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Solutions to Homework 07
1.
a.
In this case we have transmitted a 0 (i.e.
v
=

1).
The receiver makes an error if the
received signal is greater than 0 (i.e if
v
+
N
≥
0). Then
P
[error

v
=

1]
=
P
[
Y
≥
0

v
=

1]
=
P
[
v
+
N
≥
0

v
=

1]
=
P
[

1 +
N
≥
0] =
P
[
N
≥
1]
=
1

Φ(1) =
Q
(1)
=
0
.
159
b.
Similarly, if a 1 is sent
P
[error

v
= 1]
=
P
[
Y
≤
0

v
= 1]
=
P
[
v
+
N
≤
0

v
= 1]
=
P
[1 +
N
≤
0] =
P
[
N
≤ 
1]
=
Φ(

1) = 1

Q
(

1) =
Q
(1)
=
0
.
159
Figure 1 shows the areas of interest.
0
1
1
y
Figure 1: Regions corresponding to error in reception
2.
Let
T
be the random variable denoting the lifetime of the transistor. Since
T
is gaussian,
its pdf is given by
f
T
(
t
) =
1
9
√
2
π
e
1
2
(
t

200)
2
9
2
(1)
a.
The probability that a particular transistor will last for more than 220 hours is
P
[
T >
220]
=
Q
±
220

200
9
²
1
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Q
(2
.
2)
=
1
.
39
×
10

2
b.
Here we want the probability that a particular transistor will last more than 220 hours
given that the transistor lasts for 210 hours. This conditional probability is given by
P
[
T >
220

T >
210]
=
P
[
{
T >
220
} ∩ {
T >
210
}
]
P
[
{
T >
210
}
]
=
P
[
{
T >
220
}
]
P
[
{
T >
210
}
]
=
Q
±
220

200
9
²
Q
±
210

200
9
²
≈
Q
(2
.
2)
Q
(1
.
1)
=
1
.
39
×
10

2
1
.
36
×
10

1
=
0
.
1022
3.
X
is a uniform random variable between 0 and 1. Now
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 Fall '08
 GELFAND

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