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Hahn Notes

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Lecture Note 1: Some Probability & Statistics
1
Basics
Joint Distribution For discrete random variables,
fXY (x; y) = Pr (X = x and Y = y)
Marginal Distribution Sums/integrates out all but one variable,
X
X
fX (x) =
fXY (x; y) and fY (y) =
fXY (x; y)
y
x
XX
y
fXY (x; y) = 1
x
Conditional Distribution
fXjY (xjy) =
fXY (x; y)
fY (y)
for fY (y) 6= 0
Note that for a …xed y, the conditional probability must sum to 1:
X
fXjY (xjy) = 1
x
2
Measures of Central Tendency
Mean
E [X] =
X
xfX (x)
x
Conditional Mean Mean of X given a value of Y ,
X
E [XjY = y] =
xfXjY (xjy)
x
1
3
Properties of Conditional Expectations
Law of Iterated Expectations
X
E [X] = E [E [ Xj Y ]] =
E [Xj Y = y] fY (y)
y
Noting that E [ Xj Y = y] is just a function of Y , say g (Y ), we can write the above as:
X
E [X] = E [g (Y )] =
g (y) fY (y)
y
In general,
E [ Xj Y ] = E [E [Xj Y; Z]j Y ]
Example 1
E Y 2 Y = y = y2
E XY 2 Y = y = y 2 E [Xj Y = y]
E [g (X; Y )j Y = y] = E [g (X; y)j Y = y] =
X
g (x; y) fXjY (xjy)
x
4
Best Constant Predictor
Our objective is to …nd
a)2
argmin E (Y
a
Let
E [Y ]
and note that
E (Y
a)2 = E ((Y
= E (Y
)2 + 2 (
= E (Y
)2 + 2 (
a) E [(Y
)] + (
a)2
= E (Y
)2 + 2 (
a) (E [Y ]
)+(
a)2
= E (Y
)2 + 2 (
a) 0 + (
= Var (Y ) + (
which is minimized when
a))2
)+(
=
a) (Y
a)2
)+(
a)2
a)2 ;
= E [Y ], so the mean is the best constant predictor.
2
4.1
Special Case
Identi…cation Suppose that Y
2
N ( 0;
2
) with
1
' (y; ) = p exp
2
known. Is the
!
(y
)2
2 2
0
identi…ed? Let
denote the PDF. Now, let
)2
(Y
L ( ) = log ' (Y ; ) = C
2
2
where C denotes a generic constant. If we know the exact distribution F of Y , we can calculate
the expected value of L ( ):
"
#
EF (Y
)2
(Y
)2
EF [L ( )] = EF
C
=
C
2 2
2 2
But
EF (Y
)2 = EF ((Y
0)
2
= EF (Y
0)
(
+(
It follows that
2
0 ))
2
0)
2
=
0.
Therefore,
0
+(
2
is identi…ed.
Estimation Now suppose that Y1 ; : : : ; Yn
by
iid
+(
2
0)
2
EF [L ( )] = C
is maximized at
2
0)
2
0)
= EF (Y
2
N ( 0;
=C
2
2(
0 ) (Y
2
2
known. We can estimate
n
1X
E Y =
E [Yi ] =
n i=1
n
it is an unbiased estimator of
0.
0
Con…dence Interval We need two properties of the normal distribution:
2
N( ;
Lemma 2 If Y
N (0; 1), then Pr [jY j
Corollary 1 If Y
N( ;
), then
Y
Lemma 1 If Y
2
), then Pr
N (0; 1).
1:96] = 95%.
Y
1:96 = 95%.
3
2
0)
2
1X
Y =
Yi
n i=1
Because
+(
2
0)
(
) with
0)
0
Now, note that Y is a linear combination of Y1 ; : : : ; Yn
iid
2
N ( 0;
) so we have
2
Y
N
0;
n
It follows that
Pr
Y
p0
/ n
1:96 = 95%
But
Pr
Y
p
0
Y
p
/ n
0
1:96 = Pr
/ n
1:96 p
= Pr
0
2
0
n
1:96 p
Y
0
1:96 p
Y
0
n
1:96 p
= Pr Y
= Pr
1:96 = Pr
n
n
Y + 1:96 p
n
1:96 p ; Y + 1:96 p
n
n
Y
We conclude that
Pr
and Y
5
5.1
0
1:96 pn ; Y + 1:96 pn
2
Y
1:96 p ; Y + 1:96 p
n
n
= 95%
is the 95% con…dence interval for
0.
Digression
Conditional Expectation
Our objective is to …nd some function
(X)
(X) such that
argmin E (Y
g (X))2
g( )
Let
(X)
E [Y j X]
and note that
(Y
g (X))2 = ((Y
= (Y
g (X)))2
(X)) + ( (X)
(X))2 + 2 ( (X)
g (X)) (Y
(X)) + ( (X)
and therefore,
E (Y
g (X))2 X = E (Y
(X))2 X
+ E [2 ( (X)
+ E ( (X)
4
g (X)) (Y
2
g (X)) X
(X))j X]
g (X))2
But
E [2 ( (X)
g (X)) (Y
(X))j X] = 2 ( (X)
g (X)) E [Y
(X)j X]
g (X)) (E [Y j X]
= 2 ( (X)
g (X)) (E [Y j X]
= 2 ( (X)
= 2 ( (X)
E [ (X)j X])
(X))
g (X)) 0
= 0;
and
g (X))2 X = ( (X)
E ( (X)
g (X))2
It follows that
E (Y
g (X))2 X = E (Y
(X))2 X + ( (X)
g (X))2
and therefore
E (Y
g (X))2 = E (Y
(X))2 + E ( (X)
which is minimized when
( (X)
g (X))2 = 0
or
g (X) =
(X)
Therefore,
(X) = E [Y j X]
5.2
Best Linear Predictor
Predict Y using linear functions of X. Find
( ; )
and
argmin E (Y
such that
(a + bX))2
a;b
The minimization can be solved in two steps.
Step 1 We …x b, and …nd
(b) that solves
(b)
argmin E (Y
a
5
(a + bX))2
g (X))2 ;
Step 2 We …nd
that solves
( (b) + bX))2
argmin E (Y
b
and recognize that
=
( ). It is not di¢ cult to see that
(b) = E [Y
bX] = E [Y ]
bE [X] =
b
Y
X:
Therefore, Step 2 optimization can be rewritten as
min E (Y
((
b
b
Y
X)
+ bX))2 = min E
b
where
Ye
Now, let
Y
h
i
e Ye
E X
h i
e2
E X
b0
and note that
E
We therefore have
E
Ye
e
bX
2
=E
=E
=E
Ye
Ye
Ye
h
Ye
e
b0 X
X
+ 2 (b0
e
b) X
2
+ (b0
E [X]
2
b) E
2
b)
h
h
Ye
i
2
e
E X ;
e
b0 X
which is minimized when b = b0 . Therefore, we have
= b0 =
Cov (X; Y )
;
Var (X)
and
=
( )=
Cov (X; Y )
Var (X)
Y
6
2
i
e =0
X
e
b0 X
2
e
bX
Cov (X; Y )
Var (X)
=
e + (b0
b0 X
e
b0 X
e
X
E [Y ] ;
Ye
X:
i
e + (b0
X
h i
e2
b)2 E X
Lecture Note 2: Computational Issues of OLS
6
Some Notation
Data For each individual i, we observe (yi ; xi1 ; : : : ; xik ). We observe n individuals.
Objective We want to “predict”y by xs:
yi = b1 xi1 +
+ bk xik + ri : i = 1; : : : ; n
Here, ri denotes the residual.
Vector Writing
0
1
b1
B
C
b = @ ... A ;
k 1
bk
0
1
xi1
B
C
xi = @ ... A ;
k 1
xik
we may compactly write
yi = x0i b + ri : i = 1; : : : ; n
Matrix Writing
0
1
x01
B
C
X = @ ... A ;
n k
x0n
0
1
y1
B
C
y = @ ... A ;
n 1
yn
we may more compactly write
0
1
r1
B
C
r = @ ... A :
n 1
rn
y = X b + r :
n 1
7
n kk 1
n 1
n 1
OLS
It seems natural to seek b that solves
min
b
n
X
(yi
(b1 xi1 +
+ bk xik ))2 = min
b
i=1
n
X
7
2
x0i b)
i=1
= min (y
b
(yi
Xb)0 (y
Xb)
Basic Matrix Calculus
De…nition 1 For a real valued function f : t = (t1 ; : : : ; tn ) ! f (t), we de…ne
2 @f 3
@t1
@f
7
6
= 4 ... 5 ;
@t
@f
@f
=
@t0
@f
@f
;:::;
@t1
@tn
:
@tn
Lemma 3 Let f (t) = a0 t. Then, @f (t)/ @t = a.
Lemma 4 Let f (t) = t0 At, where A is symmetric. Then, @f (t)/ @t = 2At.
Back to OLS Let
S (b)
Xb)0 (y
(y
Xb) = y 0 y
b0 X 0 y
y 0 Xb + b0 X 0 Xb
0
= y0y
2 (X 0 y) b + b0 X 0 Xb
A necessary condition for minimum is
@S (b)
=0
@b
Now using the two lemmas above,weobtain
@S (b)
=
@b
2X 0 y + 2X 0 Xb
from which we obtain
Xb)0 (y
argmin (y
Xb) = (X 0 X)
1
X 0y
b
Theorem 1 Let e
y
X (X 0 X)
1
X 0 y. Then,X 0 e = 0.
Proof.
X 0e = X 0y
Remark 1 Let b
(X 0 X)
1
X 0 X (X 0 X)
X 0 y. We then have e
1
X 0y = 0
y
X b.
Corollary 2 If the …rst column of X consists of ones, we have
Theorem 2 (X 0 X)
1
X 0 y = argminb (y
Xb)0 (y
8
Xb).
P
i ei
= 0.
Proof. Write
y
Xb+ Xb
Xb = y
=y
X (X 0 X)
=e
X b
Therefore, we have
Xb) = e
= e0 e
0
b
X b
0
b
= e0 e + X b
e0 e
This gives a proof as to why b minimizes (y
8
e
b X 0e
b
X 0y
b
X b
b =0
e0 X b
Xb)0 (y
1
b
and note that
(y
Xb
b
X b
e0 X b
0
b
X b
Xb)0 (y
0
b + b
b X 0X b
b
Xb).
Digression: Restricted Least Squares
Theorem 3 The solution to the problem
min (y
b
Xb)0 (y
is given by
b
where
Here, R is m
b
(X 0 X)
1
Xb)
s:t:
h
R0 R (X 0 X)
b = (X 0 X)
1
1
Rb = q
R0
i
1
Rb
X 0 y:
q ;
k matrix. Both R and q are known. We assume that R has the full row rank.
Proof. It is not di¢ cult to see that
Rb = R b
= Rb
= q:
R (X 0 X)
Rb
q
1
h
R0 R (X 0 X)
Suppose that some b satis…es Rb = q. Let d
Rd = R (b
b
b ) = Rb
9
1
R0
i
1
Rb
q
b . We should then have
Rb = q
q = 0:
We now show that
e0 Xd = 0;
where
e
y
Xb :
Note that
X 0 e = X 0 (y
Xb ) = X 0 e
X 0X b
=0
b
X b
b
= X 0e
h
1
b = X 0X
(X 0 X) R0 R (X 0 X)
h
i 1
1
1
= X 0 X (X 0 X) R0 R (X 0 X) R0
Rb q
h
i 1
1
= R0 R (X 0 X) R0
Rb q :
=
X 0X b
b
X 0X b
1
It therefore follows that
h
i 1
h
1
d0 X 0 e = d0 R0 R (X 0 X) R0
R b q = (Rd)0 R (X 0 X)
h
i 1
1
= 00 R (X 0 X) R0
Rb q = 0
R0
1
i
R0
1
i
and
0
e0 Xd = (d0 X 0 e ) = 0:
Therefore, we have
(y
Xb)0 (y
Xb) = (e
X (b
= (e
0
Xd) (e
= e0 e
d0 X 0 e
b ))0 (e
X (b
b ))
Xd)
e0 Xd + d0 X 0 Xd
= e0 e + d0 X 0 Xd
e0 e
and the minimum is achieved if and only if d = 0 or b = b .
Theorem 4 Let
e
y
Then,
e0 e = e0 e + R b
q
0
h
Xb :
R (X 0 X)
10
1
R0
i
1
Rb
q :
Rb
1
Rb
q
q
Proof. It is immediate from
h
i
1
R R (X X) R
Rb
h
i 1
1
1
Rb
= y X b + X (X 0 X) R0 R (X 0 X) R0
h
i 1
1
1
= e + X (X 0 X) R0 R (X 0 X) R0
Rb q
e =y
and
X
b
0
(X X)
1
0
0
1
0
q
q
X 0 e = 0:
9
Projection Algebra
Let
X (X 0 X)
P
1
X 0;
M
In
P
Theorem 5 P and M are symmetric and idempotent, i.e.,
P 0 = P;
M 0 = M;
P 2 = P;
M 2 = M:
Proof. Symmetry is immediate.
P 2 = X (X 0 X)
M 2 = (In
= In
1
X 0 X (X 0 X)
P ) (In
P
1
X = X (X 0 X)
P ) = In
P + P = In
P
1
X 0 = P;
P + PP
P = M:
Theorem 6
P X = X;
M X = 0:
Proof.
P X = X (X 0 X)
M X = (In
P)X = X
1
X 0 X = X;
PX = X
X = 0:
Theorem 7
X b = P y;
e = M y:
Note that X b is the “predicted value” of y given X and the estimator b.
11
Proof.
X b = X (X 0 X) X 0 y = P y:
e = y X b = y P y = [In P ] y = M y:
1
Theorem 8
e0 e = y 0 M y:
Proof.
e0 e = (M y)0 M y = y 0 M 0 M y = y 0 M M y = y 0 M y:
Theorem 9 (Analysis of Variance)
0
y 0 y = X b X b + e0 e:
Proof.
0
y 0 y = y 0 (P + M ) y = y 0 P y + y 0 M y = y 0 P 0 P y + e0 e = (P y)0 P y + e0 e = X b X b + e0 e:
10
Problem Set
1. It can be shown that trace (AB) = trace (BA). Using such property of the trace operator,
prove that trace (P ) = k if X has k columns. (Hint:
trace (P ) = trace X (X 0 X)
1
X 0 = trace X (X 0 X)
= trace (X 0 X)
1
1
X0
X 0 X = trace (X 0 X)
1
X 0 X = trace (I) :
What is the dimension of the identity matrix?)
2. This question is taken from Goldberger. Let
2
3
1 2
6 1 4 7
6
7
6
7
X = 6 1 3 7;
6
7
4 1 5 5
1 2
Using Matlab, calculate the following:
X 0 X;
(X 0 X)
P = X (X 0 X)
1
1
;
(X 0 X)
X 0;
P y;
12
2
6
6
6
y=6
6
4
1
14
17
8
16
3
3
7
7
7
7:
7
5
X 0;
(X 0 X)
M =I
P;
1
X 0 y;
e = M y:
11
Application of Projection Algebra: Partitioned Regression
X =
X1
n k1
n k
Note that k1 + k2 = k. Partition b accordingly:
Note that b1 is (k1
1), and b2 is (k2
b=
b1
b2
X2
n k2
!
1). Thus we have
y = X1 b1 + X2 b1 + e
We characterize of b1 as resulting from two step regression:
Theorem 10 We have
where
11.1
ye
Discussion
b1 = X
e10 X
e1
e1
X
M2 y;
1
e10 ye;
X
M2 X1 :
Theorem 11 Write m = k1 , and
X1 = [x1 ; : : : ; xm ] ;
X2 = [xm+1 ; : : : ; xk ] :
Then,
ye = M2 y
is the residual when y is regressed on X2 , and
where
e1 = M2 X1 = M2 [x1 ; : : : ; xm ] = [M2 x1 ; : : : ; M2 xm ] ;
X
M2 x1
is the residual when x1 is regressed on X2 .
We can thus summarize the characterization of b1 in the following algorithm:
Regress y on xm+1 ; : : : ; xk . Get the residual, and call it ye. (e
y captures that portion of y
not correlated with X2 , i.e., ye is y partialled out with respect to X2 )
13
e1 . (e
Regress x1 on xm+1 ; : : : ; xk . Get the residual, and call it x
x1 captures that portion of
x1 not correlated with X2 )
e2 ; : : : ; x
em .
Repeat second step for x2 ; : : : ; xm . Obtain x
e1 ; : : : ; x
em . The coe¢ cient estimate is numerically equal to b1 .
Regress ye on x
You may wonder why we need to do the regression in two steps when it can be done in one
single step. The reason is mainly computational. Notice that, in order to compute b, a k k
matrix X 0 X has to be inverted. If the number of regressors k is big, then your computer cannot
implement this matrix inversion. This point will be useful in the application to the panel data.
11.2
Proof
Lemma 5 Let
M2
I
1
X2 (X20 X2 )
X20 :
Then,
M2 e = e:
Proof. Observe that
X20 e = 0
*
X 0e =
X10
X20
e=
X10 e
X20 e
= 0:
Therefore,
M2 e = In e
X2 (X20 X2 )
1
X20 e = e:
Lemma 6
Proof. Premultiply
by M2 , and obtain
M2 y = M2 X1 b1 + e:
y = X1 b1 + X2 b1 + e
M2 y = M2 X1 b1 + M2 X2 b1 + M2 e
= M2 X1 b1 + e;
where the second equality follows from the projection algebra and the previous lemma.
Lemma 7
X10 M2 y = X10 M2 X1 b1 :
14
Proof. Premultiply
M2 y = M2 X1 b1 + e
by X10 , and obtain
X10 M2 y = X10 M2 X1 b1 + X10 e
= X 0 M2 X1 b1 ;
1
where the second equality follows from the normal equation.
12
R2
Let
X1 = [x1 ; : : : ; xm ] ;
X2 = `:
Then,
Remark 2 Note that
ye = y
e 1 = [x1
X
y `;
X2 (X20 X2 )
1
x1 `; : : : ; xm
1 0
X20 = ` (`0 `)
` = ` (n)
xm `] :
1 0
` =
1 0
``
n
and therefore
X2 (X20 X2 )
1
X20 y =
1 0
`` y = `
n
1 0
`y
n
= `y
Because
e 1 b1 + e;
ye = X
we obtain
e 1 b1 + e
ye0 ye = X
0
e 0 e = 0;
and X
1
0
b0 e 0
e 1 b1 + e = b0 X
e0 e b
X
1 1 X1 1 + 2 1 X1 e + e e
0
e0 X
e b
= b10 X
1 1 1 + e e:
Here, e0 e denotes the portation of ye0 ye “unexplained” by the variation of X1 (from its own
sample average). The smaller it is relative to ye0 ye, the more of ye0 ye is explained by X1 . From
this intuitive idea, we develop
e0 e
R2 = 1
ye0 ye
as a measure of goodness of …t.
