Digital Signal Processing Reference
In-Depth Information
Find the PDF and distribution of the random variable
Y
Y ¼
min
fX;
4
g:
(2.523)
Answer
According to (
2.523
), the variable
Y
is:
Xx<
;
4
Y ¼
(2.524)
4
x
4
:
From (
2.524
), it follows that for values of
X
equal to 1, 2, and 3,
Y ¼ X
; thus,
Y
takes the same values 1, 2, and 3 as does the variable
X
, with the corresponding
probabilities
PfY ¼
1
g¼PfX ¼
1
g¼
0
:
1
;
PfY ¼
2
g¼PfX ¼
2
g¼
0
:
3
;
PfY ¼
3
g¼PfX ¼
3
g¼
0
:
3
:
(2.525)
Similarly, from (
2.524
), for the values of
X
equal to 6 and 8, the random variable
Y
will have only one value equal to 4, with the following probability:
PfY ¼
4
g¼PfX ¼
6
gþPfX ¼
8
g:
(2.526)
Finally, from (
2.525
) and (
2.526
), we get the distribution:
F
Y
ðyÞ¼
0
:
1
uðy
1
Þþ
0
:
3
uðy
2
Þþ
0
:
3
uðy
3
Þþ
0
:
3
uðy
4
Þ
(2.527)
and the PDF
f
Y
ðyÞ¼
0
:
1
dðy
1
Þþ
0
:
3
dðy
2
Þþ
0
:
3
dðy
3
Þþ
0
:
3
dðy
4
Þ;
(2.528)
where
u
and
d
are the step and delta functions, respectively.
Exercise E.2.19
A message consisting of two pulses is transmitted over a channel.
Due to noise on the channel, each or both of the pulses can be destroyed indepen-
dently with the probability
p
. The random variable
X
represents the number of the
destroyed pulses. Find the characteristic function of the variable
X
.
Answer
The variable
X
is a discrete variable which takes values 0, 1, and 2 with the
probabilities,
2
PfX ¼
0
g¼ð
1
pÞ
;
PfX ¼
1
g¼pð
1
pÞþð
1
pÞp ¼
2
pð
1
pÞ;
PfX ¼
2
g¼p
2
:
(2.529)
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