g
þ PfX
2
¼
1
; X
1
¼
1
g¼
4
=
15
þ
1
=
15
þ
4
=
15
¼
9
=
15
:
(3.278)
From (
3.273
)to(
3.275
), we get:
EfX
1
g¼
1
PfX
1
¼
1
gþ
0
PfX
1
¼
0
gþð
1
Þ PfX
1
¼
1
g¼
1
=
3
1
=
3
¼
0
:
(3.279)
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