Theorem 12 Suppose that
X = [X1 ; X2 ] :
Then, we have
min (y
b1
X1 b1 )0 (y
X1 b1 )
15
min (y
b
Xb)0 (y
Xb) :
Proof. We can write
min (y
b1
X1 b1 )0 (y
Xb)0 (y
X1 b1 ) = min (y
b
Xb)
s:t: b2 = 0:
The latter is bounded below by
min (y
b
Xb)0 (y
Xb)
Corollary 3
1
13
minb1 (y
Problem Set
X1 b1 )0 (y
ye0 ye
X1 b1 )
1
minb (y
Xb)0 (y
ye0 ye
Xb)
:
1. You are given
0
B
B
B
B
B
y=B
B
B
B
@
1
3
5
3
7
9
4
1
C
C
C
C
C
C;
C
C
C
A
2
6
6
6
6
6
X1 = 6
6
6
6
4
X = [X1 ; X2 ] ;
1
1
1
1
1
1
1
3
5
3
2
8
4
7
3
7
7
7
7
7
7;
7
7
7
5
2
6
6
6
6
6
X2 = 6
6
6
6
4
2
4
6
1
7
3
1
3
7
7
7
7
7
7:
7
7
7
5
(a) Using Matlab, compute
b = (X 0 X)
1
X 0 y;
0
e=y
X b;
e0 e;
R2 = 1
e0 e=
X
(yi
y)2 :
i
(b) Write b = b1 ; b2 ; b3 . Try computing b1 ; b2 by using the partitioned regression
technique: First, using Matlab, compute
e 1 = M2 X1 ;
X
ye = M2 y
where M2 = I X2 (X02 X2 ) 1 X02 . (In Matlab, you would have to specify the dimension of the identity matrix. What is the dimension?) Second, using Matlab again,
compute
1
e0 X
e1
e 0 ye
X
X
1
1
Is the result of your two step calculation equal to the …rst two component of b that
you computed before?
16
(c) Using Matlab, compute
b
Xb)0 (y
argmin (y
s.t. Rb = q;
Xb)
b
where R
(1; 1; 1) and q = [1].
(d) Using Matlab, compute
Xb )0 (y
e0 e = (y
Xb ) :
(e) Using Matlab, compute
Rb
Is it equal to e0 e
e0 e?
q
0
h
0
R (X X)
1
R
0
i
1
Rb
q :
2. You are given
0
B
B
B
B
B
y=B
B
B
B
@
1
3
5
3
7
9
4
1
C
C
C
C
C
C;
C
C
C
A
2
X = [X1 ; X2 ] ;
6
6
6
6
6
X1 = 6
6
6
6
4
1
1
1
1
1
1
1
3
5
3
2
8
4
7
3
7
7
7
7
7
7;
7
7
7
5
2
6
6
6
6
6
X2 = 6
6
6
6
4
2
4
6
1
7
3
1
3
7
7
7
7
7
7:
7
7
7
5
Using Matlab, compute
b = (X 0 X)
Now, compute
M2 = I X2 (X02 X2 )
1
X02 ;
1
X 0 y;
e 1 = M2 X1 ;
X
e=y
ye = M2 y;
X b:
e0 X
e
b1 = X
1 1
1
e 0 ye;
X
1
e 1b :
e = ye X
1
Verify that b1 is numerically identical to the …rst two components of b. Also verify that
e = e.
17
Lecture Note 3: Stochastic Properties of OLS
14
Note on Variances
For scalar random variable Z:
Var (Z)
E (Z
E [Z])2 = E Z 2
(E [Z])2
For random vector
Z = (Z1 ; Z2 ; : : : ; Zk )0 ;
k 1
we have a variance - covariance matrix:
Var (Z)
E (Z E [Z]) (Z E [Z])0
2
E [(Z1 E [Z1 ]) (Z1 E [Z1 ])] E [(Z1
6 E [(Z2 E [Z2 ]) (Z1 E [Z1 ])] E [(Z2
6
=6
..
4
.
2
E [(Zk
E [Zk ]) (Z1
E [(Z1
E [(Z2
E [Z1 ]) (Zk
E [Z2 ]) (Zk
..
.
E [Z2 ])]
E [(Zk
E [Zk ]) (Zk
E [(Z1 E [Z1 ]) (Z2 E [Z2 ])]
E (Z2 E [Z2 ])2
..
.
E [(Z1
E [(Z2
E [Z1 ]) (Zk
E [Z2 ]) (Zk
..
.
E [Z1 ])] E [(Zk
2
E (Z1 E [Z1 ])
6 E [(Z2 E [Z2 ]) (Z1 E [Z1 ])]
6
=6
..
4
.
E [(Zk E [Zk ]) (Z1 E [Z1 ])]
2
Var (Z1 )
Cov (Z1 ; Z2 )
6 Cov (Z2 ; Z1 )
Var (Z2 )
6
=6
..
..
...
4
.
.
It is useful to note that
15
15.1
E [Z2 ])]
E [Z2 ])]
E [Zk ]) (Z2
E [(Zk
Cov (Zk ; Z1 ) Cov (Zk ; Z2 )
E (Z
E [Z1 ]) (Z2
E [Z2 ]) (Z2
..
.
E [Z]) (Z
E [Zk ]) (Z2
3
Cov (Z1 ; Zk )
Cov (Z2 ; Zk ) 7
7
7
..
5
.
..
..
.
.
E [Z2 ])]
Var (Zk )
E [Z])0 = E [ZZ 0 ]
E [Z] E [Z]0
Classical Linear Regression Model I
Model
y =X +"
X is a nonstochastic matrix
18
E (Zk
3
E [Zk ])]
E [Zk ])] 7
7
7
5
E [Zk ])]
3
E [Zk ])]
E [Zk ])] 7
7
7
5
E [Zk ])2
X has a full column rank (Columns of X are linearly independent.)
E ["] = 0
E [""0 ] =
15.2
2
2
In for some unknown positive number
Discussion
We will later discuss the case where X is stochastic such that (i) y = X + "; (ii) X has
a full column rank; (iii) E ["j X] = 0; (iv) E [""0 j X] = 2 In .
The third assumptions amounts to the identi…ability condition on . Suppose X does not
have a full column rank. Then, by de…nition, we can …nd some 6= 0 such that X = 0.
Now, even if " = 0, we would have
y =X =X +X =X( + )
so that we would not be able to di¤erentiate
from
+
from the data
E ["] = 0 is a harmless assumption if we believe that E ["i ] are the same regardless of i.
Write this common value as 0 . Then, we may rewrite the model as
yi =
0
xi0 +
xi1 +
1
+
k
xik + ui
for
u i = "i
0
and
xi0 = 1:
E [""0 ] = 2 In consists of two parts. First, it says that all the diagonal elements, i.e.,
variances of the error terms are equal. This is called the homoscedasticity. Second, it
says all the o¤-diagonal elements are zeros. (This will not be satis…ed in the time series
environment.)
15.3
Properties of OLS
Lemma 8 b =
+ (X 0 X)
1
X 0"
Proof.
b = (X 0 X)
= (X 0 X)
1
X 0 y = (X 0 X)
1
1
X 0 X + (X 0 X)
19
X 0 (X + ")
1
X 0" =
+ (X 0 X)
1
X 0 ":
h i
Theorem 13 Under the Classical Linear Regression Model I, we have E b =
2
(X 0 X)
1
Proof.
h i
E b =
and
+ (X 0 X)
Var b = E
1
h
X 0 E ["] =
b
1
0
= E (X X)
16
1
+ (X 0 X)
b
0
X 00 =
0
0
1
0
X "" X (X X)
i
= (X 0 X)
1
X 0 E [""0 ] X (X 0 X)
1
= (X 0 X)
1
X0
1
2
and Var b =
I X (X 0 X)
=
2
(X 0 X)
1
X 0 X (X 0 X)
=
2
(X 0 X)
1
:
1
Estimation of Variance
Lemma 9
e = M"
Proof.
e = M y = M (X + ") = M X + M " = M "
Lemma 10
trace (M ) = n
k;
trace (P ) = k:
Proof.
trace (P ) = trace X (X 0 X)
1
X 0 = trace (X 0 X)
= k;
trace (M ) = trace (In
=n
P ) = trace (In )
k:
20
trace (P )
1
X 0 X = trace (Ik )
Theorem 14 Let
s2 =
e0 e
n
k
:
Then,
E s2 =
2
Proof. Observe that
e0 e = "0 M 0 M " = "0 M " = trace ("0 M ") = trace (M ""0 ) ;
and hence,
E [e0 e] = E [trace (M ""0 )] = trace (M E [""0 ]) = trace M
= (n
2
k)
Corollary 4
In other words, s2 (X 0 X)
17
1
2
I =
2
trace (M )
:
h
i
1
E s2 (X 0 X)
= Var b
is an unbiased estimator of Var b .
Basic Asymptotic Theory: Convergence in Probability
Remark 3 A sequence of nonstochastic real numbers an converges to a, if for any > 0, there
exists N = N ( ) such that jan aj < for all n N .
De…nition 2 A sequence of random variables fzn g converges in probabilityto c, a deterministic
number, if
lim Pr [jzn cj
]=0
n!1
for any > 0. We sometimes write it as plimn!1 zn = c. For a sequence of random matrices
An , we have the convergence in probability to a deterministic matrix A if every element of An
convergences in probability to the corresponding element of A.
Remark 4 If an is a sequence of nonstochastic real numbers converging to a, and if g ( ) is
continuous at a, we have limn!1 g (an ) = g (a).
Theorem 15 (Slutzky) Suppose that plimn!1 zn = c. Also suppose that g ( ) is continuous
at c. We then have plimn!1 g (zn ) = g (c).
21
Corollary 5 If
plim z1n = c1 ;
plim z2n = c2 ;
n!1
n!1
then
plim (z1n + z2n ) = c1 + c2 ;
plim z1n z2n = c1 c2 ;
n!1
n!1
and if c2 6= 0,
z1n
c1
= :
c2
n!1 z2n
plim
Theorem 16 (LLN) Given a sequence fzi g of i.i.d. random variables such that E [jzi j] < 1,
P
we have plimn!1 z n = E [zi ], where z n = n1 ni=1 zi .
Corollary 6 Given a sequence fzi g of i.i.d. random variables such that E [jg (zi )j] < 1, we
have
n
1X
plim
g (zi ) = E [g (zi )] :
n!1 n
i=1
18
18.1
Classical Linear Regression Model III
Model
yi = x0i + "i
(x0i ; "i ) i = 1; 2; : : : is i.i.d.
xi ?"i
E ["i ] = 0, Var ("i ) =
2
E [xi x0i ] is positive de…nite. Furthermore, all of its elements are …nite.
18.2
Some Auxiliary Lemmas
Lemma 11
1X
xi x0i
n i=1
n
b=
Proof.
b = (X 0 X)
1
X 0 y = (X 0 X)
=
+
1
+
!
1
!
n
1X
x i "i :
n i=1
1
X 0 (X + ") = + (X 0 X) X 0 "
! 1
!
!
n
n
n
X
X
X
1
xi x0i
x i "i = +
xi x0i
n
i=1
i=1
i=1
22
1
!
n
1X
x i "i :
n i=1
Lemma 12
s2 =
1
n
k
Proof.
1
2
s =
n
k
=
n
X
2
n
X
e2i
=
i=1
k
n
X
n
"2i
k
n
X
Lemma 14 plimn!1
1
n
Proof.
n
X
2
"i x0i
i=1
i=1
n
Pn
i=1
Pn
i=1
k
!
x0i b
yi
1
i=1
1
n
n
X
"i x0i
i=1
b
+ b
1
2
=
!
n
k
b
1
0
n
n
X
k
i=1
1
0
n
1
n
Pn
i=1
k
xi x0i = E [xi x0i ]
x i "i = 0
n
"2i =
2
1X 2
plim
"i = E "2i =
n!1 n
i=1
n
2
:
Large Sample Property of OLS
De…nition 3 An estimator b is consistent for
Theorem 17 b is consistent for .
Proof.
2
if plimn!1 b = .
!
n
X
1
plim b = plim 4 +
xi x0i
n
n!1
n!1
i=1
! 1
n
X
1
xi x0i
= + plim
n!1 n
i=1
=
+ E [xi x0i ]
1
0
= :
23
1
xi x0i
i=1
x0i b
x0i + "i
+ b
n
X
1X
plim
xi "i = E [xi "i ] = E [xi ] E ["i ] = 0
n!1 n
i=1
Lemma 15 plimn!1
18.3
k
1
Lemma 13 plimn!1
Proof.
n
i=1
1
n
"2i
1
!3
n
X
1
x i "i 5
n i=1
!
n
X
1
plim
x i "i
n!1 n
i=1
!
b
:
2
n
X
i=1
xi x0i
!
b
Theorem 18 plimn!1 s2 =
2
Proof.
plim
n!1
n
1X 2
s = plim
"i
n
n!1 n
i=1
k
n
2
1X 0
"i x i
n i=1
n
plim 2
n!1
n!1
=
2
1X 0
"i x i
n i=1
n
2 plim
n!1
0
+ plim b
n!1
=
19
b
1X
xi x0i
n i=1
0
+ plim b
!
n
!
plim
n!1
2
!
b
plim b
n!1
!
n
1X
xi x0i plim b
n i=1
n!1
Large Sample Property of OLS with Model I?
Lemma 16 If a sequence of random variables fzn g is such that E (zn
plimn!1 zn = c.
c)2
! 0, then
Proof. Let > 0 be given. Note that
2
E (zn
c)
=
Z
c)2 dFn (z)
(z
where Fn denotes the CDF of zn . But
Z
Z
Z
2
2
(z c) dFn (z) =
(z c) dFn (z) +
(z c)2 dFn (z)
jz cj
jz cj<
Z
(z c)2 dFn (z)
jz cj
Z
Z
2
2
dFn (z) =
dFn (z) = 2 Pr [jz cj
jz cj
jz cj
It follows that
Pr [jz
cj
]
c)2
E (zn
2
24
!0
]
h We
i will now use the fact that, under the Classical Linear Regression Model I, we have
b
E
= and Var b = 2 (X 0 X) 1 . We …rst note that
E
b
2
0
b
=E
= E trace
b
= trace E
2
b
b
0
1
b
b
0
= trace Var b
=
2
0
trace @
i=1
n
1X
xi x0i
n i=1
1
n
! 11
0
n
1X
@
trace
xi x0i
n i=1
! 11
A
for all n, we can conclude that
from which we can conclude that
1
n
0
n
X
i=1
xi x0i
! 11
n
X
1
A=
xi x0i
trace @
n
n i=1
Therefore, IF we assume that
E
0
b
1
(X 0 X)
= 2 trace (X 0 X)
0
0
! 11
n
X
A = 2 trace @ n
= 2 trace @
xi x0i
= trace
b
= E trace
2
A
! 11
A
B<1
2
b
!0
plim b =
n!1
P
The only question is how we can make sure that trace n1 ni=1 xi x0i
textbooks adopt di¤erent assumptions to ensure this property.
25
1
B < 1. Di¤erent
Lecture Note 4: Statistical Inference with Normality Assumption
20
Review of Multivariate Normal Distribution
De…nition 4 A square matrix A is called positive de…nite if t0 At > 0 8t 6= 0:
Theorem 19 A symmetric positive de…nite matrix is nonsingular.
De…nition 5 (Multivariate Normal Distribution) An n-dimensional random (column) vector Z has a multivariate normal distribution if its joint pdf equals
(2 )n=2
for some
1
s
det
and positive de…nite
Z
Theorem 20 Let L be an m
0
z
n 1
1
(z
)
n 1
n n
. We then write Z
N( ; )
Z
1
2
exp
)
N ( ; ). It can be shown that
E [Z] = ;
Var (Z) = :
n deterministic matrix. Then,
N( ; )
)
LZ
Z=
Z1
Z2
N (L ; L L0 ) :
Theorem 21 Assume that
has a multivariate normal distribution. Assume that E [(Z1
and Z2 are independent.
1 )(Z2
0
2) ]
= 0. Then, Z1
Theorem 22 Suppose that Z1 ; : : : ; Zn are i.i.d. N (0; 1). Then, Z = (Z1 ; : : : ; Zn )0 = N (0; In ).
Theorem 23
Z
N (0; In )
)
Z 0Z
2
(n) :
n is called the degrees of freedom.
Theorem 24 Suppose that an n
2
Z 0 1Z
(n)
n matrix
is positive de…nite, and Z
N (0; ). Then,
Theorem 25 Suppose that an n n matrix A is symmetric and idempotent, and Z
2
Then, Z 0 AZ
(trace (A))
26
N (0; In ).
21
Classical Linear Regression Model II
In addition to the assumptions of the classical linear regression model I, we now assume that
" has a multivariate normal distribution
21.1
Sampling Property of OLS Estimator
Theorem 26
Proof. Recall that
b
N
h i
E b = ;
2
;
1
(X 0 X)
Var b =
2
:
(X 0 X)
1
even without the normality assumption. It thus su¢ ces to establish normality. Note that
b
= (X 0 X)
1
X 0 ";
a linear combination of the multivariate normal vector ". Thus, b
normal distribution.
De…nition 6 For simplicity of notation, write
V
22
2
(X 0 X)
1
b
V
;
s2 (X 0 X)
, and hence b has a
1
Con…dence Interval: Known Variance
For a given k 1 vector r, we are interested in the inference on r0 . Speci…cally, we want to
construct a 95% con…dence interval.
Theorem 27 (95% Con…dence Interval) The random interval
p
p
r0 b 1:96 r0 Vr; r0 b + 1:96 r0 Vr
contains r0 with 95% probability:
h
p
Pr r0 b 1:96 r0 Vr
i
p
r0 b + 1:96 r0 Vr = :95
r0
Proof. Because
r0 b
we have
and
Pr
"
r0 b
p
N (r0 ; r0 Vr) ;
r0
r0 Vr
r0 b
p
N (0; 1)
#
r0
r0 Vr
1:96 = :95
27
Corollary 7 Write
= ( 1; : : : ;
b = b1 ; : : : ; bk
0
k) ;
Note that
bj = a0 b;
j
0
:
= a0j ;
j
where aj is the k-dimensional vector whose jth component equals 1 and the remaining elements
all equal 0. The preceding theorem implies that
bj
p
1:96 Vjj ;
p
bj + 1:96 Vjj ;
where Vjj is the (j; j)-element of V, is a valid 95% con…dence interval for j . Noting that Vjj
is the variance of bj , we come back to the undergraduate con…dence interval:
p
estimator
1:96
variance!
23
Hypothesis Test: Known Variance
Theorem 28 (Single Hypothesis: 5% Signi…cance Level) Given
H0 : r 0 = q
vs:
HA : r0 6= q;
the test which rejects the null i¤
r0 b q
p
r0 Vr
has a size equal to 5%:
Pr
"
1:96
#
r0 b q
p
r0 Vr
1:96 = :05
Corollary 8 Suppose we want to test
H0 :
j
= 0 vs:
HA :
j
6= 0
The preceding theorem suggests that we reject the null i¤
b
pj
Vjj
1:96:
This con…rms our undergraduate training based procedure rejecting the null i¤
q
bj
variance of bj
28
1:96:
Theorem 29 (Multiple Hypotheses: 5% Signi…cance Level) Given
H0 : R
m k
=q
HA : R 6= q;
vs:
the test which rejects the null i¤
has a size equal to 5%.
Rb
0
q
Rb
1
(RVR0 )
2
:05
q
(m)
Proof. It follows easily from
and
Rb
24
Rb
q
0
N (0; RVR0 )
R
1
(RVR0 )
Rb
2
q
(m)
Problem Set
1. (In this question, you are expected to verify the theorems discussed in the class using
MATLAB. Turn in your MATLAB program along with the result.) Consider the linear
model given by
yi = 1 + 2 xi2 + i ; i = 1; : : : ; n;
where
N (0;
i
2
) i:i:d:
Let b1 and b2 denote the OLS estimators of 1 and 2 . Suppose that
and n = 9. Note that the above model can be compactly written as
2
6
6
X=6
4
1
1
..
.
1
2
..
.
