SKKU econometrics 2015, solutions PDF

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1 Chang Sik Kim Spring 2015

Introduction to Econometrics SKKU Suggested Solutions for Final Exam 1. (Total 15 points)

(a) (3 points) True. When the variances of error is known, it is possible to obtain transformed model that variance of transformed error is homoskedastic. In this case, according to Gauss-Markov Theorem, OLS esimator in the transformed regession (= GLS estimator in the original model) is BLUE. (b) (3 points) False. White estimator provides consistent estimator for variance of OLS estimator when heteroskedasticity exists. (c) (3 points) True. Suppose that a qualitative variables has ‘2’ categories, and you introduce ‘2’ dummy variables. Then, D1i + D2i = 1 ∀i This exact linear relationship causes perfect multicollinearity. (d) (3 points) True. Multicollinearity is a sample feature, so by simply increasing the size of sample, multicollinearity may be weakend. (e) (3 points) False. Durbin-Watson test statistic follows its own distribution. 2. (Total 30 points) ˆ is (a) (5 points) The formula for OLS estimator β βˆ = arg min S(b) = (y − Xb)′ (y − Xb) and the residual vector u ˆ is ˆ u ˆ = y − yˆ = y − X β = (In − PX )y where PX = X (X ′ X )−1 X ′ : (b) (5 points) u ˆ′ X = [(In − PX )y]′ X = y′ (In − PX )′ X  −1 ′ ′ = (y′ − y′ P X )X = (y′ − y′ X X ′ X X )X  ′ −1 ′ ′ ′ =yX −yX X X XX = y′ X − y′ X =0 (c) (5 points) ˆ ′X ′ + u y′ y = (X βˆ + u ˆ)′ (X βˆ + u ˆ) = ( β ˆ′ )(X βˆ + u ˆ) ′ ′ ′ ′ ˆ+u ˆu ˆ (ˆ u X = X ′u ˆ = 0) = βˆ′ X ′ X βˆ + u ˆ′ X βˆ + βˆ X u = βˆ′ X ′ X βˆ + u ˆ′ u ˆ.

2 (d) (5 points) (y − Xb)′ (y − Xb) = [y − X βˆ + X βˆ − Xb]′ [y − X βˆ + X βˆ − Xb] ˆ ′ (y − X β) ˆ + (βˆ − b)′ X ′ X ( βˆ − b) = (y − X β) =u ˆ′ u ˆ + (βˆ − b)′ X ′ X ( ˆβ − b) (e) (5 points) Classical Assumptions (A1) (A2) (A3) (A4)

E(u) = 0, i.e., Ey = Xβ , V ar(u) = 0 = E(uu′ ) = σ 2 In : homoskedasticity and no autocorrelation, rank(X) = k X is a non-stochastic matrix. ˆ is unbiased. (f) (5 points) Under the classical assumption, β

Therefore, we have

−1 ′   −1 ′ βˆ = X ′ X X y = β + X ′X Xu

and the variance of βˆ is

 −1 ′ X E(u) = β E βˆ = β + X ′ X

ˆ = E[(βˆ − E β)( ˆ βˆ − E β) ˆ ′ ] = E[(βˆ − β)( βˆ − β)′ ] V ar(β)  −1  −1 ′ ] = E[ X ′ X X u · u′ X X ′ X  ′ −1  ′ −1 ′ ′ = XX X E(uu )X X X  −1   −1 X ′ In X X ′ X = σ 2 X ′X  −1 = σ 2 X ′X ˆ can be estimated by Newey-West estimator. (Auto(g) (5 points) The variance of β correlation consistent estimator should be used for the estimation of the variance ˆ of β. 3. (Total 15 points) (a) (5 points) EβˆGLS − β = (X ′ V −1 X )−1 X ′ V −1 Eu =0 ˆGLS − β)′ ] V ar(βˆGLS ) = E[(βˆGLS − β)( β = E[(X ′ V −1 X )−1 X ′ V −1 uu′ V −1 X(X ′ V −1 X )−1 ] = (X ′ V −1 X )−1 X ′ V −1 E (uu′ )V −1 X (X ′ V −1 X )−1 = (X ′ V −1 X )−1 X ′ V −1 X (X ′ V −1 X )−1 = (X ′ V −1 X )−1

