MSPE PROGRAM
ECON 506 FALL 2010
SOLUTION TO HW3
(The problem numbers refer to numbers in 7
th
edition)
3.148
Let denote q = 1p.
2
)
1
(
)
(
t
t
dt
d
qe
pe
t
m

=
.
At
t
= 0, this is 1/
p
=
E
(
Y
).
4
2
)
1
(
)
)(
1
(
2
)
1
(
)
(
2
2
t
t
t
t
t
t
dt
d
qe
qe
qe
pe
pe
qe
t
m





=
.
At
t
= 0, this is (1+
q
)/
p
2
.
Thus,
V
(
Y
) = (1+
q
)/
p
2
– (1/
p
)
2
=
q
/
p
2
.
3.153
a.
Binomial with
n
= 5,
p
= 1/3.
b.
If
m
(
t
) is multiplied top and bottom by ½, this is a geometric mgf with
p
= ½.
c.
Poisson with λ = 2.
3.155
Differentiate to find the necessary moments:
a.
E
(
Y
) = 7/3.
b.
V
(
Y
) =
E
(
Y
2
) – [
E
(
Y
)]
2
= 6 – (7/3)
2
= 5/9.
c.
Since
).
(
)
(
tY
e
E
t
m
=
Y
can only take on values 1, 2, and 3 with probabilities 1/6, 2/6,
and 3/6.
3.161
.
)
(
)
(
)
(
)
(
)
(
1
)
1
(
*
*
t
qe
p
t
tY
t
Y
t
tY
Y
t
m
e
e
e
E
e
E
e
E
t
m




=
=
=
=
=
3.167
a.
The value 6 lies (11–6)/3 = 5/3 standard deviations below the mean.
Similarly, the
value 16 lies (16–11)/3 = 5/3 standard deviations above the mean.
By Tchebysheff’s
theorem, at least 1 – 1/(5/3)
2
= 64% of the distribution lies in the interval 6 to 16.
b.
By Tchebysheff’s theorem, .09 = 1/
k
2
, so
k
= 10/3.
Since σ = 3,
k
σ = (10/3)
3
= 10 =
C
.
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3.170
Similar to Ex. 3.167: the interval (.48, 52) represents two standard deviations about the
mean.
Thus, the lower bound for this interval is 1 – ¼ = ¾.
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 Fall '08
 Staff
 Normal Distribution, Standard Deviation, Variance, Probability theory, standard deviations, dy

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