1 9
3
7
7
7
5
and
1
=
2
(a) Suppose that xi2 = i. Show that
b2
29
N
1;
=
2
=
2
=1
(1)
y=X +
for
1
1
60
:
:
(b) Show that
P b2
1
1:96 p
60
2
b2 + 1:96 p1
= :95
60
Let u(1) ; : : : ; u(1000) denote 1,000 independent N (0; 2 In ) random vectors. Notice
that y in (1) has the same distribution as y (j) given by
y (j)
X + u(j) :
(j)
(j)
Let b2 denote the OLS estimator of 2 of the above model. b2 are i.i.d. random
variables which has the same distribution as b2 . Let
(
(j)
b(j) + 1:96 p1
1 if b2
1:96 p160
2
2
60
D(j) =
0 otherwise
Show that we have
E[D(j) ] = :95
Argue that
1 X
D(j)
1000 j=1
1000
:95
Verify that this indeed is the case by generating 1,000 independent uj from the
computer. (This type of experiment is called the Monte Carlo.)
25
Con…dence Interval: Unknown Variance
Lemma 17
e0 e
2
2
Proof. Let
u
Then,
e0 e
2
= u0 M u
"
2
(n
k) :
N (0; In ) :
(trace (M )) =
Lemma 18 b and e0 e are independently distributed.
2
(n
k) :
Proof. It su¢ ces to prove that b
and e are independently distributed. But
!
b
(X 0 X) 1 X 0 "
(X 0 X) 1 X 0
=
=
"
M"
M
e
30
has a multivariate normal distribution because it is a linear combination of ". It thus su¢ ces
to prove that b
and e are uncorrelated. Their covariance equals
h
i
h
i
1
1
0
0
0 0
0
b
E
e = E (X X) X "" M = (X 0 X) X 0 E [""0 ] M
1
= (X 0 X)
Theorem 30
r0 b r0
p
b
r0 Vr
Proof. Notice that
r
0
We also have
b
2 0
N 0;
0
r (X X)
1
t (n
e0 e
2
2
Now observe that b
(X 0 X)
1
(X 0 M ) = 0:
r0 b
q
r0 (X 0 X)
N (0; 1) :
1
r
k)
and e0 e are independently distributed so that
It thus follows that
. q
0 b
r0 (X 0 X)
r
q
e0 e
(n k)
2
Because
p
(n
2
I M=
k)
)
r
2
X0
e0 e/ (n
r0 b
q
r0 (X 0 X)
1
r
=p
r0 b
e0 e/ (n
q
k) r0 (X 0 X)
we obtain the desired conclusion.
1
r
?
e0 e
2
:
r0 b
r0 b
q
= q
1
0
0
k) r (X X) r
s r0 (X 0 X)
1
r
=q
r0 b
r0 s2 (X 0 X)
1
r
t (n
1
r
r0 b
= p
r0
b
r0 Vr
Theorem 31 (95% Con…dence Interval) The random interval
p
p
0b
0b
0
b
b ;
r
t:025 (n k)
r Vr; r + t:025 (n k)
r0 Vr
where t:025 (n k) denotes the upper 2.5 percentile of the t (n
with 95% probability:
p
h
0b
b
Pr r
t:025 (n k)
r0 Vr
r0
r0 b + t:025 (n
31
k) :
k) distribution, contains r0
k)
p
b
r0 Vr
i
= :95
Proof. It follows easily from
r0 b
p
Theorem 32
1
Rb
m
Proof. Observe that
Rb
0
R
0
R
n h
R
r0
b
r0 Vr
1
b 0
RVR
2
1
(X 0 X)
being a function of b, is independent of
k) s2
(n
t (n
=
2
i
k) :
Rb
R0
R
o
e0 e
1
F (m; n
Rb
2
2
R
(n
2
k)
(m) ;
k) :
It thus follows that
Rb
0
R
R (X 0 X)
1
1
R0
Rb
2
m,
R
1
k) s2
(n
n
k
1
(n
2
F (m; n
k)
Because
Rb
=
Rb
0
R
R (X 0 X)
1
1
R0
Rb
2
R
0
R (X 0 X)
1
1
R0
s2
0
1
1
b 0
Rb R
RVR
Rb
m
we obtain the desired conclusion.
m,
R
Rb
R
R
;
=
n
m
=
k
1
Rb
m
k) s2
2
R
0
Rs2 (X 0 X)
Corollary 9
Pr
1
Rb
m
R
0
b 0
RVR
1
Rb
Remark 5 Suppose you want to test
H0 : R = q
vs:
R
F:05 (m; n
k) = :05
HA : R 6= q
We would then reject the null under 5% signi…cance level i¤
1
Rb
m
q
0
b 0
RVR
1
Rb
32
q
F:05 (m; n
k)
1
R0
1
Rb
R
Remark 6 Observe that the statistics in Theorems 29 and 32 are identical except that
replaced by s2 in Theorem 32.
Theorem 33
1
Rb
m
0
q
1
b 0
RVR
Rb
q =
2
is
e0 e)/ m
(e0 e
;
e0 e/ (n k)
where e is the residual from the restricted least squares
Xb)0 (y
min (y
b
subject to
Xb)
Rb = q
Proof. Note that
1
Rb
m
q
0
b
RVR
Recall that
1
0
Rb
q =
e e = e e + Rb
0
Therefore,
1
Rb
m
q
0
0
b 0
RVR
Rb
1
0
q
Rb
h
0
q
1
1
R0
s2
0
R (X X)
q =
R (X 0 X)
1
(e0 e
R
0
i
1
Rb
e0 e)/ m
s2
Rb
q
m
q :
(e0 e
e0 e)/ m
e0 e/ (n k)
=
Corollary 10 Suppose that the …rst column of X consists of 1s. You would want to test
H0 :
2
=
=
= 0:
k
(What are R and q?) We then have
1
Rb
m
q
0
b 0
RVR
1
Rb
q =
R2 / (k 1)
:
(1 R2 )/ (n k)
Proof. We need to obtain e0 e …rst. Note that the constrained least squares problem can
be written as
X
X
min
(yi b1 xi1 0 xi2
0 xik )2 = min
(yi b1 )2 :
b1
b1
i
i
We know that the solution is given by c1 = y. In other words, we have
X
b = (y; 0; : : : ; 0) ;
e0 e =
(yi y)2 = ye0 ye:
i
Our test statistic thus equals
(e
y 0 ye e0 e)/ (k 1)
:
e0 e/ (n k)
33
Note now that
e0 e
ye0 ye
R2 = 1
The test statistic is thus equal to
R2 / (k 1)
:
(1 R2 )/ (n k)
Remark 7 The statistic
R2 / (k 1)
(1 R2 )/ (n k)
is the “F -statistic” reported by many popular softwares.
26
Con…dence Interval for Mean: To be read, not to be
taught in class
Consider the estimation of from n i.i.d. N ( ; 2 ) random variables U1 ; : : : ; Un . It has been
argued before that this is a special example of the classical linear regression model:
y=X +
where
0
1
U1
B
C
y = @ ... A ;
Un
The m.l.e. b equals
0
1
1
B C
X = @ ... A ;
1
(`0 `)
0
B
[email protected]
= ;
1 0
1
`y =n
X
U1
Un
Ui = U
i
the sample average! We also know that the distribution of b equals
N
;
Thus, the 95% con…dence interval for
or
2
(`0 `)
1
=N
;
2
n
is
b
1:96 p ; b + 1:96 p
n
n
U
1:96 p ; U + 1:96 p
n
n
34
1
..
.
1
C
A
What if we are not fortunate enough to know 2 ? Here, we can make use of the fact that U
P
2
and i Ui U are independent of each other. Because of this independence, we know that
U
r
p
P
i
n
2
(Ui
=
U)
2 (n 1)
p U
n p
It thus follows that we can use
U
as the 95% con…dence interval for
27
t:975 (n
t(n
s2
1):
s
1) p
n
in this case.
Problem Set
In this problem set, you are expected to read and replicate results of Mankiw, Romer, and Weil
(1992, QJE ). Use MATLAB.
1. Select observations such that the “nonoil” variable is equal to 1 and discard the rest of
the observation. (How many observations do you have now?) For each country i, create
yi = ln (GDP per working-age person in 1985)
xi1 = 1
xi2 = ln (I / GDP)
xi3 = ln (growth rate of the working age population between 1960 and 1985 + g + )
assuming that g + = :05.
2. Assume that
yi = xi1
where "i
i:i:d:
N (0;
2
1
+ xi2
2
+ xi3
3
+ "i
). Compute the OLS estimator b = b1 ; b2 ; b3
0
for
= ( 1;
2;
0
3) .
Compute the sum of squared residuals e0 e. Compute an unbiased estimator s2 of 2 .
Compute an estimator of the variance-covariance matrix of b. Compute the standard
deviations of b1 , b2 , and b3 . Present 95% con…dence intervals for 1 , 2 , and 3 .
3. Because R2 monotonically increases as more regressors are added to the model, some
other measure has been developed. The adjusted R2 is computed as
2
R =1
sample size 1
1
sample size number of regressors
2
Compute R2 and R .
35
R2
4. You want to estimate the OLS estimator under the restriction that
possibility is to rewrite the model as
yi = xi1
1
+ xi2
xi3
2
2
+ "i = xi1
1
+ (xi2
xi3 )
2
3
2.
=
One
+ "i
and consider the OLS of yi on xi1 and xi2 xi3 . Compute the OLS estimator b1 ; b2
of ( 1 ; 2 ) this way. (Note that this trick is an attempt to compute the restricted least
squares of ( 1 ; 2 ; 3 ) = ( 1 ; 2 ; 2 ) as b1 ; b2 ; b2 .) Also compute the sum of
squared residuals e0 e for this restricted model.
5. The restricted least squares problem in the previous question can be written as
min (y
c
Xc)0 (y
subject to
Xc)
Rc = q
What is R? What is q? In your class, it was argued that the solution to such problem is
given by
h
i 1
b (X 0 X) 1 R0 R (X 0 X) 1 R0
Rb q ;
where b is the OLS estimator for the unrestricted model. Is it equal to b1 ; b2 ;
which you obtained in the previous question? We also learned in class that
i 1
0h
1
e0 e
e0 e = R b q
R (X 0 X) R0
Rb q :
Subtract e0 e from e0 e . Is the di¤erence equal to R b
6. You can test the restriction that
3
=
2
q
0
R (X 0 X)
or r0 = q for
r0 = (0; 1; 1)
1
R0
1
Rb
b2
0
,
q ?
q=0
by computing the t-statistic
q
r0 b
q
s2 r0 (X 0 X)
(2)
1
r
Compute the t-statistic in (2) by MATLAB, and implement the t-test under 5% signi…cance level.
7. You can test the restriction that
3
=
2
or R = q for
R = (0; 1; 1)
q=0
by computing the F -statistic
Rb
q
0
R (X 0 X)
1
R0
1
Rb
q
m
(3)
s2
What is m in this case? Compute the F -statistic in (3) by MATLAB, and implement the
F -test under 5% signi…cance level.
36
8. Compute the square of the t-statistic in (6) and compare it with the F -statistic in (7).
Are they equal to each other? In fact, if R consists of a single row so that r0 = R, then
the square of the t-statistic as in (2) is numerically equivalent to the F -statistic as in (3).
Provide a theoretical proof to such equality.
9. It was shown that the F -statistic computed as in (3) is numerically equal to
e0 e/ (sample size
e0 e)/ m
(e0 e
number of regressors in the unrestricted model)
Compute the F -statistic this way, and see if it is indeed equal to the value you computed
in (h).
10. You can test the restriction in a slightly di¤erent way. Rewrite the model as
yi = xi1
1
+ (xi2
xi3 )
2
+ xi3 (
3
+
2)
+ "i
This could be understood as a regression of yi on (1; xi2 xi3 ; xi3 ). If the restriction is
correct, the coe¢ cient of xi3 in this model should be equal to zero. Therefore, if the
estimated coe¢ cient of xi3 is signi…cantly di¤erent from zero, we can understand it as an
evidence against the restriction, and reject it. Compute the regression coe¢ cient. Compute the t-statistic. Would you accept the restriction or reject it (under 5% signi…cance
level)?
11. Based on their knowledge of capital’s share in income, Mankiw, Romer and Weil (1992)
entertained the multiple hypothesis that ( 2 ; 3 ) = (0:5; 0:5). This could be written as
0 1 0
0 0 1
2
4
1
2
3
3
5=
0:5
0:5
Compute the corresponding F -statistic. Would you accept the null under 5% signi…cance
level?
37
Lecture Note 5: Large Sample Theory I
28
Basic Asymptotic Theory: Convergence in Distribution
De…nition 7 A sequence of distributions Fn (t) converges in distribution to a distribution F (t)
if
lim Fn (t) = F (t)
n!1
at all points of continuity of F (t). With some abuse of terminology, if a sequence of random
vectors zn , whose cdf are Fn (t), converges in distribution to a random variable z with cdf F (t)
if Fn (t) converges in distribution to a distribution F (t). We will sometimes write
d
zn ! z
and call F (t) the limiting distribution of zn .
Theorem 34 A sequence of random variables Yn converges to a constant c if and only if it
converges in distribution to a limiting distribution degenerate at c.
Proof. Suppose that plimn!1 Yn = c. Let Fn ( ) denote the c.d.f. of Yn . It su¢ ces to show
that
lim Fn (y) = 0 if y < c;
= 1 if y > c
n
Assume that y < c. Let = c
Fn (y) = Pr [Yn
y > 0 We then have
y] = Pr [Yn
Now assume that y > c. Let = y
1
Fn (y) = Pr [Yn > y] = Pr [Yn
c
y
c] = Pr [Yn
c
]
Pr [jYn
cj
]!0
c.
c>y
c] = Pr [Yn
c> ]
Pr [jYn
cj > ] ! 0
so that
lim F (y) = 1
n
Now suppose that
lim Fn (y) = 0 if y < c;
n
We have
Pr [jYn
cj > ]
proving the theorem.
Pr [Yn
c
h
] + Pr Yn
38
c>
= 1 if y > c
2
i
= Fn (c
)+1
Fn c +
2
!0
d
Theorem 35 (Continuous Mapping Theorem) If zn ! z and if g ( ) is a continuous funcd
tion, then g (zn ) ! g (z)
d
Theorem 36 (Transformation Theorem) If an ! a and plim bn = b with b constant, then
d
d
d
an + bn ! a + b, an bn ! ab. If b 6= 0, thenan / bn ! a/ b.
Theorem 37 (Central Limit Theorem) Given an i.i.d. sequence zi of random vectors with
E [zi ] = …nite and Var (zi ) = …nite positive de…nite, we have
p
d
n (z
) ! N (0; ) :
Remark 8 Consider n i.i.d. random variables with unknown mean
2
. By the law of large numbers, we have
X
n 1
Xi2 ! E Xi2 ; X ! E [Xi ]
and unknown variance
i
It follows that
^2 = n
1
X
Xi2
i
With the Slutsky Theorem, we obtain
p X
n
2
2
X !
! N (0; 1)
^
In particular, we obtain
Pr
1:96 < n1=2
X
^
< 1:96 ! :95
so that we can use
^
^
1:96 p ; X + 1:96 p
n
n
as the asymptotic 95% con…dence interval for
even when Xi does not necessarily have a
normal distribution.
X
Theorem 38 If
d
and z2;n ! z;
plim z1;n = c;
n!1
then
d
z1;n + z2;n ! c + z;
d
z1;n z2;n ! c z:
Theorem 39 (Delta Method) Suppose that
p
d
n (zn c) ! N (0;
2
)
Also suppose that g ( ) is continuously di¤erentiable at c. We then have
p
d
n (g (zn ) g (c)) ! N (0; g 0 (c)2 2 )
39
Sketch of Proof. We have
g (c) = g 0 (e
c) (zn
g (zn )
c)
for some e
c between zn and c. Because zn ! c in probability, and because g 0 ( ) is continuous,
we have g 0 (e
c) ! g 0 (c). Writing
p
p
n (g (zn ) g (c)) = g 0 (e
c)
n (zn c)
we obtain the desired conclusion.
Theorem 40 (Multivariate Delta Method) Suppose that
p
d
c) ! N (0; )
n (zn
Also suppose that g ( ) is continuously di¤erentiable at c. We then have
p
d
g (c)) ! N (0; G G0 )
n (g (zn )
where
@g (c)
@z 0
G
29
Classical Linear Regression Model III
29.1
Model
yi = x0i + "i
(x0i ; "i ) i = 1; 2; : : : is i.i.d.
xi ?"i
2
E ["i ] = 0, Var ("i ) =
E [xi x0i ] is positive de…nite. Furthermore, all of its elements are …nite.
29.2
Some Auxiliary Lemmas
Lemma 19
1X
xi x0i
n i=1
n
b=
Lemma 20
s2 =
1
n
k
n
X
i=1
"2i
2
1
n
k
+
n
X
i=1
"i x0i
!
!
b
1
!
n
1X
x i "i :
n i=1
+ b
40
1
0
n
k
n
X
i=1
xi x0i
!
b
:
Lemma 21 plimn!1
1
n
Lemma 22 plimn!1
1
n
Lemma 23 plimn!1
1
n
Lemma 24
Pn
i=1
Pn
i=1
Pn
i=1
xi x0i = E [xi x0i ]
x i "i = 0
"2i =
p
2
1X
d
n
xi "i ! N 0;
n i=1
n
2
E [xi x0i ] :
Proof. Let zi = xi "i . We have
E [zi ] = E [xi "i ] = E [xi ] E ["i ] = 0;
Var (zi ) = E "2i xi x0i = E "2i
E [xi x0i ] =
2
E [xi x0i ] :
Lemma 25 limn!1 t:975 (n) = 1:96
30
Large Sample Property of OLS
if plimn!1 b = .
De…nition 8 An estimator b is consistent for
Theorem 41 b is consistent for .
Theorem 42 plimn!1 s2 =
Theorem 43
p
Proof. We have
p
But we have
2
d
n b
! N 0;
1X
xi x0i
n i=1
n
n b
=
1X
xi x0i
n i=1
n
plim
n!1
and
!
2
1
(E [xi x0i ])
!
1
!
n
1 X
p
x i "i :
n i=1
1
1 X
d
p
xi "i ! N 0;
n i=1
= [Exi x0i ]
1
n
41
2
:
Exi x0i :
Theorem 44 Suppose that g : Rk ! R is continuously di¤erentiable at
@g (c)
:
@c0
(c)
Then,
p
n g b
such that
d
2
g ( ) ! N 0;
1
( ) (E [xi x0i ])
( )0 :
Proof. Delta Method.
p
b is a valid approximate 95% con…dence interval for r0 .
Theorem 45 r0 b 1:96 r0 Vr
Proof. We have
p
Now, notice that
n r0 b
r0
d
! N 0;
v "
#
u
n
u
X
1
plim tr0
xi x0i
n i=1
n!1
and
2
r0 [Exi x0i ]
1
r :
q
1
1
r
r0 [Exi x0i ]
1
r0 [Exi x0i ]
r=
p
plim s2 = :
n!1
Thus, we have
It thus follows that
p
n r
0b
v "
#
u
n
u
X
1
plim s tr0
xi x0i
n
n!1
i=1
r
0
v "
#
u
n
. u
X
1
str0
xi x0i
n i=1
1
r=
1
r=q
and
lim Pr r
n!1
0b
q
1:96 s2 r0 (X 0 X)
1
r
r
q
0
r
0b
r0 b
r:
r0
s2 r0 (X 0 X)
d
1
r
! N (0; 1)
q
+ 1:96 s2 r0 (X 0 X)
1
r = :95
Theorem 46 Suppose that g : Rk ! R is continuously di¤erentiable at . Then,
r
0
b V
b
b
g b
1:96
is a valid approximate 95% con…dence interval for g ( ).
