UCSD ECE 153
Handout #10
Prof. YoungHan Kim
Thursday, October 23, 2008
Solutions to Homework Set #3
(Prepared by TA Halyun Jeong)
1. Read Sections 6.5.1, 8.6.1–8.6.2 in the text. Try to work on all examples.
2.
Coin with random bias.
You are given a coin but are not told what its bias (probability
of heads) is. You are told instead that the bias is the outcome of a random variable
P
∼
Unif[0
,
1]. To get more information about the coin bias, you ﬂip it independently
10 times. Let
X
be the number of heads you get. Thus
X
∼
B(10
, P
). Assuming that
X
= 9, ﬁnd and sketch the
a posteriori
probability of
P
, i.e.,
f
P

X
(
p

9).
Solution:
In order to ﬁnd the conditional pdf of P, apply Bayes’ rule for mixed random
variables to get
f
P

X
(
p

x
) =
p
X

P
(
x

p
)
R
1
0
p
X

P
(
x

p
)
f
P
(
p
)
dp
f
P
(
p
)
.
Now it is given that
X
= 9, thus for 0
≤
p
≤
1
f
P

X
(
p

9) =
p
9
(1

p
)
R
1
0
p
9
(1

p
)
dp
=
p
9
(1

p
)
1
110
= 110
p
9
(1

p
)
.
Figure 1 compares the unconditional and the conditional pdfs for
P
. It may be seen
that given the information that 10 independent tosses resulted in 9 heads, the pdf is
shifted towards the value
9
10
.
3.
Signal or no signal (from Spring 2008 midterm).
Consider a communication system
that is operated only from time to time. When the communication system is in the
“normal” mode (denoted by
M
= 1), it transmits a random signal
S
=
X
with
X
=
(
+1
,
with probability 1
/
2
,

1
,
with probability 1
/
2
.
1
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4.5
f
P
(p)
f
PX
(p9)
Figure 1: Comparison of
a priori
and
a posteriori
pdfs of P
When the system is in the “idle” mode (denoted by
M
= 0), it does not transmit any
signal (
S
= 0). Both normal and idle modes occur with equal probability. Thus
S
=
(
X,
with probability 1
/
2
,
0
,
with probability 1
/
2
.
The receiver observes
Y
=
S
+
Z,
where the ambient noise
Z
∼
Unif[

1
,
1] is indepen
dent of
S
.
(a) Find and sketch the conditional pdf
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 Winter '09
 Eggers
 Probability theory, joint PDF, fY M, TA Halyun Jeong

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