7.
This is similar to Exercise 1.
It is simplest to prove it in that framework.
Since the model has only a
dummy variable, we can use the same log likelihood as in Exercise 1.
But, in this exercise, there are no
observations in the cell (
y
=1,
x
=0).
The resulting log likelihood is, therefore,
ln
L
=
Σ
0,0
lnProb[
y
=0,
x
=0] +
Σ
0,1
lnProb[
y
=0,
x
=1] +
Σ
1,1
lnProb[
y
=1,
x
=1]
or
ln
L
=
n
3
lnProb[
y
=0,
x
=0] +
n
2
lnProb[
y
=0,
x
=1] +
n
1
lnProb[
y
=1,
x
=1].
Now, let
δ
=
α
+
β
.
The log likelihood function is ln
L
=
n
3
ln(1 
F
(
α
)) +
n
2
ln(1 
F
(
δ
)) +
n
1
ln
F
(
δ
).
For
estimation, let
A
=
F
(
α
) and
D
=
F
(
δ
).
We can estimate
A
and
D
, then
α
=
F
1
(
A
)
and
β
=
F
1
(
D
) 
α
.
The
first order condition for estimation of A is
∂
ln
L
/
∂
A
=

n
3
/(1 
A
)
=
0,
which obviously has no solution.
If
A
cannot be estimated then
α
cannot either, nor, in turn, can
β
.
This applies to both probit and logit models.
8.
We’ll do this more generally for any model F(
α
).
Since the ‘model’ contains only a constant, the log
likelihood is logL =
Σ
0
log[1F(
α
)] +
Σ
1
logF(
α
) = n
0
log[1F(
α
)]+n
1
logF(
α
) . The likelihood equation is
∂
logL/
∂α
=
Σ
0
[f(
α
)/[1F(
α
)] +
Σ
1
f(
α
)/F(
α
) = 0 where f(
α
) is the density (derivative of F(
α
) so that at the
solution, n
0
f(
α
)/[1F(
α
)] = n
1
f(
α
)/F(
α
). Divide both sides of this equation by f(
α
) and solve it for F(
α
) =
n
1
/(n
0
+n
1
), as might be expected.
You can then insert this solution for F(
α
) back into the log likelihood,
and (2328) follows immediately.
9.
Look at the two cases.
Neither case has an estimator which is consistent in both cases.
In both cases,
the unconditional fixed effects effects estimator is inconsistent, so the rest of the analysis falls apart.
This
is the incidental parameters problem at work.
Note that the fixed effects estimator is inconsistent because
in both models, the estimator of the constant terms is a function of 1/T.
Certainly in both cases, if the fixed
effects model is appropriate, then the random effects estimator is inconsistent, whereas if the random
effects model is appropriate, the maximum likelihood random effects estimator is both consistent and
efficient.
Thus, in this instance, the random effects satisfies the requirements of the test.