42
Proof. We have
p
n g b
d
2
g ( ) ! N 0;
and
( ) (E [xi x0i ])
1
( )0 ;
p
plim s2 = :
n!1
Because
is continuous and plim b = , we should have
b =
plim
n!1
from which we obtain
v
" n
#
u
u
X
1
b
xi x0i
plim t
n
n!1
i=1
1
b =
It thus follows that
p
n g b
and
"
lim Pr g b
n!1
v
u
. u
g( )
st
31
b
r
t:025 (n
1X
xi x0i
n i=1
n
b (X 0 X)
1:96 s2
Theorem 47 r0 b
"
k)
Problem Set
#
( )
q
( ) (E [xi x0i ])
1
b =r
b
1
g( )
s2
1
g b
( )0 ;
g( )
b (X 0 X)
r
g b + 1:96 s2
1
d
b
0
! N (0; 1)
b (X 0 X)
p
b is a valid approximate 95% con…dence interval for r0 .
r0 Vr
Mankiw, Romer, and Weil (1992, QJE ) considered the regression
yi = xi1
with the restriction
3
=
2,
1
+ xi2
2
+ xi3
3
+ "i
where
yi = ln (GDP per working-age person in 1985)
xi1 = 1
xi2 = ln (I / GDP)
xi3 = ln (growth rate of the working age population between 1960 and 1985 + g + )
In other words, they regressed yi on xi1 and xi2 xi3 . They noted that the coe¢ cient of xi2
in this restricted regression is an estimator of 1 , where is capital’s share in income.
43
xi3
1
b
#
= :9
1. In Table 1, Mankiw, Romer, and Weil (1992) report an estimator b of implied by the
OLS coe¢ cient of xi2 xi3 . Con…rm their …ndings with the data set provided for the
three samples.
2. Mankiw, Romer, and Weil (1992) also report the standard deviation of b. Using deltamethod, con…rm their results.
44
Lecture Note 6: Large Sample Theory II
32
32.1
Linear Regression with Heteroscedasticity: Classical
Linear Regression Model IV
Model
yi = x0i + "i
(x0i ; "i ) i = 1; 2; : : : is i.i.d.
E ["i j xi ] = 0
2
Var ( "i j xi )
(xi ) not known
E [xi x0i ] is positive de…nite.
We are going to consider the large sample property of
!
! 1
n
n
X
X
1
1
0
b= +
xi xi
x i "i
n i=1
n i=1
and the related inference.
32.2
Some Useful Results
Lemma 26
1X
x i "i = 0
plim
n!1 n
i=1
n
Proof.
1X
xi "i = E [xi "i ] = E [E [xi "i j xi ]] = E [xi E ["i j xi ]] = E [xi 0] = 0:
n!1 n
i=1
n
plim
Lemma 27
1X 2 0
plim
"i xi xi = E xi x0i "2i :
n
n!1
i=1
n
45
Lemma 28
p
1X
d
n
xi "i ! N 0; E "2i xi x0i
n i=1
n
:
Proof. Let
zi
x i "i :
We have
Var (zi ) = E "2i xi x0i :
E [zi ] = 0;
Apply CLT.
Lemma 29 Under reasonable conditions, we have
1X 2 0
ei xi xi = E "2i xi x0i
n!1 n
i=1
n
plim
Proof. For simplicity, I will assume that xi is a scalar. It su¢ ces to prove that
1X 2 2
xi ei
plim
n!1 n
i=1
n
x2i "2i = 0:
We have
2"i xi b
= "2i
so that
e2i
"2i =
2"i xi b
1X 2 2
x e
n i=1 i i
n
2
2 "i x i b
2
+ jxi j2 b
1X 2 2
x e
n i=1 i i
2
+ x2i b
+ x2i b
2 j"i j jxi j b
Thus,
2
xi b
e2i = "i
:
2
+ x2i b
n
"2i
"2i
2X
j"i j jxi j3
n i=1
n
!
! 0:
1X
jxi j4
n i=1
n
b
+
if
E j"i j jxi j3 < 1;
46
E jxi j4 < 1:
!
b
2
32.3
Large Sample Property
Theorem 48
plim b = :
n!1
Proof.
1X
xi x0i
n i=1
n
plim
n!1
!
1
1X
x i "i
n i=1
n
Theorem 49
!
1X
xi x0i
n i=1
n
= plim
n!1
p
where
n b
[Exi x0i ]
But we have
n b
plim
n!1
and
n
plim
n!1
1
1
E "2i xi x0i [Exi x0i ]
n
1X
xi x0i
n i=1
!
!
1
:
!
n
X
1
p
x i "i :
n i=1
1
= [Exi x0i ]
1 X
d
p
xi "i ! N 0; E "2i xi x0i
n i=1
1
n
Theorem 50 Let
Then,
1X
x i "i
n i=1
d
n
1X
xi x0i
n i=1
=
1
! N (0; ) ;
Proof. We have
p
!
"
n
X
bn = 1
xi x0i
n i=1
#
1
"
n
1X 2 0
e xi xi
n i=1 i
#"
plim b n = :
:
n
1X
xi x0i
n i=1
#
1
:
n!1
Theorem 51 Let
We then have
c
W
" n
#
X
1b
xi x0i
n =
n
i=1
r0 b r0
p
r0 c
Wr
1
"
n
X
i=1
e2i xi x0i
#"
! N (0; 1)
47
n
X
i=1
xi x0i
#
1
!
= E [xi x0i ]
1
0 = 0:
Proof.
Corollary 11 r0 b
r0 b r0
p
r0 c
Wr
r0 b
= q
p
r0
1 0b
r nr
n
=
n r0 b r0
q
r0 b n r
! N (0; 1)
p
1:96 r0 c
Wr is a valid approximate 95% con…dence interval for r0 .
Remark 9 For many practical purposes, we may understand c
W as equal to Var b
33
Problem Set
Mankiw, Romer, and Weil (1992, QJE ) considered the regression
yi = xi1
with the restriction
3
=
2,
1
+ xi2
2
+ xi3
3
+ "i
where
yi = ln (GDP per working-age person in 1985)
xi1 = 1
xi2 = ln (I / GDP)
xi3 = ln (growth rate of the working age population between 1960 and 1985 + g + )
In other words, they regressed yi on xi1 and xi2 xi3 . They noted that the coe¢ cient of xi2
in this restricted regression is an estimator of 1 , where is capital’s share in income.
xi3
1. Using White’s formula, construct a 95% con…dence intervals for the coe¢ cients of xi1 and
xi2 xi3 .
2. In Table 1, Mankiw, Romer, and Weil (1992) report an estimator b of implied by the
OLS coe¢ cient of xi2 xi3 . Con…rm their …ndings with the data set provided for the
three samples.
3. Mankiw, Romer, and Weil (1992) also report the standard deviation of b. Combining
White’s formula with delta-method, construct a 95% con…dence interval for .
48
Lecture Note 7: IV
34
Omitted Variable Bias
Suppose that
yi = xi + wi + "i :
We do not observe wi . Our object of interest is .
34.1
Bias of OLS
What will happen if we regress yi on xi alone?
P
P
P
b = Pi xi yi = + Pi xi wi + Pi xi "i = +
2
2
2
i xi
i xi
i xi
1
n
Letting
P
xi wi
Pi 2 +
1
i xi
n
P
x i "i
Pi 2 !
i xi
1
n
1
n
+
E [xi wi ]
E [x2i ]
E [xi wi ]
;
E [x2i ]
we have
plim b =
Because
+
:
P
xw
Pi i 2 i ! ;
i xi
we can interpret as the probability limit of the OLS when wi is regressed on xi : Unless wi
and xi are uncorrelated, will be nonzero, and OLS will be biased.
34.2
IV Estimation
Suppose that we also observe zi such that
E [zi wi ] = 0
E [zi "i ] = 0
E [zi xi ] 6= 0
Note that we can regard
ui
wi + "i
the “error term”in the regression of yi on xi . Also note that
E [zi ui ] = E [zi wi ] + E [zi "i ] = 0
49
Consider
bIV
P
zy
Pi i i ;
i zi xi
which is obtained by replacing xi by zi in the numerator and the denominator of the OLS
formula. Observe that
P
P
z
(x
+
u
)
zi ui
E [zi ui ]
i
i
i
i
bIV =
P
= + Pi
! +
= :
E [zi xi ]
i zi xi
i zi xi
We thus have
35
plim bIV = :
Problem Set
Suppose that
yi = x0i + wi0 + "i :
We do not observe wi . Our object of interest is . We assume that
E [xi "i ] = 0;
E [wi "i ] = 0
What is the probability limit of the OLS estimator of if we regress yi on xi alone? Hint:
!
! 1
n
n
X
1
1X
xi x0i
xi yi
bOLS =
n i=1
n i=1
! 1
!
! 1
!
n
n
n
n
X
X
X
X
1
1
1
1
= +
xi x0i
xi wi0
+
xi x0i
x i "i
n i=1
n i=1
n i=1
n i=1
When is the OLS estimator consistent for ?
36
Errors in Variables
For simplicity, we assume that every random variable is a zero mean scalar random variable.
Suppose that
+ "i
yi = xi
We do not observe xi . Instead, we observe a proxy
xi = xi + ui
50
36.1
Bias of OLS
What would happen if we regress yi on xi ?
Condition 1 E ["i ui ] = 0; E ["i xi ] = 0; E [ui xi ] = 0
Observe that
yi = xi
+ ("i
ui ) :
We thus have
P
b = Pi xi yi =
2
i xi
But,
+
P
i
xi ("i
P
ui )
2
i xi
!
+
E [xi ("i
ui )]
:
2
E [xi ]
E x2i = E xi 2 + E u2i
and
E [xi ("i
We thus have
bias toward zero.
36.2
plim b =
ui )] = E [(xi + ui ) ("i
E [u2i ]
=
E [xi 2 ] + E [u2i ]
ui )] =
E u2i :
E [xi 2 ]
;
E [xi 2 ] + E [u2i ]
IV estimation
Suppose that we also observe
zi = xi + vi ;
where
Condition 2 vi is independent of xi ; "i ; ui
Consider
bIV
P
zy
Pi i i ;
i zi xi
which is obtained by replacing xi by zi in the numerator and the denominator of the OLS
formula. Observe that
P
P
z
[x
+
("
u
)]
zi ("i
ui )
E [zi ("i
ui )]
i
i
i
i
bIV = i
P
= + i P
! +
:
E [zi xi ]
i zi xi
i zi xi
But,
E [zi ("i
ui )] = E [(xi + vi ) ("i
We thus have
plim bIV = :
51
ui )] = 0:
37
37.1
Simultaneous Equation: Identi…cation
A Supply-Demand Model
qd =
1
p+
qs =
1
p
d
y+
2
+
(Demand)
s
x+
3
(Supply)
q = qd = qs
1
1
p
q
1
1
1
=
1
p
q
1
1
0
2y
2 1y
1
+
3
+
3
d
1 3x
1
+
s
d
+
s
1
y
x
3x
+
d
y
x
0
2
1
1
=
0
1
1
1
0
2
+
(Equilibrium)
1
1
s
=0
1
1
1
d
1
s
1
:
Important observation: Both p; q in equilibrium will be correlated with d ; s .
What happens in the “demand”regression? The probability limit will equal
E [p2 ] E [py]
E [py] E [y 2 ]
1
E [pq]
E [yq]
=
E p
E
=
E [p2 ] E [py]
E [py] E [y 2 ]
+
2
But
d
1
h
d 2
i
E
1
1
1
E p
E y
d
d
d s
6= 0
in general!
37.2
General Notation
Individual observation consists of
(wi0 ; zi0 )
Here, wi denote the vector of endogenous variables, and zi denote the vector of exogenous
variables. We assume that there is a linear relationship:
w0 = z 0 B + 0 ;
where
denotes the vector of “errors”. Our assumption is that z is uncorrelated with :
E [z 0 ] = 0:
52
37.3
Identi…cation
Suppose that we know the exact population joint distribution of (w0 ; z 0 ). Can we compute
and B from this distribution? In many cases, we are not interested in the estimation of the
whole system. Rather we are interested in the estimation of just one equation. Assume without
loss of generality that it is the …rst equation. Write
w1 =
Here, I assume that
0
w +
0
1
w1
w = @ w A;
w0
0
1
z +
:
z
z0
z=
:
Our restriction that E [z 0 ] = 0 implies that
E zw1 = E [zw 0 ] + E [zz 0 ]
= [E [zw 0 ] ; E [zz 0 ]]
A necessary condition for the identi…cation of and
the dimension of x is bigger than that of ( 0 ; 0 )0 :
dim (z)
from this system of linear equations if
dim ( ) + dim ( ) = dim (w ) + dim (z )
But
dim (z) = dim (z ) + dim z 0 :
We thus have
dim z 0
37.4
dim (w ) :
General Identi…cation
We have
y = x0 + u;
where the only restriction given to us is that
E [zu] = 0:
Identi…cation:
E [zy] = E [zx0 ]
A necessary condition for identi…cation is
dim (z)
dim ( )
Remark 10 You would like to make sure that the rank of the matrix E [zx0 ] is equal to dim ( )
as well, but it is not very important at the …rst year level. It will become important later.
53
37.5
Estimation with ‘Exact’Identi…cation
When dim (z) = dim ( ), we say that the model is exactly identi…ed. We can then see that the
matrix E [zx0 ] is square, and that
= (E [zx0 ])
1
E [zy]
By exploiting the law of large numbers, we can construct a consistent esitmator of :
! 1
!
! 1
!
n
n
n
n
X
X
X
X
1
1
zi x0i
zi yi =
zi x0i
zi yi = bIV !
n i=1
n i=1
i=1
i=1
The IV estimator is usually written in matrix notations:
38
bIV = (Z 0 X)
1
Z 0y
Asymptotic Distribution of IV Estimator
Theorem 52 Suppose that zi is independent of "i . Then
where
2
"
= Var ("i ).
p
n bIV
! N 0;
2
"
(E [zi x0i ])
1
E [zi zi0 ] (E [xi z
Proof. Problem Set.
39
Problem Set
All questions here are taken from Greene. ys denote endogenous variables, and xs denote
exogenous variables.
1. Consider the following two-equation model:
y1 =
1 y2
+
11 x1
+
21 x2
+
31 x3
+ "1
y2 =
2 y1
+
12 x1
+
22 x2
+
32 x3
+ "2
(a) Verify that, as stated, neither equation is identi…ed.
(b) Establish whether or not the following restrictions are su¢ cient to identify (or partially identify) the model:
i.
21
=
32
=0
ii.
12
=
22
=0
iii.
1
=0
iv.
2
=
v.
21
+
1
and
22
32
=0
=1
54
2. Examine the identi…ability of the following supply and demand model:
(Demand)
ln Q =
0
+
1
ln P +
2
ln (income) + "1
ln Q =
0
+
1
ln P +
2
ln (input cost) + "2
(Supply)
3. Consider a linear model
yi = x0i
+ "i
m 1
with the restriction that
E
zi "i
m 1
Derive the asymptotic distribution of IV
! 1
n
X
1
bIV =
zi x0i
n i=1
=0
1X
zi yi
n i=1
n
under the assumption that E (zi "i ) (zi "i )0 =
(b) Show that
n bIV
=
n
1X 0
zi xi
n i=1
1X 0
zi xi
n i=1
n
converges to
(E [zi x0i ])
= (Z 0 X)
1
!
!
1
1 X
p
zi "i
n i
1
1
in probability.
(c) Show that
1 X
p
zi "i
n i
converges in distribution to
N 0;
(d) Conclude that
p
n bIV
N 0;
2
"E
[zi zi0 ]
converges in distribution to
2
"
(E [zi x0i ])
55
1
Z 0y
[zi zi0 ].
2
"E
(a) Show that
p
!
E [zi zi0 ] (E [xi zi0 ])
1
!
Lecture Note 8: MLE
40
MLE
We have a collection of i.i.d. random vectors Zi i = 1; : : : ; n such that
Zi
f (z; )
Here, f (z; ) denotes the (common) pdf of Zi . Our objective is to estimate .
De…nition 9 (MLE) Assume that Z1 ; : : : ; Zn are i.i.d. with p.d.f. f (zi ;
maximizes the likelihood:
b = argmax
c
40.1
n
Y
f (Zi ; c) = argmax
c
i=1
n
X
0 ).
The MLE b
log f (Zi ; c) :
i=1
Consistency
Theorem 53 (Consistency) Assume that Z1 ; : : : ; Zn are i.i.d. with p.d.f. f (zi ;
plimn!1 b = under some suitable regularity conditions.
0 ).
Then,
Below we provide an elementary proof of consistency. Write h (Zi ; c) = log f (Zi ; c) for
simplicity of notation. We assume the following:
Condition 3 There is a unique
0
2
such that
max E [h (Zi ; c)] = E [h (Zi ;
0 )]
c2
Remark 11 Because log is a concave function, we can use Jensen’s Inequality and conclude
that
f (Zi ; c)
f (Zi ; c)
E [log f (Zi ; c)] E [log f (Zi ; 0 )] = E log
log E
f (Zi ; 0 )
f (Zi ; 0 )
But
f (Zi ; c)
E
=
f (Zi ; 0 )
and hence
Z
f (z; c)
f (z;
f (z; 0 )
E [log f (Zi ; c)]
0 ) dz
E [log f (Zi ;
=
Z
0 )]
f (z; c) dz = 1
log (1) = 0
for all c. In other words,
E [log f (Zi ; c)]
E [log f (Zi ;
for all c.
56
0 )]
Condition 4 De…ne B ( )
f 2
:j
0j
g. For each
max E [h (Zi ; c)] < E [h (Zi ;
c2B( )
Condition 5 maxc2
1
n
Pn
i=1
> 0,
0 )] :
E [h (Zi ; c)] ! 0 almost surely.
h (Zi ; c)
Sketch of Proof. Fix > 0. Let
= E [h (Zi ;
and note that
and
0 )] ;
= max E [h (Zi ; c)] :
c2B( )
< . But
1X
max
h (Zi ; c) ! max E [h (Zi ; c)]
c2B( ) n
c2B( )
i=1
n
1X
h (Zi ; c) ! max E [h (Zi ; c)]
max
c2
c2
n i=1
n
almost surely. It follows that b 62 B( ) for n su¢ ciently large. Thus,
lim b
n!1
0
almost surely
<
Since the above statement holds for every > 0, we have b !
40.2
0
almost surely.
Fisher Information with One Observation
Remark 12 Without loss of generality, we omit the i subscript in this section.
Assumption Z
f (z; ),
2
De…nition 10 (Score) s (z; )
.
@ log f (z; )/ @
Lemma 30 E [s (Z; )] = 0
Proof. Because
1=
we have
0=
Z
@f (z; )
dz =
@
Z
Z
f (z; ) dz
@f (z; )
@
f (z; )
f (z; ) dz =
Z
s (z; ) f (z; ) dz
De…nition 11 (Fisher Information)
Z
I ( ) = s (z; ) s (z; )0 f (z; )dz = E s (Z; ) s (Z; )0 :
57
Theorem 54
I( )=
Z
@ 2 log f (z; )
f (z; )dz =
@ @ 0
Proof. Because
0=
we have
Z
Z
@s (z; )
f (z; ) dz +
@ 0
@ 2 log f (Z; )
:
@ @ 0
E
s (z; ) f (z; ) dz
Z
@f (z; )
dz
@ 0
Z
Z
@f (z; )
@
@ 2 log f (z; )
0
=
f (z; ) dz + s (z; ) @
f (z; ) dz
0
@
@
f (z; )
Z 2
Z
@ log f (z; )
=
f (z; ) dz + s (z; ) s (z; )0 f (z; ) dz
@ @ 0
0=
s (z; )
Example 2 Suppose that X
N ( ; 2 ). Assume that 2 is known. The Fisher information
I ( ) can be calculated in the following way. Notice that
#
"
1
(x
)2
f (x; ) = p exp
2 2
2
so that
)2
(x
log f (x; ) = C
2 2
where C denotes the part of the log f wich does not depend on . Because
s(x; ) =
we have
I ( ) = E s (X; )2 =
Remark 13 In the multivariate case where
s (x; ) =
@ log f (x; )
@
E
2
1
4
)2 =
E (X
1
2
= ( 1 ; : : : ; K ), we let
0 @ log f (x; ) 1
and
I ( ) = E s (X; ) s (X; )0 =
x
B
@
@
1
@ log f (x; )
@ K
@ 2 log f (x; )
=
@ @ 0
58
..
