Problem Set 6
ECON 231W
Spring 2010
Suggested Solutions
1.
a. Since
Y
i
is a binary variable, we know
E
(
Y
i

X
i
) = 1
×
Pr
(
Y
i
= 1

X
i
) + 0
×
Pr
(
Y
i
= 0

X
i
) =
Pr
(
Y
i
= 1

X
i
) =
β
0
+
β
1
X
i
.
Thus:
E
(
u
i

X
i
) =
E
[
Y
i

(
β
0
+
β
1
X
i
)

X
i
] =
E
(
Y
i

X
i
)

(
β
0
+
β
1
X
i
) = 0
b. Using Equation (2.7) in the book, we have:
V AR
(
Y
i

X
i
) =
Pr
(
Y
i
= 1

X
i
)[1

Pr
(
Y
i
= 1

X
i
)] = (
β
0
+
β
1
X
i
)[1

(
β
0
+
β
1
X
i
)]
Thus:
V AR
(
u
i

X
i
) =
V AR
[
Y
i

(
β
0
+
β
1
X
i
)

X
i
] =
V AR
(
Y
i

X
i
) = (
β
0
+
β
1
X
i
)[1

(
β
0
+
β
1
X
i
)]
c.
V AR
(
u
i

X
i
) depends on the value of
X
i
, so
u
i
is heteroskedastic.
2.
a. Let
n
1
= #(
Y
= 1), the number of observations on the random variable
Y
which equals 1;
and
n
2
= #(
Y
= 2). Then #(
Y
= 3) =
n

n
1

n
2
. The joint probability distribution of
Y
1
, ..., Y
n
is:
Pr
(
Y
1
=
y
1
, ..., Y
n
=
y
n
) =
n
Y
i
=1
Pr
(
Y
i
=
y
i
) =
p
n
1
q
n
2
(1

p

q
)
n

n
1

n
2
The likelihood function is the above joint probability distribution treated as a function of the un
known coefficients (
p
and
q
).
b. The MLEs of
p
and
q
maximize the likelihood function. Let’s use the loglikelihood function:
L
=
ln[
Pr
(
Y
1
=
y
1
, ..., Y
n
=
y
n
)]
=
n
1
ln
p
+
n
2
ln
q
+ (
n

n
1

n
2
) ln(1

p

q
)
Using calculus, the partial derivatives of
L
are:
∂
L
∂p
=
n
1
p

n

n
1

n
2
1

p

q
∂
L
∂q
=
n
2
q

n

n
1

n
2
1

p

q
Setting these two equations equal to zero and solving for
p
and
q
yield the MLE of
p
and
q
:
b
p
=
n
1
n
,
b
q
=
n
2
n
1
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 Spring '10
 JoshuaKinsler
 Probability, Probability distribution, Likelihood function, Probit

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