EE 278
October 16, 2009
Statistical Signal Processing
Handout #7
Homework #3 Solutions
1.
Family planning.
Alice and Bob choose a number
X
at random with equal probability from
the set
{
2
,
3
,
4
}
. If the outcome is
X
=
x
, they decide to have children until they have a girl
or
x
children, whichever comes first.
Assume that each child is a girl with probability 1
/
2
(independent of the number of children and gender of other children). Let
Y
be the number of
children they will have.
a. Find the conditional pmf
p
Y

X
(
y

x
) for all possible values of
x
and
y
.
b. Find the pmf of
Y
.
Solution
(10 points)
a. Note that
Y
∈ {
1
,
2
,
3
,
4
}
. The conditional pmf is as follows
p
Y

X
(1

2) =
1
2
, p
Y

X
(2

2) =
1
2
,
p
Y

X
(1

3) =
1
2
, p
Y

X
(2

3) =
1
4
, p
Y

X
(3

3) =
1
4
,
p
Y

X
(1

4) =
1
2
, p
Y

X
(2

4) =
1
4
, p
Y

X
(3

4) =
1
8
, p
Y

X
(4

4) =
1
8
.
b. The pmf of
Y
is:
p
Y
(1) =
4
summationdisplay
x
=2
p
X
(
x
)
p
Y

X
(1

x
) = 1
/
2
, p
Y
(2) =
4
summationdisplay
x
=2
p
X
(
x
)
p
Y

X
(2

x
) = 1
/
3
,
p
Y
(3) =
4
summationdisplay
x
=3
p
X
(
x
)
p
Y

X
(3

x
) = 1
/
8
, p
Y
(4) =
p
X
(4)
p
Y

X
(4

4) = 1
/
24
.
2.
First available teller.
A bank has two tellers. The service times for tellers 1 and 2 are indepen
dent exponential random variables
X
1
∼
Exp(
λ
1
) and
X
2
∼
Exp(
λ
2
). You arrive at the bank
and find that both tellers are busy but nobody else is waiting to be served. You are served by
the first available teller who becomes free. What is the probability that you are served by the
teller 1?
Solution
(5 points)
The tellers’ service times are exponentially distributed, hence memoryless.
Thus the service
time distribution does not depend on my arrival time. The probability that I will be served by
the first teller is
P
{
X
1
< X
2
}
=
integraldisplay
∞
0
integraldisplay
∞
x
1
λ
1
e

λ
1
x
1
λ
2
e

λ
2
x
2
dx
2
dx
1
=
integraldisplay
∞
0
λ
1
e

(
λ
1
+
λ
2
)
x
1
dx
1
=
λ
1
λ
1
+
λ
2
.
In other words, the probability of being served first by teller
i
is proportional to the teller’s
service rate
λ
i
.
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3.
Optical communication channel.
The input signal to an optical channel is
X
=
braceleftBigg
1
with probability
1
2
10
with probability
1
2
.
The output of the channel is
Y
, and the conditional pmf of
Y
given
X
=
a
is Poisson with
intensity
a
; i.e.,
Y
 {
X
= 1
} ∼
Poisson(1) and
Y
 {
X
= 10
} ∼
Poisson(10). Show that the
MAP rule reduces to
D
(
y
) =
braceleftBigg
1
if
y < y
*
10
otherwise.
Find
y
*
and the corresponding probability of error.
Solution
(10 points)
The probability of decoding error is minimized by the MAP rule:
g
(
y
) =
braceleftBigg
1
p
X

Y
(1

y
)
> p
X

Y
(10

y
)
10
otherwise
Since the
a priori
probabilities for the two
X
values are equal, by Bayes rule the decision rule
is equivalent to the maximumlikelihood decision rule:
g
(
y
) =
braceleftBigg
1
p
Y

X
(
y

1)
> p
Y

X
(
y

10)
10
otherwise
For the optical communication channel of this problem
p
Y

X
(
y

1)
p
Y

X
(
y

10)
=
e

1
/y
!
e

10
10
y
/y
!
=
e
9

y
ln 10
>
1
⇔
9

y
ln 10
>
0
⇔
y <
9
ln10
.
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 Fall '09
 BalajiPrabhakar
 Conditional Probability, Probability, Signal Processing, Probability theory, CDF

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