.
2
6
E4
C
A:
@ 2 log f (X; )
@ [email protected] 1
@ 2 log f (X; )
@ [email protected] K
@ 2 log f (X; )
@ [email protected] 1
@ 2 log f (X; )
@ [email protected] K
3
7
5
Example 3 Suppose that X is from N ( 1 ;
log f (x;
1;
2) =
Then,
2 ).
1
log (2
2
2
so that
s (x;
1; 2)
=
@ log f [email protected]
@ log f [email protected]
2
1)
(x
2)
x
1
=
1
2 2
from which we obtain
E [ss0 ] =
2
0
1
2
2
1
2
+
(x
2
1)
2
2
2
!
0
1
2 22
In this calculation, I used the fact that E [Z 2m ] = (2m)!= (2m m!) and E [Z 2m 1 ] = 0 if Z
N (0; 1).
40.3
Random Sample
Assumption Z1 ; : : : ; Zn are i.i.d. random vectors
f (z; ).
Proposition 1 Let In ( ) denote the Fisher Information in Z1 ; : : : ; Zn . Then,
In ( ) = n I ( ) ;
where I ( ) is the Fisher Information in Zi .
40.4
Limiting Distribution of MLE
Theorem 55 (Asymptotic Normality of MLE)
p
n b
d
! N (0; I
1
( ))
Sketch of Proof. We will assume that the MLE is consistent. By the FOC, we have
0=
n @ log f Z ; b
X
i
@
i=1
0
1
n
n @ 2 log f Z ; e
X
X
i
@ log f (Zi ; ) @
A b
=
+
0
@
@
@
i=1
i=1
where the second equality is justi…ed by the mean value theorem. Here, the e is on the line
segment adjoining b and . It follows that
p
n b
=
0
1
n @ 2 log f Z ; e
X
i
@1
A
n i=1
@ @ 0
59
1
n
1 X @ log f (Zi ; )
p
@
n i=1
!
It can be shown that, under some regularity conditions,
2
e
n
1 X @ log f Zi ;
n i=1
@ @ 0
1 X @ 2 log f (Zi ; )
!0
n i=1
@ @ 0
n
in probability. Because
1 X @ 2 log f (Zi ; )
@ 2 log f (Zi ; )
!
E
n i=1
@ @ 0
@ @ 0
n
in probability, we conclude that
2
e
n
1 X @ log f Zi ;
@ 2 log f (Zi ; )
plim
=
E
=
0
0
@
@
@
@
n!1 n
i=1
We also note that
I( )
1 X @ log f (Zi ; ) d
p
! N (0; I ( ))
@
n i=1
(4)
n
(5)
by the central limit theorem. Combining (4) and (5) with Slutzky Theorem, we obtain the
desired conclusion.
Remark 14 How do we prove
2
e
n
1 X @ log f Zi ;
n i=1
@ @ 0
Here’s one way. Assume that
1 X @ 2 log f (Zi ; ) p
! 0?
n i=1
@ @ 0
n
is a scalar, so what we need to prove is
2
e
n
1 X @ log f Zi ;
n i=1
@ 2
1 X @ 2 log f (Zi ; ) p
!0
n i=1
@ 2
n
Note that we have by the mean value theorem
e
n
1 X @ log f Zi ;
n i=1
@ 2
2
0
1
ee
3
@
log
f
Z
;
n
n
i
X
C
1 X @ 2 log f (Zi ; ) B
B1
C e
=
@n
A
n
@ 2
@ 3
i=1
i=1
e
for some e in between e and . Now assume that
sup
@ 3 log f (Zi ; )
@ 3
60
M (Zi )
and that E [M (Zi )] < 1. Then we have
2
e
n
1 X @ log f Zi ;
n i=1
@ 2
n
1 X @ 2 log f (Zi ; )
n i=1
@ 2
!
n
X
1
M (Zi ) e
n i=1
!
n
1X
M (Zi ) b
n i=1
where the second inequality used the fact that e is on the line segment adjoining b and and
hence the distance between e and should be smaller than that b and . By the law of large
numbers, we have
n
1X
p
M (Zi ) ! E [M (Zi )]
n i=1
By consistency, we also have b
p
! 0. The conclusion then follows by Slutzky.
Theorem 56 (Estimation of Asymptotic Variance 1)
0
1
n @ log f Z ; b @ log f Z ; b
X
i
i
1
A
plim Vb1 plim @
n i=1
@
@ 0
n!1
n!1
Theorem 57 (Estimation of Asymptotic Variance 2)
1
0
n @ 2 log f Z ; b
X
i
1
A
plim Vb2 plim @
0
n i=1
@ @
n!1
n!1
1
=I
1
( ):
1
=I
1
( ):
Theorem 58 (Approximate 95% Con…dence Interval) For simplicity, assume that dim ( ) =
1. We have
q
q 3
2
Vb1
Vb
b + 1:96 p 1 5 = :95
lim Pr 4 b 1:96 p
n!1
n
n
q
q 3
2
b
V2
Vb
b + 1:96 p 2 5 = :95
lim Pr 4 b 1:96 p
n!1
n
n
Remark 15 Approximate 95% con…dence interval may therefore be constructed as
q
q
b
V
Vb2
b 1:96 p 1 ;
b
or
1:96 p
n
n
.
.
Many softwares usually report Vb1 n or Vb2 n, and call it the (estimated) variance. Therefore,
p
you do not need to make any adjustment for n with such output.
61
41
Latent Utility
For simplicity of notation, assume that x0i is nonstochastic. We have
U1i = x0i
1
+ u1i :
Choice 1
U0i = x0i
0
+ u0i :
Choice 0
Choice 1 is made if and only if
Ui
U1i
U0i = x0i (
0)
1
+ (u1i
u0i )
0:
or
x0i
"i
0:
Otherwise, choice 0 is made.
Example 4 Suppose (u1 ; u0 ) has a bivariate normal distribution. Then, "i has a normal distribution.
Example 5 Suppose u1 and u0 are i.i.d. with the common c.d.f. F (z) = exp [ exp ( x)].
Then,
et
Pr [" t] = t
:
e +1
(Proof omitted.)
42
Binary Response Model
Two Choices:
yi = 1 Choice 1 is made
yi = 0 Choice 0 is made
Assume that
yi = 1 , Ui = x0i
"i
0
Let
G (t)
Pr ["i
t] :
Then,
Pr [yi = 1] = Pr ["i
Example 6 G (t) =
Example 7 G (t) =
(t): Probit Model
et
et +1
=
(t): Logit Model
62
x0i ] = G (x0i ) :
Note that individual likelihood equals
y
1 yi
G (x0i ) i [1
G (x0i )]
It follows that the joint log likelihood equals
X
yi log G (x0i ) + (1
yi ) log [1
G (x0i )]
i
MLE from FOC:
X
i
G x0i b
i g x0i b
h
0b
0b
1 G xi
G xi
yi
xi = 0:
Proposition 2 The log likelihood of the Probit or Logit model is globally concave.
Proof. Exercise.
Proposition 3
p
n b
d
! N 0; I
1
( ) :
Proposition 4 The Fisher Information I ( ) from the individual observation equals
"
#
g (x0i )2
E
xi x0i :
0
0
G (xi ) [1 G (xi )]
Proof. Obvious from
@ log f (zi ; )
yi G (x0i )
=
g (x0i ) xi
@
G (x0i ) [1 G (x0i )]
and
I( )=E
@ log f @ log f
:
@
@ 0
Proposition 5
2
g x0i b
1X
h
i xi x0i = I ( ) ;
plim
n i G x0 b 1 G x0 b
i
i
h
i2
yi G x0i b
2
1X
0
plim
xi x0i = I ( ) :
i2 g xi b
2h
n i
G x0i b
1 G x0i b
63
43
Tobit Model (Censoring)
Suppose
yi = x0i + "i
"i j x i
N 0;
2
We observe (yi ; Di ; xi ),where
Di = 1 (yi > 0)
y i = Di y i
Individual Likelihood:
x0i
1 Di
1
yi
x0i
Di
= 1
x0i
1 Di
1
yi
x0i
Di
MLE is not simple because the likelihood is not concave in parameters: We have to deal with
local vs. global maxima problem!
43.1
Bias of OLS
Consider regressing yi on xi for those observations with Di = 1. For simplicity, assume that
is a scalar. In this case, we may write
P
Di x i y i
:
b= P
Di x2i
Lemma 31
plim b =
n!1
E [Pr [Di = 1j xi ] E [yi j Di = 1; xi ] xi ]
:
E [Pr [Di = 1j xi ] x2i ]
Proof. Denominator:
1X
Di x2i = E Di x2i = E E [Di j xi ] x2i = E Pr [Di = 1j xi ] x2i
plim
n!1 n
Numerator:
1X
Di xi yi = E [Di yi xi ] = E [E [Di yi j xi ] xi ] ;
n!1 n
plim
and
E [Di yi j xi ] = E [1 yi j Di = 1; xi ] Pr [Di = 1j xi ]
+ E [0 yi j Di = 0; xi ] Pr [Di = 0j xi ]
= E [1 yi j Di = 1; xi ] Pr [Di = 1j xi ]
64
Corollary 12 b is consistent only if
E [yi j Di = 1; xi ] = x0i :
Lemma 32 Suppose u
N (0; 1). Then,
E [uj u > t] =
(t)
(t)
1
Proof.
Rs
Rs
(x) dx
(x) dx
t
Pr [u sj u > t] = R 1
= t
:
1
(t)
(x)
dx
t
Thus, conditional p.d.f. of u at s given u > t equals
Rs
(x) dx
d
d
(s)
t
Pr [u sj u > t] =
=
:
ds
ds
1
(t)
1
(t)
It follows that
E [ uj u > t] =
Z
t
1
s
1
(s)
ds =
(t)
1
Z
1
(t)
1
s
(s) ds:
t
But from
d
(s) = s (s) ;
ds
we have
Z 1
Z 1
d
(s) ds =
s (s) ds =
ds
t
t
from which the conclusion follows.
(s)j1
t =
Lemma 33
x0i
E [yi j xi ; Di = 1] = x0i +
x0i
:
Proof.
x0i ]
E ["i j xi ; Di = 1] = E ["i j xi ; "i
=
E
"i
xi ;
"i
x0i
=
x0i
1
x0i
=
x0i
Corollary 13 b is inconsistent.
65
:
x0i
(t) ;
43.2
Heckman’s Two Step Estimator
For notational simplicity, I will drop xi in the conditioning event. We know that
E [ yi j Di = 1] = x0i +
where
x0i
;
(s)
:
(s)
(s)
Thus, if is known, we can estimate consistently by regressing yi on xi and
x0i
.
Observe that can be consistently estimated by the Probit MLE of D on x: we have
Di = 1 i¤ x0i
and
"i
+
"i
>0
N (0; 1) :
This suggests two step estimation:
Obtain MLE of d from the Probit model
x0i d
Regress yi on xi and
44
Sample Selection Model
Suppose
yi = x0i + ui
We observe (yi ; Di ; xi ),where
Di = 1 (zi0 + vi > 0) ;
yi = Di yi
Our goal is to estimate . We assume
ui
vi
xi
N
0
0
2
u
;
u v
2
v
u v
:
By the same reasoning as in the censoring case, yi regressed on xi for the subsample where
Di = 1 will result in an inconsistent estimator. To …x this problem, we can either rely on MLE,
or we can use the two step estimation.
Lemma 34
E [yi j Di = 1] = x0i +
66
u
zi0
:
v
Proof. Note that
E [yi j Di = 1] = E [x0i + ui j zi0 + vi > 0]
zi0 ]
= x0i + E [ui j vi >
Now recall that
u
wi = ui
vi
v
is independent of vi . We thus have
u
zi0 ] = E
E [ ui j v i >
=
vi
E
u
vi >
v
=
vi vi
E
u
v
=
zi0
vi + wi vi >
v
zi0
u
The lemma suggests that if we know
v
zi0
zi0
>
v
+ E [ wi j vi >
zi0 ]
+ E [wi ]
v
+0
v
, then
can be estimated consistently by regressing
yi on xi and
zi0 v . But v can be consistently estimated by the Probit MLE of Di on zi !
This suggests two step estimation.
Obtain MLE of
[
v
from the Probit model
Regress yi on xi and
45
zi0
[
v
Problem Set
1. Suppose that (u; v) are bivariate normal with mean equal to zero. Let
"
Cov (u; v)
u:
Var (u)
v
Show that " and u are independent of each other.
2. Recall that, if
N (0; 1), we have
E[ j
t] =
1
(t)
:
(t)
Suppose that
v
u
N
0
0
67
;
2
v
v u
v u
2
u
:
Show that
t
E [ vj u
u
t] =
v
:
t
1
u
Hint: Observe that
E [vj u
v
t] = E " +
u u
u
t =E
u
t
=
v
u
u
u
E
v
u
u
u
u
u
t
:
u
3. Suppose that
yi = x0i + vi ;
but we observe yi if and only if
zi0 + ui
0:
In other words, we observe
(yi ; Di ; xi ; zi )
for each individual, where
Di = 1 (zi0 + ui
and
yi = yi Di ;
Assume that
vi
ui
N
0
0
2
v
;
0) :
v u
2
u
v u
:
Show that
E [yi j Di = 1; xi ; zi ] =
(a) Suppose that you know
x0i
+
v
zi0
u
zi0
u
:
. Show that the OLS of yi on xi and
u
the subsample where Di = 1 yields a consistent estimator of .
(b) Suggest how you would construct a consistent estimator of
u
(zi0 u )
applied to
(zi0 u )
.
(c) Suggest a two step method to construct a consistent estimator of .
68
Lecture Note 9: Efficiency
46
46.1
Classical Linear Regression Model I
Model
y =X +"
X is a nonstochastic matrix
X has a full column rank (Columns of X are linearly independent.)
E ["] = 0
E [""0 ] =
46.2
2
In for some unknown positive number
2
Gauss-Markov Theorem
Theorem 59 (Gauss-Markov) Given the Classical Linear Regression Model I, OLS estimator is the minimum variance linear unbiased estimator. (OLS is BLUE)
Proof. First note that b is a linear combination of y using (X 0 X)
linear estimator
Cy = CX + C":
If c is to be unbiased, we should have
CX = ;
or
CX = I:
Also note that
Var (Cy) =
2
CC 0 :
Because the di¤erence
CC 0
(X 0 X)
1
= CC 0
h
=C I
= CM C 0
CX (X 0 X)
X 0C 0
i
1
0
0
X (X X) X C 0
= CM M 0 C 0
= CM (CM )0 ;
is nonnegative de…nite, the result follows.
69
1
1
X 0 . Consider any other
46.3
Digression: GLS
Theorem 60 Assume that
is a positive de…nite matrix. Then,
Xb)0
argmin (y
1
(y
Xb) = X 0
1
1
X
X0
1
y:
b
Proof. Because
is a positive de…nite matrix, there exists T such that
T T 0 = In :
Observe that
= (T )
1
(T 0 )
1
= (T 0 T )
1
1
)
= T 0 T:
We can thus rewrite the objective function as
(y
Xb)0 T 0 T (y
T Xb)0 (T y
Xb) = (T y
X b)0 (y
T Xb) = (y
X b) ;
which is minimized by
(X 0 X )
1
X 0 y = (X 0 T 0 T X)
1
X 0T 0T y = X 0
1
X
1
X0
1
y:
We keep every assumption of the classical linear regression model I except we now assume
that
Var (") = , some known positive de…nite matrix.
Consider the estimator bGLS de…ned by
bGLS
Xb)0
argmin (y
1
(y
Xb)
b
Theorem 61 Under the new assumption, bGLS is BLUE.
Proof. There exists T such that
T T 0 = In :
Now, consider the transformed model
y =X
+" ;
where
y = T y;
X = T X;
" = T"
This transformed model satis…es every assumption of Classical Linear Regression Model I:
h
i
0
E " " = E [T ""0 T 0 ] = T T 0 = In :
The BLUE for the transformed model equals
(X 0 X )
1
X 0 y = (X 0 T 0 T X)
1
X 0T 0T y = X 0
70
1
X
1
X0
1
y:
Remark 16 OLS
b=
remains to be unbiased.
1
X 0"
is unknown and has to be estimated. Let b denote some ‘good’
Remark 17 In general,
estimator. Then,
X0 b
is called the feasible GLS (FGLS).
47
+ (X 0 X)
1
1
X
X0 b
1
y
Approximate E¢ ciency of MLE: Cramer-Rao Inequality
Theorem 62 (Cramer-Rao Inequality) Suppose that Z
ased for , i.e., E [Y ] = . Then,
E (Y
)0
) (Y
I( )
f (Z; ) and Y = u (Z) is unbi1
:
Proof. Assume for simplicity that
is a scalar. We then have
Z
= u (z) f (z; ) dz
and
1=
We can rewrite them as
Z
@f (z; )
u (z)
dz =
@
E [Y ] = ;
Z
u (z) s (z; ) f (z; ) dz
E [Y s (Z; )] = 1
Now, recall that E [s (Z; )] = 0, so that
E [(Y
) s (Z; )] = E [Y s (Z; )]
E [s (Z; )] = 1:
De…ne
y
Y
;
s (Z; ) :
X
Letting
E X2
1
E [X y] = I ( )
and
"
y
X
;
we may write
y=X
71
+ ":
1
;
Observations to be made:
(1) The coe¢ cient does not depend on Y at all; it only depends on the parameter
estimated and the Fisher information I ( );
(2) X and hence U are zero mean random variables and they are not correlated:
E [X "] = E [X y]
E X2
=0
We thus have
E y2 =
2
2
E X2 + E U 2
to be
E X2 =
1
I( )
This inequality is known as the Cramer-Rao inequality, and the right hand side of this inequality
is sometimes called the Cramer-Rao lower bound.
De…nition 12 Let Y be an unbiased estimator of a parameter . Call Y an e¢ cient estimator
if and only if the variance of Y equals the Cramer-Rao lower bound.
De…nition 13 The ratio of the actual variance of some unbiased estimator and the CramerRao lower bound is called the e¢ ciency of the estimator.
Remark 18 Cramer-Rao bound is not sharp. Suppose that Xi i = 1; : : : ; n are from N ( 1 ;
Then,
!
n
0
2
In ( ) =
0 2n2
2 ).
2
which implies that the Cramer-Rao lower bound for
2
is equal to
2 22
?
n
2 22
.
n
Does there exist an
unbiased estimator of 2 with variance equal to
Note that the usual estimator S 2 =
P
P
P
2
n
1
X is unbiased and is a function of su¢ cient statistic ( ni=1 Xi ; ni=1 Xi2 ).
i=1 Xi
n 1
It can be shown (using Lehman-Sche¤é Theorem, which you can learn from any textbook on
mathematical statistics) that S 2 is the unique unbiased minimum variance estimator. Because
2
2
2
2 2
(n 1)S 2
2
= (n 21)2 2 (n 1) = n 21 , which
(n 1), we have Var (S 2 ) = (n 21)2 Var (n 1)S
2
2
is strictly larger than the Cramer-Rao bound
2 22
.
n
Theorem 63 (Asymptotic E¢ ciency of MLE)
48
p
n b
d
! N (0; I
1
( ))
E¢ cient Estimation with Overidenti…cation: 2SLS
We are dealing with
yi = x0i + "i ;
where the only restriction given to us is that
E [zi "i ] = 0:
72
We assume that
0
1
z1i
B
C
zi = @ ... A ;
zri
such that
r = dim (zi )
dim (xi ) = dim ( ) = q:
We can estimate in more than one way. For example, if dim ( ) = 1, we can construct r
di¤erent IV estimators:
P
P
zri yi
z
y
1i
i
i
b(1) = P
; : : : ; b(r) = P i
:
i z1i xi
i zri xi
How do we combine them e¢ ciently? The answer is given by
2
!
