ECE 3100
Fall 2009
Homework #7 Solutions
1.
We have
∞
∞
f
X
(
x
)
dx
=
∞
∞
λ
2
e

λ

x

dx
=
0
∞
λ
2
e

λ

x

dx
+
∞
0
λ
2
e

λ

x

dx
= 2
·
1
2
∞
0
λ
e

λ
x
dx
= 2
·
1
2
= 1
,
since
∞
0
λ
e

λ
x
dx
= 1
by the normalization property of the exponential PDF.
Since the Laplace PDF is symmetric about
x
= 0
, we have
E
[
X
] = 0
. Also,
E
[
X
2
] =
∞
∞
x
2
λ
2
e

λ

x

dx
=
∞
0
x
2
λ
e

λ
x
dx
=
2
λ
2
,
since
∞
0
x
2
λ
e

λ
x
dx
=
2
λ
2
is the second moment of the exponential PDF. Thus
Var
[
X
] =
E
[
X
2
]

(
E
[
X
])
2
=
2
λ
2

0 =
2
λ
2
.
2.
Denote by
W
the waiting time for Calamity Jane and let
N
denote the number of customers that she finds ahead of her,
N
takes the value
1
with probability
1
2
and the value
0
with probability
1
2
, we are interested in
F
W
(
x
) =
P
[
W
≤
x
]
.
Note first that if
x <
0
, then
F
W
(
x
) = 0
because the random variable
W
is nonnegative (
W
≥
0
). For
x
≥
0
, using
the total law of probability we obtain
F
W
(
x
)
=
P
[
W
≤
x
] =
P
[
W
≤
x

N
= 1]
P
[
N
= 1] +
P
[
W
≤
x

N
= 0]
P
[
N
= 0]
=
1
2
(
P
[
W
≤
x

N
= 1] +
P
[
W
≤
x

N
= 0])
.
When
N
= 0
, we have
W
= 0
and hence
P
[
W
≤
x

N
= 0] = 1
(because
x
≥
0
), when
N
= 1
, we have
W
is
exponentially distributed with parameter
λ
, consequently,
P
[
W
≤
x

N
= 1] = 1

e

λ
x
, hence we obtain that for
x
≥
0
,
F
W
(
x
) =
1
2
(2

e

λ
x
)
.
3. (a)
We calculate the CDF of
X
as follows. For
x
∈
[0
, r
]
,
F
X
(
x
) =
P
(
X
≤
x
) =
π
x
2
π
r
2
=
x
2
r
2
.
For
x <
0
,
F
X
(
x
) = 0
and
for
x > r
,
F
X
(
x
) = 1
. The PDF is obtained by differentiating the CDF. Thus
f
X
(
x
) =
2
x
r
2
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 '05
 HAAS
 Probability theory, yn, dx

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