Econometrics 710 Final Exam May 13, 2004 Sample Answers
1. The model is just-identified by the moment E(xi ei ) = 0, so the appropriate (and asymptotˆ ically efficient) estimator is OLS β = (X 0 X)−1 (X 0 Y ) which has the asymptotic distribution ´ √ ³ˆ −1 ΩQ−1 where Q = Ex x0 and Ω = Ex x0 e2 . We can estimate V n β − β →d N (0, V ), V = Q i i i i i by the White estimator à !−1 à !à !−1 n n n X X X ˆ xi xi xi xi e2 ˆi xi xi . V = n−1 n−1 n−1
i=1 i=1 i=1
An asymptotically efficient estimator of θ = β 1 β 2 is ˆ = β 1 β 2 which has the asymptotic distribuθ ˆ ˆ ´ √ ³ˆ tion n θ − θ →d N (0, h0 V h) where ∂ (β 1 β 2 ) = h= ∂β Thus an asymptotic standard error for ˆ is s(ˆ = θ θ) β2 β1 0 . . . 0
∗ ˆ∗ ˆ∗ ˆ∗ θ (b) Draw B samples with replacement from the data and construct β b and ˆb = β 1 β 2 on each. ∗ ∗ ˆ∗ θ Let q1 and q2 be the 2.5% and 97.5% sample quantiles of ˆb . These are estimates of q1 and ˆ∗ ∗ , the bootstrap quantiles of the distribution of ˆ∗ . A percentile 95% confidence interval for q2 θ ∗ , q ∗ ]. θ is [ˆ1 ˆ2 q Alternatively, another percentile interval can be constructed as follows. Let q1 and q2 be ˆ∗ ˆ∗ ∗ the 2.5% and 97.5% sample quantiles of ˆb − ˆ A percentile 95% confidence interval for θ is θ θ. ˆ − q ∗ , ˆ − q ∗ ]. [θ ˆ2 θ ˆ1 ∗ ˆ ∗ θ∗ ˆ ∗ ˆ ∗ θ (c) Draw B samples with replacement from the data and construct β b , ˆb = β 1 β 2 , and s(ˆ ) = p ∗ −1 ˆ ∗0 ˆ ∗ ˆ ∗ ˆ∗ ˆ∗ ³ n h ´V h ∗ on each. Let q1 and q2 be the 2.5% and 97.5% sample quantiles of Tb = ∗ ˆb − ˆ /s(ˆ ). These are estimates of q ∗ and q ∗ , the bootstrap quantiles of the distribution θ θ θ 1 2 ∗ . The equal-tailed percentile-t 95% confidence interval for θ is [ˆ − s(ˆ q ∗ , ˆ − s(ˆ q ∗ ]. of T θ θ)ˆ θ θ)ˆ 2 1
(a) An asymptotic 95% confidence interval for θ is ˆ ± 1.96s(ˆ θ θ).
³ ´0 p ˆ ˆ ˆˆ ˆ ˆ n−1 h0 V h where h = β 2 β 1 0 · · · 0
2. (a) Since the model is a regression, conditioning on a subset of the xi ’s does not affect the validity ˆ of the regression. The OLS estimator β remains consistent. Algebraically, we can write the
estimator as ˆ β= Thus ˆ β−β =
à n X
i=1
xi x0 1 (x1i i
> 0)
!−1
n X i=1
xi yi 1 (x1i > 0)
¢¢−1 ¡ ¡ → p E xi x0 1 (x1i > 0) E (xi ei 1 (x1i > 0)) i = 0 since E (xi ei 1 (x1i > 0)) = E (xi 1 (x1i > 0) E (ei | xi )) = 0 by the law of iterated expectations. For this to work, it is critical that the model is a regresssion rather than a projection. (In the latter case, OLS will be inconsisent for the population projection coefficient β.) Technically we also need the side condition E (xi x0 1 (x1i > 0)) > 0. This is not automatic. i A necessary condition is that x1i have positive support on the region (0, ∞). (If all x1i are negative, then the available sample is empty.) (b) If we observe the observation only if yi > 0, this the sample is truncated, not censored (Tobit) model. (What is important is not the label, but to understand that this is not the same model as the one introduced in class.) However, the basic fact remains that truncation based on the dependent variable renders naive estimation methods inconsistent. Indeed !−1 Ã n n X X −1 0 ˆ−β = xi x 1 (yi > 0) n−1 xi ei 1 (yi > 0) β n ¡ → p E 6= 0
i i=1 ¡ 0 xi xi 1 (yi
Ã
n
−1
n X i=1
xi x0 1 (x1i i
> 0)
!−1
n−1
n X i=1
xi ei 1 (x1i > 0)
> 0)
¢¢−1
i=1
E (xi ei 1 (yi > 0))
Adding the assumption that ei is independent of xi and N (0, σ 2 ), and letting zi = ei /σ ∼ N (0, 1), we find ¡ ¡ ¢ ¢ E (ei 1 (yi > 0) | xi ) = E ei 1 ei > −x0 β | xi i ¢ ¢ ¡ ¡ = σE zi 1 zi > −x0 β/σ | xi i ¢ ¡ = σλ −x0 β/σ i where λ(s) = φ(s)/Φ(s). Thus ¡ ¡ ¢¢ E (xi ei 1 (yi > 0)) = E (xi E (ei 1 (yi > 0) | xi )) = σE xi λ −x0 β/σ . i ¢¢ ¡ ¢ ¡ ¡ ¡ ¢ ¡ ¢¢ ¡ E xi x0 1 (yi > 0) = E xi x0 E 1 zi > −x0 β/σ | xi = E xi x0 Φ x0 β/σ . i i i i i ¡ ¡ ¡ ¢¢¢−1 ¡ ¡ ¢¢ ˆ E xi λ −x0 β/σ 6= β. β →p β + σ E xi x0 Φ x0 β/σ i i i 2
And
Together