! 1
!3 1
!
!
X
X
X
X
X
b2SLS = 4
xi zi0
zi zi0
zi x0i 5
xi zi0
zi zi0
i
i
h
= X 0 Z (Z 0 Z)
1
Z 0X
i
i
1
X 0 Z (Z 0 Z)
i
1
1
X
i
i
zi yi
!
Z 0y
This estimator is called the two stage least squares estimator because it is numerically equivalent
to
b = Z (Z 0 Z)
1. Regress xi on zi . Get a …tted value matrix X
2. Regress yi on x
bi , and obtain X 0 Z (Z 0 Z)
It can be shown that
for some
, and that
bn =
where ei
yi
b
V
p
1
n b2SLS
Z 0X
1
1
Z 0X
X 0 Z (Z 0 Z)
1
Z 0 y.
! N (0; )
can be consistently estimated by
2
!
!
P 2
X
X
e
1
1
i i
4
xi zi0
zi zi0
n
n i
n i
xi b2SLS is the residual. Let
2
!
P 2
X
1b
1
i ei
4 1
xi zi0
n =
n
n
n
n i
1X 0
zi zi
n i
Then a valid 95% asymptotic con…dence interval is given by
p
b2SLS 1:96 V
b
73
!3
1
1X 0 5
zi xi
n i
!
1
1
!3
X
1
zi x0i 5
n i
1
Remark 19 In the derivation of the asymptotic variance above, I assumed that zi and "i are
independent of each other.
Remark 20 Our discussion generalizes to the situation where dim ( ) > 1. We still have
h
i 1
1 0
1
0
0
X Z (Z Z) Z X
X 0 Z (Z 0 Z) Z 0 y
as our optimal estimator.
49
Why is 2SLS E¢ cient?
Consider the following seemingly unrelated problem. Write U = (U1 ; : : : ; Ur )0 . Suppose that
P
P
E [U ] = 0 and Var (U ) = . Consider w0 U = rj=1 wj Uj with rj=1 wj = 1. You would like to
minimize the variance. This problem can be written as
min w0 w
s:t: `0 w = 1
w
The Lagrangian of this problem is
1 0
w w + (1
2
`0 w)
FOC is given by
0= w
`
`0 w
0=1
which can be alternatively written as
1
w=
`
and
1 = `0
from which we obtain
w=
Now, note that
0
1 2
b(1)
B .. C 6
@ . A=4
b(r)
2
6
=4
1
n
P
i z1i
1
`0
1
n
i z1i
0
`
1
1`
xi
`
3
0
...
1
n
0
P
1
xi
..
.
1
n
74
P
i zri
P
xi
0
i zri
xi
7
5
3
7
5
1
1
2
6
4
1
n
P
i z1i
yi
3
7
..
5
.
P
1
i zri yi
n
1X
zi yi
n i
!
so that
0
p B
[email protected]
b(1)
..
.
b(r)
1
2
C 6
A=4
E [z1 x]
3
0
..
1
7
5
.
6
4
0
E [zr x]
X
1
zi "i + op (1)
= Q 1p
n i
! N 0; Q
2
"
1
1
Q
2
p1
n
p1
n
= N 0;
P
i z1i
P
"i
..
.
i zri
2
1
"Q
Q
"i
3
7
5 + op (1)
1
where
2
6
Q=4
E [z1 x]
3
0
..
.
0
E [zr x]
= E [zz 0 ]
7
5
According to the preceding analysis, the optimal combination is given by
1
`0
(
2
1
"Q
1)
Q
1
`
2
1
"Q
Q
1
1
`=
1
`0 Q
1 Q`
Q
1
Q`
Now, note that
2
6
Q` = 4
E [z1 x]
0
so that the optimal combination is
2
3
3
E [z1 x]
6
7
7
..
...
5` = 4
5 = E [zx]
.
E [zr x]
E [zr x]
0
Q (E [zz 0 ]) 1 E [zx]
E [xz 0 ] (E [zz 0 ]) 1 E [zx]
We now note that, if plim wn = w,
0
b(1)
p
B
..
wn0 n @
.
b(r)
then
1
0
p B
C
0
A = w [email protected]
b(1)
b(r)
..
.
1
C
A + op (1)
so we can use the approximate optimal combination, using
2 1P
3
0
i z1i xi
n
P
P
16
7
..
1
1
0
0
x
z
4
5
.
i
i
i
i zi zi
n
n
P
1
0
i zri xi
n
wn0 =
P
P
P
1
1
1
1
0
0
i xi zi
i zi zi
i zi xi
n
n
n
75
So the approximately optimal combination is
2 1P
i z1i xi
n
P
P
16
...
1
1
0
0
4
i xi zi
i zi zi
n
n
1
n
P
i
xi zi0
2
6
4
1
n
1
n
0
P
P
1
0
i zi zi
i z1i xi
1
n
..
i zri
i zi xi
1
n
0
P
1
n
P
.
3
0
P
i zri
xi
3
0
xi
1X
zi yi
n i
P
which is 2SLS!
50
1
7
5
1
n
1
n
=
7
5
i
P
xi zi0
i
xi zi0
P
1
n
1
n
!
0
i zi zi
P
0
i zi zi
P
1
1
n
1 1
n
i zi
P
yi
i zi xi
Problem Set
1. (From Goldberger) You are given a sample produced produced by a simultaneous equations model:
y1 =
1 y2
+
2 x1
+ "1
y2 =
3 y1
+
4 x2
+ "2
You naively regressed the endogenous variables on exogenous variables, and obtained
following OLS estimates:
y1 =
6x1 + 2x2
y2 = 3x1 + x2
Is ( 1 ; 2 ) identi…ed? Is (
identi…ed.
3;
4)
identi…ed? Compute consistent estimates of s that are
2. Consider a linear model
yi = x0i
+ "i
m 1
with the restriction that
E
zi "i
k 1
=0
where k > m. Derive the asymptotic distribution of 2SLS
2
!
! 1
!3 1
!
!
X
X
X
X
X
b2SLS = 4
xi zi0
zi zi0
zi x0i 5
xi zi0
zi zi0
i
h
= X 0 Z (Z 0 Z)
i
1
Z 0X
i
i
1
i
X 0 Z (Z 0 Z)
76
1
Z 0y
i
1
X
i
zi yi
!
under the assumption that E (zi "i ) (zi "i )0 =
2
"E
[zi zi0 ].
(a) Show that
p
2
n b2SLS
=4
1X
xi zi0
n i
!
!3 1
1X 0 5
zi xi
n i
! 1
!
1X 0
1 X
p
zi zi
zi "i
n i
n i
!
1
1X 0
zi zi
n i
!
1X
xi zi0
n i
(b) Show that
2
X
4 1
xi zi0
n i
!
1X 0
zi zi
n i
!
!3
X
1
zi x0i 5
n i
1
converges to
in probability.
h
E [xi zi0 ] (E [zi zi0 ])
(c) Show that
1
i
(E [zi x0i ])
N 0;
p
n b2SLS
N
0;
1X
xi zi0
n i
2
"E
[zi zi0 ]
converges in distribution to
2
"
h
E
[xi zi0 ] (E
1
[zi zi0 ])
(E
i
[zi x0i ])
3. The “wage.xls”…le contains three variables (from three years):
w0 earnings (in dollars), 1990
ed0 education (in years), 1990
a0 age (in years), 1990
w1 earnings (in dollars), 1991
ed1 education (in years), 1991
a1 age (in years), 1991
w2 earnings (in dollars), 1992
ed2 education (in years), 1992
a2 age (in years), 1992
77
!
E [xi zi0 ] (E [zi zi0 ])
1 X
p
zi "i
n i
converges in distribution to
(d) Conclude that
1
1
1
1X 0
zi zi
n i
1
!
1
(a) For the 1992 portion of the data, regress ln (wage)on edu, exp, (exp)2 , and a constant.
(b) Ability is an omitted variable which may create an endogeneity problem with the
education variable in our usual wage equation. It may be reasonable to assume
that lagged education (1990 and 1991) are valid instruments for education in the
1992 regression. Re-estimate the wage regression using 2SLS. Never mind that the
education variable hardly changes over time.
51
Method of Moments: Simple Example
Suppose in the population we have
E Yi
i.e., the
average
1
n
=0
1 1
1 1
is the mean of Yi . Given that the expectation E can be approximated by a sample
Pn
by b that solves
i=1 , it seems reasonable to estimate
1X
Yi
n i=1
n
or
52
b =0
X
b= 1
Yi
n i=1
n
Method of Moments: Generalization
Suppose that we are given a model that satis…es
E Yi
Xi
q q q 1
q 1
=0
Here, 0 is the parameter of interest. In order to estimate 0 , it makes sense to recall that
sample average provides an analog of population expectation. Therefore, we expect
1X
(Yi
n i=1
n
Xi )
0
by b that solves
It therefore makes sense to estimate
1X
Yi
n i=1
n
or
1X
Xi
n i=1
n
b=
!
1
1X
Yi
n i=1
n
Xi b = 0
!
=
n
X
Xi
i=1
Such estimator is called the method of moments estimator.
78
!
1
n
X
i=1
Yi
!
Theorem 64 Suppose that (Yi ; Xi ) i = 1; 2; : : : is i.i.d. Also suppose that E [Xi ] is nonsingular.
Then, b is consistent for .
Proof. Let
Ui
Yi
Xi
Note that E [Ui ] = 0 by assumption. We then have
1X
Xi
n i=1
n
b=
!
0
!
1
+ (E [Xi ])
1X
Xi
n i=1
n
1
1X
+
Ui
0
n i=1
n
(E [Ui ]) =
!
1X
Xi
n i=1
n
= +
0:
!
1
1X
Ui
n i=1
n
!
Theorem 65 Suppose that (Yi ; Xi ) i = 1; 2; : : : is i.i.d. Also suppose that E [Xi ] is nonsingular.
Finally, suppose that E [Ui Ui0 ] exists and is …nite. Then,
p
n b
! N 0; (E [Xi ])
p
1X
Xi
n i=1
1
E [Ui Ui0 ] (E [Xi0 ])
1
Proof. Follows easily from
52.1
n
n b
=
!
1
1 X
p
Ui
n i=1
n
!
Estimation of Asymptotic Variance
De…nition 14 Given a matrix A, we de…ne kAk =
p
trace (A0 A).
q
De…nition 15 If a = (a1 ; : : : ; aq ) is a q-dimensional column vector, kak = a21 +
0
Lemma 35 kA + Bk
+ a2q .
kAk + kBk
Lemma 36 kA0 k = kAk
Lemma 37 kABk
kAk kBk
bi = Yi
Theorem 66 Suppose that b is consistent for . Let U
E [Ui Ui0 ] in probability.
79
Xi b. Then,
1
n
Pn
i=1
bi U
b0 !
U
i
Proof. Note that
Xi b = Xi + Ui
bi = Yi
U
and
0
bi U
b 0 = Ui U 0 + Ui Xi b
U
i
i
Now note that
Xi b = Ui + Xi b
+ Xi b
Ui0 + Xi b
1X
1 X b b0
Ui Ui =
Ui Ui0
n i=1
n i=1
n
0
n
1X
Ui Xi b
n i=1
n
+
0
1X
Xi b
+
n i=1
n
Because
1X
Ui Xi b
n i=1
n
and
0
Ui Xi b
Xi b
we have
1X
Ui Xi b
n i=1
n
1X
Ui Xi b
n i=1
n
0
0
kUi k kXi k b
= kUi k Xi b
1X
kUi k kXi k b
n i=1
1X
kUi k kXi k
n i=1
n
0
Ui0
0
Xi b
0
Xi b
kUi k
1X
Xi b
n i=1
n
+
n
=
!
b
!0
in probability. The remaining two terms also converge to zero by similar reasoning.
Theorem 67
probability.
52.2
1
n
Pn
i=1
Xi
1
1
n
Pn
i=1
bi U
bi0
U
1
n
Pn
i=1
Xi0
1
! (E [Xi ])
E¢ ciency by GMM
Suppose that we are given a model that satis…es
E Yi
r 1
with r > q. Because there is no
Xi
r q q 1
=0
satisfying
1X
(Yi
n i=1
n
Y
X =
80
Xi ) = 0
1
E [Ui Ui0 ] (E [Xi0 ])
1
in
we will minimize
Qn ( ) = Y
X
0
Wn Y
X
and obtain
1
b = X 0 Wn X
0
X Wn Y
1
0
= + X Wn X
0
X Wn U
or
p
Assume that
n b
= X Wn X
1 X
p
Ui
n i=1
n
1
0
0
X Wn
!
E [Xi ] = G0
plim Wn = W0
E [Ui Ui0 ] = S0
It is straightforward to show that
p
d
n b
! N 0; (G00 W0 G0 )
1
G00 W0 S0 W0 G0 (G00 W0 G0 )
1
Given that Wn , hence W0 , can be chosen by the econometrician, we can ask what the optimal
choice would be. It can be shown that, if we choose W0 = S0 1 , then the asymptotic variance
is minimized. If W0 = S0 1 , then the asymptotic variance formula simpli…es to
1
G00 S0 1 G0
Remark 21 If you are curious, here’s the intuition. Consider the linear regression model in a
matrix form
y = G0 + u
where G0 is nonstochastic and E [u] = 0 and E [uu0 ] = S0 . We can minimize
G0 b)0 W0 (y
(y
G0 b)
and obtain an estimator
Because
bb = (G0 W0 G0 )
0
1
G00 W0 y
bb = (G0 W0 G0 ) 1 G0 W0 (G0 + u) = + (G0 W0 G0 )
0
0
0
we can see that bb is unbiased and has variance equal to
(G00 W0 G0 )
1
G00 W0 S0 W0 G0 (G00 W0 G0 )
1
G00 W0 u
1
By the Gauss-Markov, this estimator cannot be as good as the GLS estimator that minimizes
(y
G0 b)0 S0 1 (y
81
G0 b)
53
Method of Moments Interpretation of OLS
Suppose that
yi = x0i + "i
where E
E [xi "i ] = 0
Then, we can derive
x0i )] = E [xi yi
0 = E [xi "i ] = E [xi (yi
xi x0i ] = E [Yi
Xi ]
with
Yi = x i y i
Xi = xi x0i
Remark 22 It can be shown that
! 1
!
!
n
n
n
1X
1 X b b0
1X 0
Xi
Ui Ui
X
n i=1
n i=1
n i=1 i
54
1
1X
xi x0i
n i=1
n
=
!
1
1X
xi x0i e2i
n i=1
n
!
1X
xi x0i
n i=1
n
Method of Moments Interpretation of IV
The preceding discussion can be generalized to the following situation. Suppose that
yi = xi
+ "i
IF we can …nd zi such that
E [zi "i ] = 0;
the model can then be written as
E [zi yi
zi xi ] = 0
Write
Yi
zi yi ; Xi
zi xi
The method of moments estimator is then given by
! 1
! P
n
n
X
X
zi yi
= bIV ;
Xi
Yi = P i
z
x
i
i
i
i=1
i=1
which is consistent for by the argument in the preceding subsection.
Now assume that xi and zi are both q-dimensional, so that the model is given by
yi = x0i + "i
82
!
1
We then have
Yi
zi yi ; Xi
zi x0i
Therefore, we obtain
n
X
Xi
i=1
54.1
!
1
n
X
i=1
Yi
!
=
n
X
zi x0i
i=1
!
1
n
X
zi yi
i=1
!
= bIV
Asymptotic Distribution
As for its asymptotic distribution, we note that
E [Ui Ui0 ] = E (zi "i ) (zi "i )0 =
2
"E
[zi zi0 ]
and
E [Xi ] = E [zi x0i ]
(if we assume that z and " are independent of each other.) We then have
p
54.2
n bIV
! N 0;
2
"
(E [zi x0i ])
1
0
E [zi zi0 ] E (zi x0i )
1
Overidenti…cation: E¢ ciency Consideration
As before, we assume that the model is given by
yi = x0i + "i
Our restriction is
E [zi "i ] = 0
The only di¤erence now is that we assume
r = dim (zi ) > dim (xi ) = q
This means that we cannot …nd any b such that
1X
zi yi
n i=1
n
x0i b = 0
Given that equality is impossible, we do the next best thing: We minimize
!0
!
n
n
1X
1X
0
0
Qn ( ) =
zi (yi xi ) Wn
zi (yi xi )
n i=1
n i=1
for some weighting matrix Wn , which is potentially stochastic.
83
Example 8 Take
1X 0
zi zi
n i
Wn =
!
1
Then it can be shown that the solution is equal to
2
!
! 1
X
X
1
1
4
xi zi0
zi zi0
n i
n i
2
!
!
X
X
=4
xi zi0
zi zi0
i
i
!3
X
1
zi x0i 5
n i
!3
1
X
zi x0i 5
1
1X
xi zi0
n i
1
X
i
xi zi0
i
h
= X 0 Z (Z 0 Z)
1
!
!
1X 0
zi zi
n i
X
zi zi0
i
Z 0X
i
1
!
!
1
1
1X
zi yi
n i
X
zi yi
i
X 0 Z (Z 0 Z)
1
!
!
Z 0 y = b2SLS
According to the previous discussion, if we are to develop an optimal GMM, we need to use
the weight matrix that converges to the inverse of
S0 = E (zi "i ) (zi "i )0 = E "2i zi zi0
Suppose that "i happens to be independent of zi . If this were the case, we have
S0 = E "2i zi zi0 = E "2i E [zi zi0 ] =
2
"E
[zi zi0 ]
which can be estimated consistently by
Therefore, we would want to minimize
1X
zi (yi
n i=1
n
!0
x0i )
1X 0
b"2
zi zi
n i
!
= b"
1
1X 0
zi zi
n i
1X
zi (yi
n i=1
!0
1X
zi (yi
n i=1
n
2
b"2
n
x0i
)
!
x0i )
1X 0
zi zi
n i
!
1
1X
zi (yi
n i=1
n
x0i
!
)
Note that the b" 2 does not a¤ect the minimization. Therefore, the GMM estimator e¤ectively
minimizes
!0
! 1
!
n
n
X
X
1X
1
1
zi (yi x0i )
zi zi0
zi (yi x0i )
n i=1
n i
n i=1
This is 2SLS!
84
55
GMM - Nonlinear Case
Suppose that we are given a model
E [h (wi ; )] = 0
where h is a r 1 vector, and is a q 1 vector with r > q. Method of moments estimation is
P
impossible because there is in general no that solves n1 ni=1 h (Wi ; ) = 0. We minimize
!0
!
!
!
X
X
X
X
1
1
1
1
Qn ( ) =
h (wi ; ) Wn
h (wi ; ) =
h (wi ; )0 Wn
h (wi ; )
n i
n i
n i
n i
instead. We will derive the asymptotic distribution under the assumption that b is consistent.
55.1
Asymptotic Distribution
The FOC is given by
1
0
0
b
C
B 1 X @h wi ;
[email protected]
A Wn
n i
@
1X
h wi ; b
n i
!
10
b
@h
w
;
X
i
1
A Wn
[email protected]
n i
@ 0
0
1X
h wi ; b
n i
!
Using the mean value theorem, write the last term as
1
0
e
@h
w
;
i
1X
1X
1X
A b
h wi ; b =
h (wi ; ) + @
n i
n i
n i
@ 0
We then have
0
1 X @h
@
0=
n i
0
1 X @h
[email protected]
n i
from which we obtain
p
n b
=
0
wi ; b
@
0
wi ; b
@
0
10
A Wn
10
!
