EE 511 Solutions to Problem Set 3
1. (a)
F
Y
(
y

X
= 1
/
4) =
P
[
Y
≤
y

X
= 1
/
4] =
P
[
X
+
N
≤
y

X
= 1
/
4] =
P
[1
/
4 +
N
≤
y

X
= 1
/
4] =
P
[
N
≤
y

1
/
4

X
= 1
/
4].
Since
X
and
N
are independent, we have
P
[
N
≤
y

1
/
4

X
= 1
/
4] =
P
[
N
≤
y

1
/
4] and
F
Y
(
y

X
= 1
/
4) =
F
N
(
y

1
/
4)
.
Therefore,
f
Y
(
y

X
= 1
/
4) =
f
N
(
y

1
/
4)
i.e., uniform over (1/4,3/4).
Similarly, we get
f
Y
(
y

X
=

1
/
4) =
f
N
(
y
+ 1
/
4)
i.e., uniform over (3/4,1/4).
Now,
f
Y
(
y
) =
P
[
X
= 1
/
4]
f
Y
(
y

X
= 1
/
4) +
P
[
X
=

1
/
4]
f
Y
(
y

X
=

1
/
4)
1
f
(yX = 1/4)
Y
1/4
3/4
1
f
(yX = 1/4)
Y
f
(y)
Y
1/4
1/4
3/4
1/4
0.4
1
0.6
3/4
3/4
(b) Let
C
denote the event that we make a correct decision. We have
P
[
C
] =
Z
∞
∞
P
[
C

Y
=
y
]
f
Y
(
y
)
dy
Since the integrand in the above equation is always positive, maximizing
P
[
C
] is the same as
maximizing
P
[
C

Y
=
y
] for each
y
. Therefore, the optimal rule is to choose
Decision
= 1
/
4
if
P
[
X
= 1
/
4

Y
=
y
]
> P
[
X
=

1
/
4

Y
=
y
]
Decision
=

1
/
4
if
P
[
X
=

1
/
4

Y
=
y
]
> P
[
X
= 1
/
4

Y
=
y
]
This can also be written as
Decision
= 1
/
4
if
f
Y
[
y

X
= 1
/
4]
P
[
X
= 1
/
4]
f
Y
(
y
)
>
f
Y
[
y

X
=

1
/
4]
P
[
X
=

1
/
4]
f
Y
(
y
)
Decision
=

1
/
4
if
f
Y
[
y

X
=

1
/
4]
P
[
X
=

1
/
4]
f
Y
(
y
)
>
f
Y
[
y

X
= 1
/
4]
P
[
X
= 1
/
4]
f
Y
(
y
)
i.e.,
Decision
= 1
/
4
if
f
Y
[
y

X
= 1
/
4]
P
[
X
= 1
/
4]
> f
Y
[
y

X
=

1
/
4]
P
[
X
=

1
/
4]
Decision
=

1
/
4
if
f
Y
[
y

X
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 TU
 XY sexdetermination system

Click to edit the document details