3. (a) We know that E µ = µ and thus µ is unbiased for µ. However, ˆ = g(ˆ ) with g(x) = 1/x is a ˆ ˆ θ µ nonlinear function of µ, so will be a biased estimator. ˆ (b) The function g(x) is strictly convex. Thus by Jensen’s inequality, E ˆ = Eg(ˆ ) < g (E µ) = θ µ ˆ g(µ) = θ. The inequality is strict since V ar(ˆ ) > 0. Thus ˆ has upwards bias. µ θ (c) A second-order Taylor series expansion is (in general) 1 µ µ g(ˆ ) ' g(µ) + g 0 (µ) (ˆ − µ) + g00 (µ) (ˆ − µ)2 µ 2 Since g(x) = x−1 , it follows that g 0 (x) = −x−2 and g00 (x) = 2x−3 . Thus the expansion can be written as ˆ ' θ − µ−2 (ˆ − µ) + µ−3 (ˆ − µ)2 µ µ θ Taking expectations, E ˆ − θ ' −µ−2 E (ˆ − µ) + µ−3 E (ˆ − µ)2 = µ−3 V ar (ˆ − µ) = θ µ µ µ σ2 µ3 n
since µ is a sample mean. This is a positive number, suggesting a positive bias, and is ˆ consistent with the implication of Jensen’s ienquality from part 2. This expression can be used to suggest the magnitude of the bias. Note that it depends on the mean and variance (µ and σ 2 ) as well as the sample size n. (d) Yes, the nonparametric bootstrap can be used here. Sampling from the observations with ∗ θ µ replacement, on each sample estimate µ∗ and ˆ = 1/ˆ . The bootstrap estimate of bias is ˆ PB ˆ∗ ˆ θ − θ. B −1
b=1 b
4. The three statistics are numerically identical, so the dispute is an illusion. The GMM Distance statistic equals the Wald statistic in linear models with linear restrictions. The GMM Distance statistic is the difference between the GMM criterion evaluated under the null and alternative hypotheses, e.g. D = J0 − J1 . Since the model is just-identified, J1 = 0. Thus D = J0 . The statistic J0 is the test for overidentifying restrictions for the null model. Thus the three statistics are numerically identical. ˆ 5. Let P = X(X 0 X)−1 X 0 and P Z = Z. The 2SLS estimator is ˆ β = ¡ 0 ¢−1 ¡ 0 ¢ Z X(X 0 X)−1 X 0 Y Z X(X 0 X)−1 X 0 Z ¢−1 ¡ 0 ¢ ¡ Z PY = Z 0P Z ¢−1 ¡ 0 ¢ ¡ = Z 0P P Z Z PY ³ ´−1 ³ ´ ˆ ˆ ˆ = Z0Z Z 0Y
ˆ ˆ so the OLS regression of Y on Z yields the 2SLS slopes β. However, the residuals from the latter
3
regression are ˆˆ e = Y − Zβ ˜ ˆ = Y − P Zβ ˆ 6= Y − Z β = e ˆ The correct 2SLS standard errors are calculated using the residuals e, not e. Thus the “two-stage” ˆ ˜ procedure produces the correct estimates but not the correct standard errors. This result does not depend upon whether or not the model is just identified or how homoskedasticity is treated. 6. The Monte Carlo procedure (as described) appears correct, but the conclusion is incomplete. (Side note: this is a Monte Carlo experiment, not a bootstrap procedure.) Note that the stated conclusion is that the test is oversized. This is a concrete statement about the true probability of Type I error. Specifically, let p = P (Tn > 7.815) . This can be any number. The test is properly sized if p = .05, undersized if p < .05 and oversized if p > .05. The stated conclusion is therefore equivalent to the rejection of the hypothesis that p = .05. The evidence in favor of this conclusion is that p = .07. This is a point estimate, and has a ˆ √ p sampling distribution. By the CLT, B (ˆ − p) →d N (0, p(1 − p)) as B → ∞ Furthermore, under √ p the null hypothesis of p = .05, B (ˆ − .05) →d N (0, (.05) (.95)). Thus an appropriate test of H0 : p = .05 is to reject for large values of the t-ratio p − .05 ˆ t= p (.05) (.95)/B p In this particular case, B = 200 and (.05) (.95)/200 =p .015, so t = 1.33. Alternatively, we can ˆ ˆ calculate standard errors for p using the formula s(ˆ) = p(1 − p)/B = .018, yielding the similar ˆ p t-ratio 1. 11. In either case, we cannot reject H0 at conventional significance levels. Based on this reasoning, it is incorrect to claim that the test must be oversized. A constructive recommendation would be to increase B. Alternatively, we can construct a confidence interval for the true unknown p. A 95% interval is p ±1.96s(ˆ) = .07±1.96 (.018) = [0.035, 0.105]. The true p lies in this interval with 95% probability. ˆ p Since the set includes .05, it is incorrect to conclude that the test is oversized. Another question might be: “If B = 200 is too small, how large should it be?”. One simple answer is to ask how large should B be to reject at the 5% level the hypothesis that p = .05 when we observe p = .07. This is equivalent to finding a B so that ˆ .07 − .05 p > 1.96 (.05) (.95)/B B> µ 1.96 .02 ¶2 (.05) (.95) = 457
or
This is the smallest B for which p = .07 allows us to reject p = .05 ˆ
4