ORIE 360/560 Practice Final Solutions
Fall, 2007
Problem 1.
(a)
EX
=
E
[
E
[
X

Y
]] =
1
2
E
[
X

Y
= 1] +
1
2
E
[
X

Y
= 2] =
3
4
.
Var
X
=
E
[
X
2
]

E
2
[
X
] =
E
[
E
[
X
2

Y
]]

9
16
=
1
2
E
[
X

Y
= 1] +
1
2
E
[
X

Y
= 2]

9
16
.
Now, if
X
∼
U
(0
, a
), then
EX
2
=
Z
a
0
1
a
x
2
dx
=
a
2
3
,
so
Var
X
=
1
2
1
3
+
4
3

9
16
=
5
6

9
16
=
40
48

27
48
=
13
48
.
(b)
P
[
X >
1] =
P
[
Y
= 2
, X >
1] =
P
[
Y
= 2]
P
[
X >
1

Y
= 2] =
1
2
·
1
2
=
1
4
.
Problem 2.
(a) For a given suit, there are
(
13
k
)(
39
13

k
)
ways to have
k
cards of that suit (and hence 13

k
cards of the other three). Therefore, the probability of getting a hand with one suit of at least nine cards is
4
13
X
k
=9
(
13
k
)(
39
13

k
)
(
52
13
)
.
(b) For a given pair of suits, there are
(
26
13
)

2 ways to choose 13 cards from those two suits (not including
the two hands consisting of only cards from one suit or only cards from the other). There are
(
4
2
)
ways to
choose two suits. Hence, the answer is
4
2
(
26
13
)

2
(
52
13
)
.
Problem 3.
(a) We have
EZ
=
E
[4
X

3
Y
+ 2] = 4
·
5

3
·
(

2) + 2 = 28
.
(b) We have
Cov(
X, Z
) = Cov(
X,
4
X

3
Y
+ 2)
= Cov(
X,
4
X
) + Cov(
X,

3
Y
) + Cov(
X,
2)
= 4 Var
X
+ 0 + 0
= 4
.
(c) Recall that when you condition on
Y
,
Y
behaves as a constant. Hence,
Var[
Z

Y
] = Var[4
X

3
Y
+ 2

Y
]
= Var[4
X

Y
]
= Var 4
X
(by independence of
X
and
Y
)
= 16
.
Problem 4.
(a) We have
F
X
(
x
) = 0 if
x
≤
0,
F
X
(
x
)
≥
1 if
x >
1, and for 0
< x <
1,
F
X
(
x
) =
Z
x
0
θt
θ

1
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 Fall '07
 EHRLICHMAN
 Derivative, Likelihood function, Playing card, Likelihoodratio test, Suit, PRACTICE FINAL SOLUTIONS

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