3 (b) (5 points) ˜ = βˆGLS + EAy Eβ = β + EA(Xβ + u) = β + AXβ + E(Au) = β + AXβ =β if and only if AX = 0 (c) (5 points) β˜ − β = (X ′ V −1 X )−1 X ′ V −1 y + Ay − β = (X ′ V −1 X )−1 X ′ V −1 u + AXβ + Au   = (X ′ V −1 X )−1 X ′ V −1 + A u Since AX = 0    ′ E(β˜ − β)( β˜ − β)′ = (X ′ V −1 X )−1 X ′ V −1 + A E[uu′ ] (X ′ V −1 X )−1 X ′ V −1 + A

= (X ′ V −1 X)−1 + AX(X ′ V −1 X )−1 + (X ′ V −1 X )−1 X ′ A′ + AV A′ = (X ′ V −1 X)−1 + AV A′

Since AX = 0

To show that βˆGLS is the best linear unbiased estimator since AV A′ > 0. 4. (Total 15 points) (a) (5 points) Note that

and therefore (b) (5 points) From (a),

(c) (5 points)

Pn ui ¯ β = β + Pni=1 X i=1 i   E β¯ = β.

 2 Pn Pn   E(( i=1 nσ 2 ui ) 2 ) i=1 E u i ¯ = V ar β = Pn = . P P 2 ( ni=1 Xi ) ( ni=1 Xi )2 ( i=1 Xi )2

Pn ui 0 →p =0 β¯ − β = Pni=1 X C i=1 i Pn under the assumption that n1 i=1 Xi → C for some constant C. Or   nσ 2 V ar β¯ = Pn →0 ( i=1 Xi )2

as n → ∞. (You don’t need to give all the details, and full credit will be given if you state the unbiasedness and V ar β¯ → 0.

4 5. (Total 10 points) e2 is linear in Y because it is a weighted sum of Y1 and Y1 . Note that (a) β   1 E βe2 = E (β1 + β2 X1 + e1 − (β1 + β2 X2 + e2 )) X1 − X2 1 E (β2 (X1 − X2 ) + e1 − e2 ) = X1 − X2 1 = β2 + E (e1 − e2 ) = β2 X1 − X2 e2 is unbiased. under the classical assumptions. β

(b) Gauss-Markov theorem : Given the classical assumptions, the OLS estimators are BLUE(Best Linear Unbiased Estimator ), i.e., they are the most efficient in e2 is LINEAR and UNBIASED, by the class of linear unbiased estimators. Since β b e2 . Gauss-Markov theorem we know that β2 is better than β

6. (Total 10 points)

(a) (5 points) Set up a regression ln Yi = β1 + β2 ln X2i + β3 ln X3i + ei where β1 = ln(α), and then we have β2 = 1 − β3 which implies ln Yi = β1 + (1 − β3 ) ln X2i + β3 ln X3i + ei = β1 + ln X2i + β3 (ln X3i − ln X2i ) + ei . ⇒ ln Yi − ln X2i = β1 + β3 (ln X3i − ln X2i ) + ei ⇒ ln



Yi X2i





X3i = β1 + β3 ln X2i



+ ei .

Now the number of unknown parameters to estimate is 2 instead of 3. (b) (5 points) Restricted Model: Unrestricted Model:

RSS = 0.6 2

R = 0.95

n = 210 T SS = 100

.

We can use Wald test as Wn =

(RSSR − RSSU ) /J 6−5 = 41.4 = 5/207 RSSU / (n − K)

Under any significance level (1%, 5%, 10%), we reject the null hypothesis since Wn follows χ2 (1) .

5 7. (Total 15 points) (a) (5 points) Due to V ar(ei ) = σ 2 X i2 , transformed simple regession model would be Yi 1 ei = β1 + β2 + Xi Xi Xi ⇒

Y i∗ = β1 X i∗ + β2 + e∗i

(b) (5 points) ∀i, E(e∗i ) = 0, V ar(e∗i ) = X12 × σ 2 Xi2 = σ 2 i Transformed model satisfies ‘Classical Assumption(A.2 - A.4)’. According to Gauss-Markov Theorem, OLS estimator from transformed model (=GLS estimator) is BLUE. (c) (5 points) Σ(Xi − X)(Yi − Y ) b βb2 OLS = , β1 OLS = Y − βb2 OLS X Σ(Xi − X)2 ⇔ Σ(Xi∗ − X ∗ )(Y i∗ − Y ∗ ) b βb1 GLS = , β2 GLS = Y ∗ − βb1 GLS X ∗ Σ(X ∗i − X ∗ )2