1X
h (wi ; )
n i
1
e
@h
w
;
X
i
A Wn @ 1
A b
n i
@ 0
0
20
10
0
13
b
e
X @h wi ;
[email protected] 1 X @h wi ; A
@1
A7
W
4
5
n
0
0
n i
@
n i
@
10
b
@h
w
;
X
i
@1
A Wn
n i
@ 0
85
!
1 X
p
h (wi ; )
n i
1
(6)
Assuming that
b
1 X @h wi ;
n i
@ 0
e
1 X @h wi ;
n i
@ 0
1 X @h (wi ; ) p
!0
n i
@ 0
1 X @h (wi ; ) p
!0
n i
@ 0
we can infer that
0
10
1
0
b
e
@h
w
;
@h
w
;
X
X
i
i
p
@1
A Wn @ 1
A!
G00 W0 G0
n i
@ 0
n i
@ 0
where
G0 = E
@h (wi ; )
@ 0
W0 = plim Wn
It follows that
20
10
0
13 0
10
b
e
b
@h
w
;
@h
w
;
@h
w
;
X
X
X
i
i
i
1
p
[email protected] 1
A Wn @ 1
A7
A Wn !
(G00 W0 G0 )
4
[email protected]
0
0
0
n i
@
n i
@
n i
@
1
G00 W0
(7)
Because E [h (wi ; )] = 0, we have by the CLT that
1 X
d
p
h (wi ; ) ! N (0; S0 )
n i
(8)
where
S0 = E h (wi ; ) h (wi ; )0
It follows that
0
10
b
@h
w
;
X
i
@1
A Wn
n i
@ 0
!
1 X
d
p
h (wi ; ) ! N (0; S0 )
n i
Combining (6) –(8), we conclude that
p
n b
d
! N 0; (G00 W0 G0 )
86
1
G00 W0 S0 W0 G0 (G00 W0 G0 )
1
55.2
Optimal Weight Matrix
Given that Wn , hence W0 , can be chosen by the econometrician, we can ask what the optimal
choice would be. It can be shown that, if we choose W0 = S0 1 , then the asymptotic variance
is minimized. If W0 = S0 1 , then the asymptotic variance formula simpli…es to
1
G00 S0 1 G0
55.3
Two Step Estimation
How do we actually implement the above idea? The trick is to recognize that the asymptotic
variance only depends on the probability limit of Wn and that Wn is allowed to be stochastic.
Let’s assume that there is a consistent estimator . We can then see that S0 can be
estimated by noting
1X
1X
S0 = E h (wi ; ) h (wi ; )0 = plim
h (wi ; ) h (wi ; )0 = plim
h (wi ; ) h (wi ; )0
n i
n i
Remark 23 The last equality requires some justi…cation, i.e., we need to show that
1X
1X
p
h (wi ; ) h (wi ; )0
h (wi ; ) h (wi ; )0 ! 0
n i
n i
We have seen in the discussion of MLE how this can be done.
Therefore, if we choose our weight matrix to be
Wn =
1X
h (wi ;
n i
) h (wi ;
0
)
!
1
then we are all set.
Question is where we …nd such a . We usually …nd it by the preliminary GMM, that
minimizes
!
!
!
!
X
X
X
1X
1
1
1
h (wi ; )0 In
h (wi ; ) =
h (wi ; )0
h (wi ; )
n i
n i
n i
n i
although we can choose any other weight matrix.
So, here’s the summary:
1. Minimize
1X
h (wi ; )0
n i
!
An
!
1X
h (wi ; )
n i
for arbitrary positive de…nite An . Call the minimizer
Wn =
1X
h (wi ;
n i
87
) h (wi ;
, and let
! 1
)0
2. Minimize
55.4
1X
h (wi ; )0
n i
!
!
1X
h (wi ; )
n i
Wn
Estimation of Asymptotic Variance
Suppose that we estimated the optimal GMM estimator b. How do we estimate the asymptotic
variance? Noting that the asymptotic variance is
G00 S0 1 G0
1
b0 Sb 1 G
b
G
1
we can estimate it by
where
b
1 X @h wi ;
b
G=
n i
@ 0
1X
Sb =
h wi ; b h wi ; b
n i
88
0
Lecture Note 10: Hypothesis Test
56
Elementary Decision Theory
We are given a statistical model with an observation vector X whose distribution depends on a
parameter , which would be understood to be a speci…cation of the true state of nature. The
ranges over a known parameter space .
The decision maker has available a set A of actions, which is called the action space. For
example, A = f0; 1g where 0 is “accept H0 ”and 1 is “reject H0 ”.
We are given a loss function l ( ; a). In the case of testing, we may take l ( ; a) = 0 if the
decision is correct and l ( ; a) = 1 if it is incorect.
We de…ne a decision rule to be a mapping from the sample space to A. If X is observed,
then we take the action (X). Our loss is then the random variable l ( ; (X)). We de…ne the
risk to be
Z
R ( ; ) = E [l ( ; (X))] = l ( ; (x)) f (xj ) dx
where f (xj ) denotes the pdf of X.
A procedure improves a procedure 0 i¤ R ( ; ) R ( ; 0 ) for all (with strict inequality
for some ). If is admissible, then no other 0 improves .
We may want to choose such that the worst possible risk sup R ( ; ) is minimized. If
is such that
sup R ( ; ) = inf sup R ( ; )
then the
is called the minimax procedure.
The Bayes risk is
Z
r( ) = R( ; ) ( )d
R
starts from weights ( ) such that
( ) d = 1. Viewing
( ), we can write
r ( ) = E [R ( ; )]
as a random variable with pdf
A Bayesian would try to minimize the Bayes risk. Because
Z
Z Z
r( ) = R( ; ) ( )d =
l ( ; (x)) f (xj ) ( ) dxd
Z Z
f (xj ) ( )
=
l ( ; (x))
d f (x) dx
f (x)
where
f (x) =
Z
f (xj ) ( ) d
89
we can write
r ( ) = E [r ( j X)]
where
r ( j x) =
Z
f (xj ) ( )
d =
l ( ; (x))
f (x)
= E [ l ( ; (x))j X = x]
Z
l ( ; (x)) f ( j x) d
Therefore, if we choose (x) such that r ( j x) is minimized, then the Bayes risk is minimized.
(If is discrete, we should replace the integral by summation.)
In a very simple testing context, the Bayes procedure with 0-1 loss would look like the
following. Let a0 denote the action of accepting H0 , and let a1 denote the action of accepting
H1 . Then the Bayes rule says that we should choose a0 if r (a0 j x) < r (a1 j x), and a1 if
r (a1 j x) < r (a0 j x). Let denote f 0 ; 1 g, where 0 is “H0 is correct”and 1 is “H1 is correct”.
Because
X
X
f (xj j ) ( j )
r ( j x) =
l ( j ; (x)) f ( j j x) =
l ( j ; (x))
f (x)
j=0;1
j=0;1
=
l ( 0 ; (x)) f (xj
0)
( 0 ) + l ( 1 ; (x)) f (xj
f (x)
1)
( 1)
we have r (a0 j x) < r (a1 j x) if and only if
l ( 0 ; a0 ) f (xj
0)
( 0 ) + l ( 1 ; a0 ) f (xj
f (x)
0)
( 0 )+l ( 1 ; a0 ) f (xj
1)
( 1)
<
l ( 0 ; a1 ) f (xj
0)
( 0 ) + l ( 1 ; a1 ) f (xj
f (x)
1)
( 1)
or
l ( 0 ; a0 ) f (xj
1)
( 1 ) < l ( 0 ; a1 ) f (xj
0)
( 0 )+l ( 1 ; a1 ) f (xj
1)
( 1)
or
f (xj
or
57
1)
( 1 ) < f (xj
f (xj
f (xj
0)
( 0)
1)
( 1)
<1
0) ( 0)
Tests of Statistical Hypothesis
Example 9 Assume that X1 ; : : : ; Xn N ( ; 10). We are going to consider the null hypothesis
H0 : 2 M0 against the alternative H1 : 2 M1 , where M0 and M1 are some subsets of the
one dimensional Euclidean space. We believe that 2 M0 or 2 M1 . We can either accept
the null or reject the null. When we reject the null even when the null is true, we are making
a Type I error. When we accept the null even when the alternative is true, we are making a
Type II error. Our objective is to make a decision in such a way as to minimize the probability
of either error.
90
Our decision will hinge on the realization of X1 ; : : : ; Xn . Suppose that we are going to reject
the null if (X1 ; : : : ; Xn ) 2 C, which implicitly will lead us to accept the alternative. The set C
is sometimes called the critical region.
De…nition 16 A Statistical Hypothesis is an assertation about the distribtion of one or more
random variables. If the statistical hypothesis completely speci…es the distribution, it is called a
simple statistical hypothesis; if it does not, it is called a composite statistical hypothesis.
Example 10 Assume that X1 ; : : : ; Xn
N ( ; 10). A hypothesis H0 :
hypothesis, whereas H0 :
75 is a composite hypothesis.
= 75 is a simple
De…nition 17 A test of a statistical hypothesis is a rule which, when the experimental sample values have been obtained, leads to a decision to accept or to reject the hypothesis under
consideration.
De…nition 18 Let C be that subset of the sample space which, in accordance witha prescribed
test, leads to the rejection of the hypothesis under consideration. Then C is called the critical
region of the test.
De…nition 19 The power function of a test of a statistical hypothsis H0 against H1 is the
function, de…ned for all distributions under consideration, which yields the probability that the
sample point falls in the critical region C of the test, that is, a function that yields the probability
of rejecting the hypothesis under consideration. The value of the power function at a parameter
point is called the power of the test at that point.
De…nition 20 Let H0 denote a hypothesis that is to be tested against an alternative H1 in
accordance with a prescribed test. The signi…cance level of the test is the maximum value of the
power function of the test when H0 is true.
Example 11 Suppose that n = 10. Suppose that H0 :
75 and H1 : > 75. Notice that we
are dealing with composite hypotheses. Suppose that we adopted a test procedure in which we
reject the null only when X > 76:645 In this test, the critical region C equals
(X1 ; : : : ; X10 ) : X > 76:645
The power function of the test is
p( ) = P
=1
X > 76:645 = Pr [N ( ; 1) > 76:645] = Pr [Z > 76:645
(76:645
]
);
where Z N (0; 1) and is the c.d.f. of N (0; 1). Notice that this power function is increasing
in . In the set f :
75g, the power function is maximized when = 75, at which it equals
1
(1:645) = :05 Thus, the signi…cance level of the test equals 5%.
91
58
Additional Comments about Statistical Tests
Suppose that H0 : = 75. If the alternative takes the form H1 : > 75 (or H1 : < 75 ), we
call such an alternative a one-sided hypothesis. If instead, if takes the form H1 : 6= 75, we
call it a two-sided hypothesis.
Example 12 Now suppose that the pair of hypotheses is H0 : = 0 vs. H1 :
that 2 and n are known. A commonly used test rejects the null if and only if
X
p 0
= n
>
0.
Assume
c
for some c, where c is chosen such that the signi…cance level of the test equals some prechosen
value. Assume that it has been decided to set the signi…cance level at 5%, a common choice.
Noting that
X
p 0 N (0; 1) ;
= n
it su¢ ces to …nd c such that
Pr [Z c] = 5%
where Z is a standard normal random varaible. From the normal distribution table, it can easily
be seen that we want to set
c = 1:645:
Example 13 What would happen if we now have H1 : 6= 0 . A common test rejects the null
if and only if
X
p 0
c;
= n
where c is again chosen so as to have the signi…cance level of the test equal to some prechosen
value. Assuming that we want to have the signi…cance level of the test equal to 5%, it can easily
be seen that we want to set
c = 1:96:
Example 14 What would happen if we now have the same two sided alternative, but do not
know 2 ? A common procedure is to reject the null i¤
X
p 0
s= n
c;
where s is the sample standard deviation:
s2 =
Because the ratio has a t (n
t (n 1) distribution.
1
n
1
X
Xi
X
2
:
i
1) distribution, we want to set c equal to 97.5th percentile of the
92
59
Certain Best Tests
Remark 24 The term “test” and “critical region” can be used interchangeably: a test speci…es
a critical region; but it can also be said that a choice of a critical region de…nes a test.
Let f (x; ) denote the p.d.f. of a random vector X. Consider the two hypothesis H0 :
vs. H1 : = 1 . We have = f 0 ; 1 g.
=
0
De…nition 21 Let C denote the subset of the sample space. Then C is called the best critical
region of size for testing H0 against H1 if, for every subset A of the sample space such that
Pr [X 2 A; H0 ] = ; (a) Pr [X 2 C; H0 ] = ; (b) Pr [X 2 C; H1 ]
Pr [X 2 A; H1 ]. In e¤ect,
the best critical region maximizes the power of the test while keeping the signi…cance level of the
test equal to .
Theorem 68 (Neyman-Pearson Theorem) Let X denote a random vector with p.d.f. f (x; ) =
L ( ; x). Assume that H0 : = 0 and H1 : = 1 . Let
C
L ( 1 ; x)
L ( 0 ; x)
x:
k ;
where k is chosen in such a way that
P [X 2 C; H0 ] = :
Then, C is the best critical region of size
for testing H0 against the alternative H1 .
Proof. Assume that A is another critical region of size . We want to show that
Z
Z
f (x; 1 ) dx:
f (x; 1 ) dx
A
C
By using the familiar indicator function notation, we can rewrite the inequality as
Z
(IC (x) IA (x)) f (x; 1 ) dx 0:
It su¢ ces to show that
(IC (x)
IA (x)) f (x;
1)
k (IC (x)
Notice that, if it (9) holds, it follows that
Z
(IC (x) IA (x)) f (x; 1 ) dx
But since
Z
(IC (x)
IA (x)) f (x;
0 ) dx
=
Z
k
Z
f (x;
C
IA (x)) f (x;
(IC (x)
0 ) dx
IA (x)) f (x;
Z
A
93
f (x;
0 ) dx
(9)
0) :
=
0 ) dx:
= 0;
we have the desired conclusion. Notice that IC
obviously have
(IC (x)
When IC
IA = 1, we have f (x;
(IC (x)
When IC
IA (x)) f (x;
IA =
1)
k (IC (x)
kf (x;
IA (x)) f (x;
1, we have f (x;
(IC (x)
1)
1)
IA equals 1, 0, or -1. When IC
0)
1)
0) :
IA (x)) f (x;
0) :
so that
k (IC (x)
< kf (x;
IA (x)) f (x;
IA = 0, we
so that
0)
IA (x)) f (x;
1)
k (IC (x)
IA (x)) f (x;
1) :
IA (x)) f (x;
1)
k (IC (x)
IA (x)) f (x;
0) :
We thus conclude that
(IC (x)
Example 15 Consider X1 ; : : : ; Xn i.i.d. N ( ;
H0 : = 0 vs. H1 : = 1 with 0 < 1 . Now,
p
1=
2
L ( 1 ; x1 ; : : : ; x n )
=
p
L ( 0 ; x1 ; : : : ; x n )
1= 2
"
= exp
2
2
X
The best critical region C takes the form
"
!
X
exp
xi ( 1
i
2
). We assume that
P
n
exp[
i
P
n
exp[
!
xi (
0) =
i
2
(xi
2
1)
=2 2 ]
(xi
2
0)
=2 2 ]
0) =
1
n
2
2
0
i
2
n
2
1
=2
2
0
2
#
2
1
is known. We have
=2
2
#
:
k
for some k. In other words, C takes the form
x
c
for some c. We can …nd the value of c easily from the standard normal distribution table.
60
Uniformly Most Powerful Test
In this section, we consider the problem of testing a simple null against a composite alternative.
Note that a composite hypothesis may be viewed as a collection of simple hypotheses.
94
De…nition 22 The critical region C is a uniformly most powerful critical region of size for
testing a simple H0 against a composite H1 if the set C is a best critical region of size for
testing H0 against each simple hypothesis in H1 . A Test de…ned by this critical region is called
a uniformly most powerful test with signi…cance level .
Example 16 Assume that X1 ; : : : ; Xn are i.i.d. N (0; ) random variables. We want to test
H0 : = 0 against H1 : > 0 . We …rst consider a simple alternative H1 : = 00 where
00
> 0 . The best critical region takes the form
k
p
1= 2
p
1= 2
00
0
P
n
exp [
i
x2i =2 00 ]
P
n
=
2
0
i xi =2 ]
exp [
n=2
0
exp
00
"
00
2
0
00 0
X
i
x2i
#
P
In other words, the best critical region takes the form i x2i c for some c, which is determined
by the size of the test. Notice that the same argument holds for any 00 > 0 . It thus follows
P
that i x2i c is the uniformly most powerful test of H0 against H1 !
Example 17 Let X1 ; : : : ; Xn i.i.d. N ( ; 1). There exists no uniformly most powerful test of
H0 : = 0 against H1 : 6= 0 . Consider 00 6= 0 . The best critical region for testing = 0
against = 00 takes the form
p
P
n
00 2
1= 2
exp
) =2
i (xi
p
k
P
n
0 )2 =2
(x
exp
1= 2
i
i
or
"
exp (
Thus, when
00
00
0
)
X
00 2
xi
n ( )
0 2
( )
#
=2
i
k
> 0 , the best critical region takes the form
X
xi c;
i
and when
00
< 0 , it takes the form
X
xi
c:
i
It thus follows that there exists no uniformly most powerful test.
61
Likelihood Ratio Test
We can intuitively modify and extend the notion of using the ratio of the likelihood to provide a
method of constructing a test of composite null against composite alternative, or of constructing
a test of a simple null against some composite alternative where no uniformly most powerful
test exists.
95
Idea: Suppose that X
f (x; ) = L ( ; x). The random vector X is not necessarily one
dimensional, and the parameter is not necessarily one dimensional, either. Suppose we are
given two sets of parameter ! and where !
. We are given
2 !;
H0 :
2
H1 :
!
The likelihood ratio test is based on the ratio
sup
sup
L ( ; x)
2! L ( ; x)
2
If this ratio is bigger than k, say, the null is rejected. Otherwise, we do not reject the null. The
number k is chosen in such a way that the size of the test equals some prechosen value, say .
Example 18 Suppose that X1 ; : : : ; Xn are i.i.d. N ( ; 2 ). We do not know
is positive. We have
H0 : = 0; H1 : 6= 0:
2
except that it
Formally, we can write
!=
2
;
:
2
= 0;
>0
and
=
;
2
1<
:
2
< 1;
>0 :
Now, we have to calculate
sup L
( ;
;
2
; x1 ; : : : ; xn = sup
2 )2!
( ;
2 )2!
2
2
and
sup L
( ;
;
2
; x1 ; : : : ; x n =
2 )2
sup
( ;
2 )2
exp
" P
n=2
1
2
exp
" P
n=2
1
2
i (xi
2
i (xi
2
In these calculations, we can consider maximizing
log L =
instead.
For (10), we set
n
log (2 )
2
n
log
2
2
2
1 X
2
= 0 and di¤erentiate with respect to
n
1 X 2
+
x =0
2 2 2 2 i i
96
(xi
)2
i
2
obtaining
)2
#
;
(10)
)2
#
:
(11)
2
2
We thus have
sup L
2
;
1
; x1 ; : : : ; xn = L 0; n
2 )2!
