UCSD ECE153
Handout #37
Prof. YoungHan Kim
Thursday, June 3, 2010
Solutions to Final (Spring 2008)
1.
Coin with random bias (20 points).
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]. Assume
P
does not change during the sequence of tosses.
(a) What is the probability that the Frst three ±ips are heads?
(b) What is the probability that the second ±ip is heads given that the Frst ±ip is heads?
Solution:
(a) We have
P
{
Frst three ±ips are heads

P
=
p
}
=
p
3
.
Hence, by the law of total probability,
P
{
Frst three ±ips are heads
}
=
i
1
0
p
3
dp
= 1
/
4
.
(b) Since
P
{
the Frst ±ip is heads
}
=
i
1
0
pdp
= 1
/
2
and
P
{
Frst two ±ips are heads
}
=
i
1
0
p
2
dp
= 1
/
3
,
we have
P
{
second ±ip heads

Frst ±ip heads
}
= 2
/
3
.
2.
Estimation (20 points).
Let
X
1
and
X
2
be independent identically distributed random vari
ables. Let
Y
=
X
1
+
X
2
.
(a) ²ind
E
[
X
1

X
2

Y
].
(b) ²ind the minimum mean squared error estimate of
X
1
given an observed value of
Y
=
X
1
+
X
2
. (Hint: Consider
E
[
X
1
+
X
2

X
1
+
X
2
].)
Solution:
(a) By symmetry,
E
[
X
1

Y
] =
E
[
X
2

Y
]. Hence,
E
[
X
1

X
2

Y
] = 0.
(b) On one hand, we have
E
[
X
1
+
X
2

X
1
+
X
2
] =
X
1
+
X
2
=
Y
. On the other hand, we have
E
[
X
1
+
X
2

X
1
+
X
2
] =
E
[
X
1

Y
] +
E
[
X
2

Y
] = 2
E
[
X
1

Y
]
.
Hence, the MMSE estimate of
X
1
given
Y
is
E
[
X
1

Y
] =
Y
2
.
1
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Stationary process (20 points).
Consider the Gaussian autoregressive random process
X
k
+1
=
1
3
X
k
+
Z
k
,
k
= 0
,
1
,
2
,...,
where
Z
0
,Z
1
,Z
2
,...
are i.i.d.
∼
N
(0
,
1).
(a) Find the distribution on
X
0
that makes this a stationary stochastic process.
(b) What is the resulting autocorrelation
R
X
(
n
)?
Solution:
(a) Consider
X
0
∼
N
(0
,
9
/
8). It is easy to see that
X
n
∼
N
(0
,
9
/
8), which implies that
X
n
is stationary (why?).
(b) We have
X
n
=
1
3
X
n

1
+
Z
n

1
=
p
1
3
P
2
X
n

2
+
1
3
Z
n

2
+
Z
n

1
=
p
1
3
P
n
X
0
+
n
s
k
=1
p
1
3
P
n

k
Z
k

1
.
Hence,
R
X
(
n
) =
EX
0
X
n
=
p
9
8
Pp
1
3
P
n
for
n
≥
0, and in general
R
X
(
n
) =
EX
0
X
n
=
p
9
8
Pp
1
3
P

n

.
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 Spring '11
 Eggers
 Probability, Probability theory, Stationary process, Sx, Minimum mean square error

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