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Finite-Sample Properties of OLS 71
where ε i here is defined as log( A (^) i )−E[log( A (^) i )] and α 0 = E[log( A (^) i )]. Suppose, in addition to total costs, output, and factor prices, we had data on factor inputs. Can we estimate α’s by applying OLS to this log-linear relationship? Why or why not? Hint: Do input levels depend on ε i? Suggest a different way to estimate α’s. Hint: Look at input shares.
P R O B L E M S E T F O R C H A P T E R 1
A N A L Y T I C A L E X E R C I S E S
- (Proof that b minimizes SSR ) Let b be the OLS estimator of β. Prove that, for any hypothetical estimate, ˜β, of β,
( y − X ˜β) ′^ ( y − X ˜β) ≥ ( y − Xb ) ′^ ( y − Xb ).
In your proof, use the add-and-subtract strategy: take y − X ˜β, add Xb to it and then subtract the same from it. It produces the decomposition of y − X ˜β:
y − X ˜β = ( y − Xb ) + ( Xb − X ˜β).
Hint: ( y − X ˜β) ′^ ( y − X ˜β) = [( y − Xb ) + X ( b − ˜β)]′^ [( y − Xb ) + X ( b − ˜β)]. Using the normal equations, show that this equals
( y − Xb ) ′^ ( y − Xb ) + ( b − ˜β) ′^ X ′^ X ( b − ˜β).
- (The annihilator associated with the vector of ones) Let 1 be the n -dimensional column vector of ones, and let M (^) 1 ≡ I n − 1 ( 1 ′^ 1 ) −^1 1 ′^. That is, M (^) 1 is the anni- hilator associated with 1. Prove the following: (a) M (^) 1 is symmetric and idempotent. (b) M (^) 1 1 = 0. (c) M (^) 1 y = y − y ¯ · 1 where
y ¯ =
n
∑ n
i = 1
yi.
M (^) 1 y is the vector of deviations from the mean. (d) M (^) 1 X = X − 1 x ¯′^ where x ¯ = X ′^ 1 / n. The k -th element of the K × 1 vector x^ ¯ is (^1) n
∑ (^) n i = 1 x^ ik^.
72 Chapter 1
- (Deviation-from-the-mean regression) Consider a regression model with a con- stant. Let X be partitioned as
( n^ X × K ) =
[
n^1 × 1
. X 2
n ×( K − 1 )
]
so the first regressor is a constant. Partition β and b accordingly:
β =
[
β 1 β (^2)
]
← scalar ← ( K − 1 ) × 1
, b =
[
b 1 b (^) 2
]
Also let ˜ X (^) 2 ≡ M (^) 1 X (^) 2 and y ˜ ≡ M (^) 1 y. They are the deviations from the mean for the nonconstant regressors and the dependent variable. Prove the following: (a) The K normal equations are
y ¯ − b 1 − x ¯′ 2 b (^) 2 = 0
where x ¯ 2 = X ′ 2 1 / n ,
X ′ 2 y − n · b 1 · ¯ x 2 − X ′ 2 X (^) 2 b (^) 2 = 0 (( K − 1 )× 1 )
(b) b (^) 2 = (˜ X ′ 2 ˜ X (^) 2 ) −^1 ˜ X ′ 2 y ˜. Hint: Substitute the first normal equation into the other K − 1 equations to eliminate b 1 and solve for b (^) 2. This is a generalization of the result you proved in Review Question 3 in Section 1.2.
- (Partitioned regression, generalization of Exercise 3) Let X be partitioned as
( n^ X × K ) =^
[
X 1
( n × K (^) 1 )
. X 2
( n × K (^) 2 )
]
Partition β accordingly:
β =
[
β (^1) β (^2)
]
← K 1 × 1
← K 2 × 1
Thus, the regression can be written as
y = X (^) 1 β 1 + X (^) 2 β 2 + ε.
Let P (^) 1 ≡ X (^) 1 ( X ′ 1 X (^) 1 ) −^1 X ′ 1 , M (^) 1 ≡ I − P (^) 1 , ˜ X (^) 2 ≡ M (^) 1 X (^) 2 and y ˜ ≡ M (^) 1 y. Thus, y ˜ is the residual vector from the regression of y on X (^) 1 , and the k -th column of ˜ X (^) 2 is the residual vector from the regression of the corresponding k -th column of
74 Chapter 1
Hint: Apply the general decomposition formula (1.2.15) to the regression in (c) to derive
y^ ˜′^ y ˜ = b ′ 2 ˜ X ′ 2 ˜ X (^) 2 b (^) 2 + e ′^ e.
