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Econ 371
Problem Set #1
Answer Sheet
2.1 In this question, you are asked to consider the random variable
Y
, which denotes the number of “heads” that
occur when two coins are tossed.
a. The ﬁrst part of the question asks you to derive the probability distribution for
Y
. For a discrete
random variable, this is just a listing of all the possible outcomes and the probability that they can
occur. Assuming that the two coins are fair, there are only four possible results when we ﬂip the two
coins:
•
Heads,Heads (
Y
= 2)
•
Heads,Tails (
Y
= 1)
•
Tails,Heads (
Y
= 1)
•
Tails,Tails (
Y
= 0)
which happen with equal probability. The corresponding probability distribution is given by the following
table.
outcome (number of heads)
Y
= 0
Y
= 1
Y
= 2
probability
0.25
0.50
0.25
b. The cumulative probability distribution function for
Y
corresponds to just computing the probabilities
of
Y
being less than or equal to a given value
outcome (number of heads)
Y <
0
0
≤
Y <
1
1
≤
Y <
2
Y >
2
cumulative probability
0
0.25
0.75
1
c. Finally, you are asked to derive the mean and variance of
Y
. These are given by:
E
(
Y
) =
k
X
i
=1
Pr
(
Y
=
y
i
)
y
i
=
k
X
i
=1
p
i
y
i
= (0
.
25)
·
0 + (0
.
50)
·
1 + (0
.
25)
·
2
= 1
and
V ar
(
Y
) =
E
[(
Y

μ
Y
)
2
] =
k
X
i
=1
p
i
(
y
i

μ
Y
)
2
= (0
.
25)
·
(0

1)
2
+ (0
.
50)
·
(1

1)
2
+ (0
.
25)
·
(2

1)
2
= (0
.
25)
·
1 + (0
.
50)
·
0 + (0
.
25)
·
1
= 0
.
5
2.3 In this question, you are asked to compute various characteristics of two random variables,
W
and
V
that
are functions of the random variables in Table 2.2 (
X
and
Y
). Speciﬁcally, we have:
W
= 3 + 6
X
V
= 20

7
Y
a. First, you are asked to compute
E
(
W
) and
E
(
V
). From equation (2.12) in the text, we know that:
E
(
W
) = 3 + 6
E
(
X
)
(1)
1
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View Full Document but, using Table 2.2 and our formula for computing means, we have that
E
(
X
) =
k
X
X
i
=1
p
i
x
i
= (0
.
30)
·
0 + (0
.
70)
·
1
= 0
.
7
Substituting this into (1) yields.
E
(
W
) = 3 + 6
E
(
X
) = 3 + 6(0
.
7) = 7
.
2
(2)
Similarly,
E
(
Y
) =
k
Y
X
i
=1
p
i
x
i
= (0
.
22)
·
0 + (0
.
78)
·
1
= 0
.
78
,
so that
E
(
V
) = 20

7
E
(
Y
) = 20

7(0
.
78) = 14
.
54
.
(3)
b. In part b you are asked to compute the corresponding variances for
W
and
V
. Equation (2.13) in the
text tells us that:
σ
2
W
= 6
2
·
σ
2
X
= 36
σ
2
X
σ
2
V
= (

7)
2
·
σ
2
Y
= 49
σ
2
Y
Turning to our formulas for variances, however, we know that:
V ar
(
X
) =
k
X
X
i
=1
p
i
(
x
i

μ
X
)
2
= (0
.
30)
·
(0

0
.
7)
2
+ (0
.
70)
·
(1

0
.
7)
2
= (0
.
30)
·
(0
.
49) + (0
.
70)
·
(0
.
09)
= 0
.
21
So that
σ
2
W
= 6
2
·
σ
2
X
= 36(0
.
21) = 7
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This note was uploaded on 08/09/2009 for the course ECON 120B taught by Professor Jeon during the Spring '08 term at UCSD.
 Spring '08
 Jeon
 Econometrics

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