8. (Total 15 points) (a) (5 points)

V = Euu′ = σv2Σ where  1 φ ···  φ 1 · ·· 1  Σ=  . . . .. .. 1 − φ2  .. n−1 n−2 φ φ ··· since cov (ut , ut−k ) =

n−1

φ φn−2 ... 1



  , 

φ|k| σ 2. 1 − φ2 v

(b) (5 points) Infeasible GLS estimator will be (No formal proof necessary)  −1 ′ 2 −1 βˆIF GLS = X ′ σ 2 Σ−1 X Xσ Σ y  ′ −1 −1 ′ −1 = XΣ X X Σ y.

(c) (5 points) Feasible GLS estimator will be −1  ˆβF GLS = X ′Σ ˆ −1 X ˆ −1 y, X′Σ where



 ˆ = Σ  

1 b φ .. . bn−1 φ

φb 1 ...

··· ··· .. .

φbn−2 · · ·

bn−1 φ bn−2 φ ... 1



  . 

Here φb is the OLS estimator of AR(1) model with OLS residuals.

(1)

(2)

6 9. (50 points) (a) (5 points) Mean level of employment equals 689,000. (in column (1)) (b) (5 points) Estimated coeffecient of W is -0.41(< 0, negative), and it is significant. This result contradicts the statement that “there is no trade-off between high wages and full employment.” (c) (5 points) Coefficient of Overall demand(Y) becomes smaller and statistically insignificant. The coefficient of Y becomes insignificant by adding P and A, and the coefficients on P and A are positive. These results imply that these variables are negatively correlated with Y. Also, model in column (3) has insignificant tstatistics on many coefficient (tY = 1.6, tP = 1, tA = 1.05) while it has large Fstatistics (F =

RSS(1) −RSS(3) /4 RSS(1) /105−5

= 12.5). So, there exists multicollinearity problem.

(d) (5 points) When it compares to base (L=0, before 1982), 25,000 people lost their jobs due to minimum wage law. (e) i. (5 points) H0 : βL = 0, H1 : βL = 6 0 ) = P (Z > |1.6666|) = Φ(−1.6666) + 1 − Φ(1.6666) p-value = P (Z > − 0.25 0.15 = 2 × (1 − 0.9515) = 0.097 ii. (5 points) H0 : βL ≤ 0, H1 : βL > 0 p-value = P (Z > −1.66) = P (Z < 1.66) = 0.9515 iii. (5 points) Due to different null hypothesis, they have different critical region and p-value. iv. (5 points) Power = P(reject H0 |H1 is true) Assume α = 0.05, when H0 : βL = 0, H1 : βL = −0.2, 96×0.15+0.2 ) + P (Z ≥ −0.25+1.96×0.15+0.2 ) = 0.0628 power = P (Z ≤ −0.25−1.0.15 0.15 when H0 : βL ≤ 0, H1 : βL = −0.2, ×0.15+0.2 ) = 0.0948 power = P (Z ≥ −0.25+1.645 0.15 Power function 0.5

One sided power function Two sided power function

0.45 0.4 0.35 power

0.3 0.25 0.2 0.15 0.1 0.05 −0.375

−0.25 µ

−0.2

−0.125

a

(f)

i. (5 points) Overall demand has positive effect on employment. In case of wage, it gives different effect on employment. L=0(before 1982, ∄ legistration) b = 6.92 − 0.43W + 0.62Y E

0

7 L=1(after 1982, ∃ legistration) b = (6.92 − 0.20) + (−0.43 + 0.05)W + 0.62Y E

(need to consider dummy L, and slople dummy W × L) Therefore wage affects employment differently according to dummy variable L. 399 = 412 = 0.9684 ii. (5 points) R2 = ESS T SS 2 b i − E) b 2 T SS = Σ(Ei − E) implies variation of employment and ESS = Σ( E indicates explained variation of employment by regressors. In this model, R2 is close to unity, so regressors explained a significant proportion of the variation of employment....