ISYE 6414  Spring 2009
Solution Lecture 1  Prerequisites
Problem 1.
Q
n
i
=1
c
·
e
x
i
=
c
n
e
(
∑
n
i
=1
x
i
)
Problem 2.
•
The most likely sum of two rolled sixsided dice is 7
•
The most likely sum of three rolled sevensided dice is 12
Problem 3.
Cov
(
aX
+
b, Y
) =
E
[(
aX
+
b

E
(
aX
+
b
)) (
Y

E
(
Y
))] =
=
E
[(
aX
+
b

aE
(
X
)

b
) (
Y

E
(
Y
))] =
E
[(
a
(
X

E
(
X
))) (
Y

E
(
Y
))] =
=
E
[(
X

E
(
X
)) (
Y

E
(
Y
))] =
aCov
(
aX
+
b, Y
)
Problem 4.
Cov
n
X
i
=1
X
i
,
n
X
j
=1
Y
j
=
E
n
X
i
=1
X
i

E
n
X
i
=1
X
i
!!
n
X
j
=1
Y
j

E
n
X
j
=1
Y
j
=
=
E
n
X
i
=1
X
i

n
X
i
=1
E
(
X
i
)
!
n
X
j
=1
Y
j

n
X
j
=1
E
(
Y
j
)
=
=
E
n
X
i
=1
(
X
i

E
(
X
i
))
n
X
j
=1
(
Y
j

E
(
Y
j
))
=
=
n
X
i
=1
E
(
X
i

E
(
X
i
))
n
X
j
=1
(
Y
j

E
(
Y
j
))
=
=
n
X
i
=1
n
X
j
=1
E
[(
X
i

E
(
X
i
)) (
Y
j

E
(
Y
j
))]
Problem 5.
By definition we know that
λ
(
x
)
≥
0;
x
∈ <
.
Also:
R
∞
∞
λ
(
x
)
dx
=
R
∞
x
0
λ
(
x
)
dx
=
R
∞
x
0
f
(
x
)
dx
1

F
(
x
0
)
=
1

F
(
x
0
)
1

F
(
x
0
)
= 1.
Now suppose
X
∼
f
(
x
) =
e

x
·
1
(0
,
∞
)
. Then:
P
(
X < x

X
≥
x
0) =
P
(
x
0
< X
∩
X
≤
x
)
P
(
X
≥
x
0
)
=
P
(
x
0
≤
X
≤
x
)
P
(
X
≥
x
0
)
=
e

x
0

e

x
e

x
0
= 1

e

(
x

x
0
)
⇒
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 Fall '08
 Staff
 Maximum likelihood, Estimation theory, Bias of an estimator, i=1, yj

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