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THE UNIVERSITY OF HONG KONG
DEPARTMENT OF MATHEMATICS
MATH2603 Probability Theory
Solution to Assignment 2 (Updated Version)
Problem 1
(1) By Hint
(2)
E
[
x
] =
∑
∞
n
=1
nP
(
x
=
n
)
=
∑
∞
n
=1
∑
n
i
=1
P
(
x
=
n
)
=
∑
∞
i
=1
∑
∞
n
=
i
P
(
x
=
n
)
=
∑
∞
i
=1
P
(
X
≥
i
)
=
∑
∞
i
=0
P
(
X > i
)
For the continuous case, we have
If
E
[
X
]
<
∞
,
X
nonnegative, then
E
[
X
] =
Z
∞
0
P
(
X > x
)
dx
Proof:
Let
G
(
x
) =
P
(
X > x
) = 1

P
(
X
≤
x
) = 1

F
(
x
)
E
[
X
] =
R
∞
0
xf
(
x
)
dx
=

R
∞
0
xdG
(
x
)
=

(
xG
(
x
)

∞
0

R
∞
0
G
(
x
)
dx
)
=

xG
(
x
)

∞
0
+
R
∞
0
G
(
x
)
dx
Claim:
lim
x
→∞
xG
(
x
) = 0
Proof:
xG
(
x
)
≤
Z
∞
x
tf
(
t
)
dt
lim
x
→∞
xG
(
x
)
≤
lim
x
→∞
Z
∞
x
tf
(
t
)
dt
= 0
since
E
(
X
)
<
∞
Problem 45
Let
N
denote the number of collisions, and let
X
denote the
number of nonempty boxes, then
N
=
r

X
Then let
I
i
=
±
1
if box
i
is nonempty
0
otherwise
X
=
I
1
+
I
2
+
...
+
I
k
E
[
X
] =
E
[
I
1
] +
E
[
I
2
] +
...
+
E
[
I
k
]
=
∑
k
i
=1
P
(box
i
is nonempty)
=
∑
k
i
=1
(1

P
(box
i
is empty))
=
∑
k
i
=1
(1

(1

p
i
)
r
)
1
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View Full Document E
(
N
) =
r

E
[
X
] =
r

k
+
k
X
i
(1

p
i
)
r
Alternatives:
Let
N
=
N
1
+
N
2
+
...
+
N
k
, where
N
i
is the number
of collisions in box
i
.
More discussion:
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This note was uploaded on 01/02/2012 for the course MATH 2603 taught by Professor Han during the Spring '10 term at HKU.
 Spring '10
 Han
 Probability

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