( ;
X
x2i ; x1 ; : : : ; xn
i
=
1
P
2
=
i
e
P
2
i
P 2
x
P i 2i
2 i xi =n
n=2
exp
x2i =n
n=2
1
:
x2i =n
For (11), we set the partial derivatives with respect to
1 X
(xi
) = 0;
2
2
and
!
equal to zero:
i
and
n
1 X
+
(xi
2 2 2 2 i
We then obtain
sup L
( ;
;
2
; x1 ; : : : ; xn = L x; n
)2 = 0:
1
2 )2
X
(xi
x)2 ; x1 ; : : : ; xn
i
e 1
P
2
x)2 =n
i (xi
=
Thus, the likelihood ratio test would be based on
!n=2
P 2
=n
x
= P i i 2
:
x) =n
i (xi
!n=2
!
:
It is important to note that the numerator and the denominator are constrained mle’s of 2 .
This important observation will be used often in the classical linear regression hypothesis testing
setup.
The likelihood ratio test would reject the null i¤ is bigger than certain threshhold. It is
equivalent to the rejection when
P 2
x =n
P i i 2
x) =n
i (xi
is big. Because
X
X
x2i =n =
(xi x)2 =n + x2 ;
i
i
the test is equivalent to the rejection when
qP
i
is big, which has a jt (n
x
(xi
x)2 = (n
1)j distribution!
97
1)
62
Asymptotic Tests
Suppose that we have Xi
i:i:d:
f (x; ), where dim ( ) = k. Let
b = argmax
c
n
X
log f (Xi ; c)
i=1
denote the MLE. We would like to test H0 : = against H1 : 6= .
The …rst test is Wald Test. Tentatively assume that k = 1. Recall that
p
It follows that we should have
d
n b
! N 0; I ( )
n b
I( )
2
d
!
1
2
(1)
In most cases, I ( ) is a continuous function in , so I b
1
will be consistent for I ( ) 1 , and
2
n b
Under the null, we should have
1
1
I b
!
d
2
(1)
d
2
(1)
2
n b
!
1
I b
When k is an arbitrary number, we have the generalization
n b
0
d
b
I b
2
!
(k)
The Likelihood Ratio Test is based on the result that
2
n
X
i=1
log f Xi ; b
n
X
!
!
d
2
(k)
!
!
d
2
(k)
!
!
d
2
(k)
log f (Xi ; )
i=1
Under the null, we should have
2
n
X
i=1
log f Xi ; b
n
X
log f (Xi ;
i=1
The Score Test is based on the result that
!0
n
1 X
s (Xi ; ) I ( ) 1
n i=1
98
n
X
i=1
)
s (Xi ; )
so that
n
X
1
n
!0
s (Xi ;
)
i=1
n
X
1
I( )
!
s (Xi ;
)
i=1
d
!
2
(k)
This test is sometimes called the Lagrange Multiplier Test because, if we want to maximize
Pn
= , and if we use the Lagrangian
i=1 log f (Xi ; ) subject to the constraint
n
X
0
log f (Xi ; )
(
)
i=1
the …rst order conditions are
n
X
s (Xi ; ) =
i=1
so
=
=
n
X
s (Xi ;
)
i=1
63
63.1
Some Details
Wald Test of Linear Hypothesis
p
Let b be such that n b
We are interested in testing
where R is m
k with m
n Rb
r
0
h
d
! N (0; V ) for some V , which is consistently estimated by Vb .
H0 : R
r=0
HA : R
r 6= 0
k. The test statistic is then given by
RVb R0
i
1
Rb
r = Rb
Its asymptotic distribution under the null is
63.2
2
r
0
(m).
Wald Test of Noninear Hypothesis
We are interested in testing
H0 : h ( ) = 0
HA : h ( ) 6= 0
99
R
1b
V
n
1
R0
Rb
r
where dim (h) = m
k. The test statistic is then given by
i 1
0h
0
bVb R
b0
b 1 Vb
nh b R
h b =h b R
n
b is a consistent estimator of
where R
1
b0
R
h b
@h ( )
:
@ 0
Its asymptotic distribution under the null is 2 (m). Common sense suggests that we can take
R=
b=
R
@h b
@
0
p
The idea can be understood by Delta method: Because n b
have
p
d
n h b
h ( ) ! N (0; RV R0 )
and therefore, we should have
p
under H0 .
63.3
d
! N (0; V ), we should
d
nh b ! N (0; RV R0 )
LR Test
We now assume that b is MLE, i.e., V = I ( ) 1 . We are interested in testing H0 : h ( ) = 0
vs HA : h ( ) 6= 0, where dim (h) = m
k. We assume that H0 can be equivalently written
H0 : = g ( ) for some , where dim ( ) = k m = dim ( ) dim (h).
P
The LR test requires calculation of restricted MLE e, i.e., the maximizer of ni=1 log f (Xi ; c)
subject to the restriction h (c) = 0. Given the alternative characterization, it su¢ ces to maxiP
mize ni=1 log f (Xi ; g (a)) without any restriction. Let
e = arg max
e = g (e)
We now calculate the LR test statistic by
2
n
X
i=1
log f Xi ; b
We note that
n
X
log f (Xi ;
n
X
i=1
0)
log f Xi ; e
=
i=1
n
X
i=1
+
1
2
n
X
!
log f (Xi ; g (a))
i=1
=2
n
X
i=1
log f Xi ; b
n
X
log f (Xi ; g (e))
i=1
1
0
n @ log f X ; b
X
i
A
log f Xi ; b + @
0
@
i=1
!
n
X @ 2 log f (Xi ; )
0
b
b
0
0
0
@
@
i=1
100
!
0
b
for some
in between
zero. We therefore have
2
n
X
i=1
0
n
X
log f Xi ; b
and b. Because of the obvious FOC, the second term on the right is
log f (Xi ;
!
0)
i=1
= b
0
=
p
=
p
=
p
2
n
X
log f (Xi ; g (e))
i=1
log f (Xi ; g (
J(
0)
!
=
@ log f (Xi ; g (
@
=E
p
0
n (e
0 )) @
b
0)
0
(J (
log f (Xi ; g (
@ 0
0 ))
p
n (e
0 ))
But because
@ log f (Xi ; g (
@ 0
0 ))
=
@ log f (Xi ; g (
@ 0
0 )) @g
(
@
0)
0
@ log f (Xi ; g (
@ 0
we have J ( 0 ) = G0 I ( 0 ) G.
Now recall that
p
Similarly, we have
p
n (e
n b
0)
0
= I ( 0)
1
1 X
p
s (Xi ;
n
0)
+ op (1)
1 X 0 @ log f (Xi ; g ( 0 ))
p
G
+ op (1)
@
n
1 X
1
= [G0 I ( 0 ) G] G0 p
s (Xi ; 0 ) + op (1)
n
=J(
0)
1
101
0
!
n
p
1 X @ 2 log f (Xi ; )
n b
0
n i=1
@ @
0
p
n b
(I ( 0 ) + op (1))
0
0
p
n b
+ op (1)
I ( 0)
0
0
n b
!
0
0
n b
0 ))
i=1
where
n b
)
0
@ @
i=1
Likewise, we have
n
X
n
X
@ 2 log f (Xi ;
0
0 ))
G
0)
+op (1)
0
Therefore, we conclude that
2
n
X
i=1
=
=
=
where
log f Xi ; b
1 X
p
s (Xi ; 0 )
n
1 X
p
s (Xi ; 0 )
n
1 X
p
s (Xi ; 0 )
n
1 X
p
s (Xi ; 0 )
n
n
X
i=1
0
log f Xi ; e
!
1 X
p
s (Xi ; 0 )
n
0
1 X
1
G [G0 I ( 0 ) G] G0 p
s (Xi ;
n
I ( 0)
0
0
1
I ( 0)
h
1
1
G [G0 G]
1 X
p
s (Xi ;
n
and we write
1
G [G0 I ( 0 ) G]
1
0)
G0
i
G0
0)
+ op (1)
1 X
p
s (Xi ;
n
1 X
p
s (Xi ;
n
0)
0)
+ op (1)
+ op (1)
d
! N (0; I ( 0 ))
= I ( 0 ) to avoid confusion with an identity matrix. It follows that
!
n
n
X
X
d
! 2 (m)
2
log f Xi ; b
log f Xi ; e
i=1
i=1
An intuition can be given in the following way. Now consider
u0
1
G [G0 G]
1
G0 u
where u
N (0; ). Let T be a square matrix such that T T 0 = , and write "
Note that we can without loss of generality write u = T ". We then have
u0
1
where X = T 0 G is a k
G [G0 G]
(k
1
G 0 u = "0 T 0
1
G [G0 G]
= "0 T 0
1
= "0 Ik
X [X 0 X]
T
1
G [G0 G]
102
1
G0 u
G0 T "
T 0 G [G0 T T 0 G]
1
X0 "
m) matrix. It follows that
u0
1
2
(m)
1
G0 T "
N (0; Ik ).
63.4
LM Test
The score test is based on
1 X
1 X
p
s Xi ; e = p
s (Xi ;
n i=1
n i=1
n
1 X @s (Xi ;
n i=1
@ 0
n
1 X
s (Xi ;
=p
n i=1
n
0)
+
n
1 X
=p
s (Xi ;
n i=1
!
p
0)
p
I ( 0 ) n (g (e)
g(
0)
p
I ( 0 ) G n (e
0)
0)
I ( 0 ) G [G0 I ( 0 ) G]
n
1 X
s (Xi ;
=p
n i=1
)
n e
0 ))
+ op (1)
+ op (1)
1 X
p
s (Xi ;
n
n
h
G [G0 G]
= I
1
G0
i
1 X
p
s (Xi ;
n i=1
n
0
1
G0
!
0)
0)
+ op (1)
+ op (1)
It follows that
1 X
p
s Xi ; e
n i=1
n
1 X
p
s (Xi ;
n i=1
n
=
1 X
p
s (Xi ;
n i=1
n
=
!0
1 X
p
s Xi ; e
n i=1
n
1
!0
0)
!0
0)
h
G [G0 G]
I
h
1
1
G0
G [G0 G]
1
!
i
G0
i
1
h
1
G [G0 G]
I
1 X
p
s (Xi ;
n i=1
n
G0
!
0)
i
1 X
p
s (Xi ;
n i=1
n
!
0)
+ op (1)
It follows that the score test is asymptotically equivalent to the LR test.
64
Problem Set
1. Let X1 ; : : : ; X10 be a random sample from N (0; 2 ). Find a best critical region of size
= 5% for testing H0 : 2 = 1 against H1 : 2 = 2. Is this a best critical region for
testing against H1 : 2 = 4? Against H1 : 2 = 12 > 1?
2. Let X1 ; : : : ; Xn be a random sample from a distribution with pdf f (x) = x 1 , 0 < x < 1,
zero elsewhere. Show that the best critical region for testing H0 : = 1 against H1 : = 2
takes the form
(
)
n
Y
(x1 ; : : : ; xn ) :
xi c
i=1
3. Let X1 ; : : : ; Xn be a random sample from N ( ; 100). Find a best critical region of size
= 5% for testing H0 : = 75 against H1 : = 78.
103
+ op (1)
65
Test of Overidentifying Restrictions
Suppose that we are given a model
E [h (wi ; )] = 0
where h is a r 1 vector, and is a q 1 vector with r > q. We ask if there is such a
begin with. To be more precise, our null hypothesis is
H0 : There exists some
such that E [h (wi ; )] = 0
In order to test this hypothesis, a reasonable test statistic can be based on
!0
!
X
X
1
1
h wi ; b
Wn
h wi ; b
Qn b =
n i
n i
p
where Wn ! S0 1 and b solves
min Qn ( ) =
!0
1X
h (wi ; ) Wn
n i
!
1X
h (wi ; )
n i
for the same Wn . It can be shown that
d
nQn b !
2
r q
The proof is rather lengthy, and consists of several steps:
Step 1 Because S0 is a positive de…nite matrix, there exists T0 such that
T0 S0 T00 = Ir
Observe that
S0 = (T0 )
1
(T00 )
1
= (T00 T0 )
1
)
S0 1 = T00 T0
Now, write
nQn b =
=
=
!0
!
1 X
1 X
p
h (wi ; ) Wn p
h (wi ; )
n i
n i
!0
!
1 X
1 X
0
0 1
1
p
h (wi ; ) T0 (T0 ) Wn T0 T0 p
h (wi ; )
n i
n i
!0
!
X
1 X
1
1
T0 p
h (wi ; )
(T00 ) Wn T0 1
T0 p
h (wi ; )
n i
n i
104
to
Note that
1
(T00 )
Wn T0
1 p
! (T00 )
1
S0 1 T0
1
= Ir
Therefore, we can see that the nQn b has the same limit distribution as
!0
1 X
T0 p
h (wi ; )
n i
!
1 X
T0 p
h (wi ; )
n i
Step 2 In general, we learned that
20
10
13
0
b
e
@h
w
;
@h
w
;
X
i
i
p
6 1X
A Wn @ 1
A7
n b
= [email protected]
5
0
0
n i
@
n i
@
10
b
@h
w
;
X
i
A Wn
@1
n i
@ 0
0
1
!
1 X
p
h (wi ; )
n i
20
10
0
13 0
10
b
e
b
@h
w
;
@h
w
;
@h
w
;
X
X
X
i
i
i
1
p
[email protected] 1
A Wn @ 1
A7
A Wn !
(G00 W0 G0 )
[email protected]
4
0
0
0
n i
@
n i
@
n i
@
and
p
p
n
n b
=
n
!
1 X
p
h (wi ; )
n i
20
10
0
13 0
10
b
e
b
@h
w
;
@h
w
;
@h
w
;
X
X
X
i
i
i
1
6 1
A Wn @ 1
A7
A Wn
= [email protected]
[email protected]
0
0
n i
@
n i
@
n i
@ 0
! G00 S0 1 G0
G00 W0
1 X
d
p
h (wi ; ) ! N (0; S0 )
n i
Using that Wn ! S0 1 , we can conclude that
where
1
1
G00 S0 1 in probability.
Step 3 We now consider the distribution of
1
0
e
@h
w
;
X
X
X
i
1
1
1
A b
p
h wi ; b = p
h (wi ; ) + @ p
@ 0
n i
n i
n i
0
1
e
@h
w
;
X
X
i
p
1
1
A n b
=p
h (wi ; ) + @
0
n i
@
n i
105
(12)
As before, we will assume that
e
1 X @h wi ;
n i
@ 0
1 X @h (wi ; ) p
!0
n i
@ 0
which will imply that
e
1 X @h wi ;
p
! G0
0
n i
@
Combined with (12), we can conclude that
1 X
p
h wi ; b = [I
n i
!
1 X
p
h (wi ; )
n i
n]
such that
0
0
1 20
10
13 0
10
e
b
e
b
@h
w
;
@h
w
;
@h
w
;
@h
w
;
X
X
X
X
i
i
i
i
1
1
@1
A6
A Wn @ 1
A7
A Wn
[email protected]
[email protected]
n =
0
0
0
n i
@
n i
@
n i
@
n i
@ 0
! G0 G00 S0 1 G0
1
G00 S0 1 in probability.
Write
0
1
= G0 G00 S0 1 G0
G00 S0 1
By Slutsky, we conclude that
1 X
d
h wi ; b ! N (0; T0 (I
T0 p
n i
Because
T0 (I
0 ) S0
(I
0
0
0 ) T0
= T0 S0 T00
= Ir
T0
0
0 S0 T0
T0 G0 G00 S0 1 G0
0 ) S0
T0 S0
1
+ T0 G0 G00 S0 1 G0
= Ir
1
106
0 S0
0 0
0 T0
G00 T00
1
G00 T00
1
T0 G0 G00 S0 1 G0
G00 T00
1
G00 T00
G00 T00
T0 G0 G00 S0 1 G0
we can further conclude that
1 X
d
h wi ; b ! N 0; Ir
T0 p
n i
+ T0
G00 S0 1 S0 S0 1 G0 G00 S0 1 G0
T0 G0 G00 S0 1 G0
+ T0 G0 G00 S0 1 G0
= Ir
1
0 0
0 T0
G00 S0 1 S0 T00
1
T0 S0 S0 1 G0 G00 S0 1 G0
0
0
0 ) T0 )
(I
1
G00 T00
T0 G0 G00 S0 1 G0
1
G00 T00
(13)
Step 4 This step will be taken care of by analogy. We now imagine a GLS model
y = G0 + "
where "
N (0; S0 ). Because
T0 S0 T00 = Ir
the transformed model
y =G
,
+"
T0 y = T0 G0 + T0 "
is such that "
N (0; Ir ).
The residual vector
e = Ir
G
(G )0 G
1
(G )0
then has the distribution
N 0; Ir
G
1
(G )0 G
(G )0 = N 0; Ir
T0 G0 (T0 G0 )0 T0 G0
1
(T0 G0 )0
Because
Ir
G
(G )0 G
1
(G )0 = Ir
1
T0 G0 (T0 G0 )0 T0 G0
(T0 G0 )0
= Ir
T0 G0 (G00 T00 T0 G0 )
1
G00 T00
= Ir
T0 G0 G00 S0 1 G0
1
G00 T00
we conclude that
e
N 0; Ir
T0 G0 G00 S0 1 G0
1
G00 T00
Now compare (14) with (13). We can see that the limit distribution of T0 p1n
(14)
P
i
h wi ; b is
identical to the distribution of e . Also recall that nQn b has the same limit distribution as
!0
1 X
T0 p
h (wi ; )
n i
!
1 X
T0 p
h (wi ; )
n i
which suggests that the limit distribution of nQn b should be identical to the distribution of
(e )0 e . But we already know that
(e )0 e
66
2
r q
Hausman Test of OLS vs IV
Suppose
yi = xi
+ "i ;
(i = 1; : : : ; n):
107
Also suppose that we observe zi such that E [zi "i ] = 0. Either H0 : E [xi "i ] = 0 or HA :
E [xi "i ] 6= 0. Under H0 , both OLS bOLS and IV bIV are consistent:
H0 : plim bOLS = plim bIV
Furthermore, bOLS is e¢ cient. Under HA , only …xed e¤ects estimator are consistent:
HA : plim bOLS 6= plim bIV
Therefore, a reasonable test of H0 can be based on the di¤erence bOLS bIV . If the di¤erence
is large, there is an evidence supporting HA . If small, there is an evidence supporting H0 . For
p
this purpose, we need to establish the asymptotic distribution of n bOLS bIV under the
null. Under H0 , we would have
p
p
In general, because
n bOLS
p
n bIV
n bOLS
Vara bOLS
bIV
! N 0; Vara bOLS
bIV
! N 0; Vara bIV
! N 0; Vara bOLS
= Vara bOLS + Vara bIV
bIV
2 Cov a bOLS ; bIV ;
we need to …gure out Cov a bOLS ; bIV , to implement such a test.
Hausman’s intuition is that, if bOLS is e¢ cient under H0 , then we must have
Cov a bOLS ; bIV
= Vara bOLS :
Suppose otherwise. Consider an arbitrary linear combination
bOLS + (1
) bIV ;
whose asymptotic distribution is given by
p
n
bOLS + (1
) bIV
! N 0;
2
Vara bOLS + (1
)2 Vara bIV
+ 2 (1
Consider minimizing the asymptotic variance. First order condition is given by
or
Vara bOLS
1
=
(1
) Vara bIV
+ (1
Vara bOLS
Cov a bOLS ; bIV
Vara bOLS + Vara bIV
108
2 ) Cov a bOLS ; bIV
2 Cov a bOLS ; bIV
=0
:
) Cov a bOLS ; bIV
In order for bOLS to be e¢ cient, we had better have
1
=0
or
Vara bOLS
Cov a bOLS ; bIV :
Therefore, the asymptotic variance calculation simpli…es substantially: Because bOLS is
e¢ cient under the null, we should have
Vara bOLS
bIV
= Vara bOLS +Vara bIV
Therefore, a test can be based on
r
p
n bOLS
d a bIV
Var
bIV
2 Cov a bOLS ; bIV
d a bOLS
Var
109
N (0; 1) :
= Vara bIV
Vara bOLS :
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