Digital Signal Processing Reference
In-Depth Information
Example 3.2.2
Find the joint distribution in Example 3.1.2.
Solution
All outcomes have the same probability of occurrence, resulting in:
PfX
1
¼
1
; X
2
¼
1
g¼PfX
1
¼
0
; X
2
¼
0
g¼PfX
1
¼
1
; X
2
¼
0
g
¼ PfX
1
¼
0
; X
2
¼
1
g¼
1
=
4
:
(3.19)
From (
3.19
), we have:
F
X
1
X
2
ðx
1
; x
2
Þ¼þ
1
4
½uðx
1
1
Þuðx
2
1
Þþuðx
1
1
Þuðx
2
Þþuðx
1
Þuðx
2
1
Þ
þ uðx
1
Þuðx
2
Þ:
=
(3.20)
Example 3.2.3
Find the joint distribution in Example 3.1.3.
Solution
All outcomes have the same probability of occurrence. However, the
probabilities of the values of the random variables are:
Pfx
1
¼
1
; x
2
¼
1
g¼Pf
H
;
T
gþPf
T
;
H
g¼
1
=
4
þ
1
=
4
¼
1
=
;
2
Pfx
1
¼
1
; x
2
¼
0
g¼Pf
H
;
H
g¼
1
=
4
;
Pfx
1
¼
0
; x
2
¼
1
g¼Pf
T
;
T
g¼
1
=
4
:
(3.21)
The joint distribution is:
F
X
1
X
2
ðx
1
; x
2
Þ¼
1
=
2
uðx
1
1
Þuðx
2
1
Þþ
1
=
4
uðx
1
1
Þuðx
2
Þþ
1
=
4
uðx
1
Þuðx
2
1
Þ:
(3.22)
Example 3.2.4
Find the probability that a random point (
x
1
,
x
2
) will fall in the area
A
, as shown in Fig.
3.6
.
Solution
Area
A
is defined as:
fx
1
a; x
2
cgfx
1
a; x
2
bg:
(3.23)
Fig. 3.6
Area
A
in Example 3.2.4
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