IE 300
/
GE 331 Spring 2009
Homework #4 Solutions
Question 1
Associate a success with a paper that receives a grade that has not been received before. Let
X
i
be
the number of papers between the
i
th success and the (
i
+
1)st success. Then we have
X
=
1
+
∑
5
i
=
1
X
i
and hence:
E
[
X
]
=
1
+
5
X
i
=
1
E
[
X
i
]
.
After receiving
i

1 di
ﬀ
erent grades so far (
i

1 successes), each subsequent paper has probability
(6
i
)
/
6 of receiving a grade that has not been received before. Therefore, the random variable
X
i
is
geometric with parameter
p
i
=
(6

i
)
/
6, so
E
[
X
i
]
=
6
/
(6

i
). It follows that
E
[
X
]
=
1
+
5
X
i
=
1
6
6

i
=
1
+
6
5
X
i
=
1
1
i
=
14
.
7
.
Question 2
Let
X
be the waiting time and
Y
be the number of customers found. For
x
<
0, we have
F
X
(
x
)
=
0,
while for
x
≥
0,
F
X
(
x
)
=
P
(
X
≤
x
)
=
1
2
P
(
X
≤
x

Y
=
0)
+
1
2
P
(
X
≤
x

Y
=
1)
.
Since,
P
(
X
≤
x

Y
=
0)
=
1
,
P
(
X
≤
x

Y
=
1)
=
1

e

λ
x
,
we obtain
F
X
(
x
)
=
1
2
(2

e

λ
x
)
,
if
x
≥
0
,
0
,
otherwise.
CDF has a discontinuity at
x
=
0.
X
is neither discrete nor continuous.
1
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View Full DocumentQuestion 3
(a)
We ﬁrst calculate the CDF of
X
. For
x
∈
[0
,
r
], we have
F
X
(
x
)
=
P
(
X
≤
x
)
=
π
x
2
π
r
2
=
±
x
r
²
2
.
For
x
<
0, we have
F
X
(
x
)
=
0, and for
x
>
r
, we have
F
X
(
x
)
=
1. By di
ﬀ
erentiating, we obtain the
PDF
f
X
(
x
)
=
2
x
r
2
,
if 0
≤
x
≤
r
,
0
,
otherwise.
We have
E
[
X
]
=
Z
r
0
2
x
2
r
2
dx
=
2
r
3
.
Also,
E
[
X
2
]
=
Z
r
0
2
x
3
r
2
dx
=
r
2
2
.
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 Spring '09
 NegarKayavash
 Probability theory, CDF, r2 r2 r2

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