Then use (b). (f) Consider the following four regressions: (1) regress y ˜ on X (^) 1. (2) regress y ˜ on ˜ X (^) 2. (3) regress y ˜ on X (^) 1 and X (^) 2. (4) regress y ˜ on X (^) 2. Let SSR (^) j be the sum of squared residuals from regression j. Show: (i) SSR (^) 1 = ˜ y ′^ y ˜. Hint: y ˜ is constructed so that X ′ 1 ˜ y = 0 , so X (^) 1 should have no explanatory power. (ii) SSR (^) 2 = e ′^ e. Hint: Use (c). (iii) SSR (^) 3 = e ′^ e. Hint: Apply the Frisch-Waugh Theorem on regression (3). M (^) 1 y ˜ = ˜ y. (iv) Verify by numerical example that SSR (^) 4 is not necessarily equal to e ′^ e.
- (Restricted regression and F ) In the restricted least squares, the sum of squared residuals is minimized subject to the constraint implied by the null hypothesis R β = r. Form the Lagrangian as
L = 1 2
( y − X ˜β) ′^ ( y − X ˜β) + λ ′^ ( R ˜β − r ),
where λ here is the # r -dimensional vector of Lagrange multipliers (recall: R is # r × K , ˜β is K × 1, and r is # r × 1). Let ̂β be the restricted least squares estimator of β. It is the solution to the constrained minimization problem. (a) Let b be the unrestricted OLS estimator. Show:
̂ β = b − ( X ′^ X ) −^1 R ′^ [ R ( X ′^ X ) −^1 R ′^ ]−^1 ( Rb − r ), λ = [ R ( X ′^ X ) −^1 R ′^ ]−^1 ( Rb − r ).
Hint: The first-order conditions are X ′^ y − ( X ′^ X )̂ β = R ′^ λ or X ′^ ( y − X ̂ β) = R ′^ λ. Combine this with the constraint R ̂ β = r to solve for λ and ̂β.
Finite-Sample Properties of OLS 75
(b) Let εˆ ≡ y − X ̂ β, the residuals from the restricted regression. Show:
SSR (^) R − SSRU = ( b − ̂β) ′^ ( X ′^ X )( b − ̂β) = ( Rb − r ) ′^ [ R ( X ′^ X ) −^1 R ′^ ]−^1 ( Rb − r ) = λ ′^ R ( X ′^ X ) −^1 R ′^ λ = ˆε ′^ P εˆ,
where P is the projection matrix. Hint: For the first equality, use the add- and-subtract strategy:
SSR (^) R = ( y − X ̂ β) ′^ ( y − X ̂ β) = [( y − Xb ) + X ( b − ̂β)]′^ [( y − Xb ) + X ( b − ̂β)].
Use the normal equations X ′^ ( y − Xb ) = 0. For the second and third equalities, use (a). To prove the fourth equality, the easiest way is to use the first-order condition mentioned in (a) that R ′^ λ = X ′^ εˆ. (c) Verify that you have proved in (b) that (1.4.9) = (1.4.11).
- (Proof of the decomposition (1.2.17)) Take the unrestricted model to be a regression where one of the regressors is a constant, and the restricted model to be a regression where the only regressor is a constant. (a) Show that (b) in the previous exercise is the decomposition (1.2.17) for this case. Hint: What is ̂β for this case? Show that SSR (^) R =
i ( yi^ −^ y ¯)^2 and ( b − ̂β) ′^ ( X ′^ X )( b − ̂β) =
i (^ y ˆ^ −^ ¯ y )^2. (b) ( R^2 as an F -ratio) For a regression where one of the regressors is a con- stant, prove that
F =
R^2 /( K − 1 )
( 1 − R^2 )/( n − K )
- (Hausman principle in finite samples) For the generalized regression model, prove the following. Here, it is understood that the expectations, variances, and covariances are all conditional on X. (a) Cov(̂ β (^) GLS , b − ̂β (^) GLS ) = 0. Hint: Recall that, for any two random vectors x and y ,
Cov( x , y ) ≡ E
[(
x − E( x )
y − E( y )
) ′ ]
