Ec402 Econometrics. Suitable for all candidates. Summer 2012 (part of) examination. Instructions to candidates. Time allowed: 3 hours

Transcription

1 Summer 2012 (part of) examination Ec402 Suitable for all candidates Instructions to candidates Time allowed: 3 hours This paper contains THREE sections. Answer all questions in Section A, ONE question from Section B, and ONE question from Section C. Question 1 carries 60% of the overall mark; Sections B and C each carry 20% of the overall mark. You are supplied with: Dickey-Fuller Statistical Tables Murdoch and Barnes Statistical Tables (2nd-4th editions) Table A5 Durbin-Watson d-statistic Calculators are not allowed in this examination.

2 Summer 2050 examination by Danny Quah LSE March 2011 Section A This section carries 60% of the total marks. Answer all questions. 1. ANS 2. ANS 3. ANS 4. ANS 5. ANS 6. (8 points) Consider the covariance stationary AR(1) process X t = ρx t ½ +ǫ t ρ < ½ ǫ w.n. (¼ ½)º (a) Calculate the covariogram G X (s), s = ¼ ±½ ±¾ º º º terms of ρ. (b) Given observations on X t, all integer t, a research assistant accidentally runs the regression X t = βx t+½ +ν t (i.e., instead of regressing X t on the lag X t ½, he regresses X t on future X t+½ ). Prove that OLS consistently estimates β = ρ. (c) What is the relation between ν and ǫ? Show that ν is serially uncorrelated. ANS in c LSE 2012/Ec402 Page 2

3 (a) Notice EX t = ¼. Then [( ) ¾ ] G X (¼) = E ρ s ǫ t s = ρ ¾s = (½ ρ ¾ ) ½ [( )( )] G X (½) = E ρ s ǫ t s ρ s ǫ t ½ s = (½ ρ ¾ ) ½ ρ [( )( )] G X (¾) = E ρ s ǫ t s ρ s ǫ t ¾ s = (½ ρ ¾ ) ½ ρ ¾ º Continuing this we see G X (s) = (½ ρ ¾ ) ½ ρ s s = ¼ ±½ ±¾ º º º º (b) The OLS estimator converges to G X (¼) ½ G X (½), i.e., ρ (not ρ ½ ). (c) To see the relation between ν and ǫ, write ν t = X t ρx t+½ = ρ j ǫ t j ρ ρ j ǫ t+½ j j=¼ j=¼ = ρǫ t+½ +(½ ρ ¾ ) ρ j ǫ t j º j=¼ Thus, the relationship is a dynamic one involving past, current, and future values. Since ν is a moving average of ǫ s, it seems ν cannot be white noise. However, let s check: Eν t ν t ½ = (½ ρ ¾ )ρ+(½ ρ ¾ ) ¾ ρ ρ ¾j j=¼ = (½ ρ ¾ )ρ+(½ ρ ¾ ) ¾ ρ (½ ρ ¾ ) ½ = ¼ c LSE 2012/Ec402 Page 3

4 Eν t ν t ¾ = (½ ρ ¾ )ρ ¾ +(½ ρ ¾ ) ¾ ρ ¾ j=¼ ρ ¾j = (½ ρ ¾ )ρ ¾ +(½ ρ ¾ ) ¾ ρ ¾ (½ ρ ¾ ) ½ = ¼ and continuing this calculation establishes Eν t ν t j = ¼ for all j ¼. Alternatively, we can calculate the covariogram E[ν t ν t s ] = E[(X t ρx t+½ )(X t s ρx t+½ s )] For s ½, so that = G X (s) ρg X (s+½) ρg X (s ½)+ρ ¾ G X (s) = (½+ρ ¾ )G X (s) ρg X (s+½) ρg X (s ½)º G X (s+½) = ρg X (s) = ρ ¾ G X (s ½) E[ν t ν t s ] = (½+ρ ¾ )G X (s) ρ ¾ G X (s) ρρ ½ G X (s) = ¼ for all s ½. Hence, apparently paradoxically, ν a moving average of w.n. is actually itself also w.n. 7. (8 points) Suppose u t = ǫ t +θǫ t ½ ǫ w.n. (¼ ½)º (a) Derive the covariogram of u in terms of θ. (b) Suppose θ = ½. Establish that u has covariogram sum equal to 0. (c) Consider the regression model: y t = X t β+u t where u t is as above, with θ = ½, and where X t = ½, i.e., the regressors are just a constant. What are the properties of the OLS estimator for β? c LSE 2012/Ec402 Page 4

5 ANS (a) Notice Eu t = ¼. Straightforwardly, G u (¼) = E[u ¾ t] = ½+θ ¾ G u (½) = E[u t u t ½ ] = θ G u (s) = E[u t u t s ] = ¼ for all s ¾. (b) Direct calculation gives s= G u (s) = ½+θ ¾ + ¾θ so that when θ = ½, the covariogram sum equals ½ + ½ ¾ = ¼. (c) The OLS estimator β T satisfies β T β ¼ = ( T ½ T X ¾ t ) ½ ( T ½ T ) X t u t T = T ½ u t = (ǫ T ǫ ¼ )/T which converges in probability to 0. Thus, OLS is consistent. However, apart from that inherited from ǫ trivially, the OLS estimator has no non-degenerate distribution, regardless of scaling. c LSE 2012/Ec402 Page 5

6 Section B This section carries 20% of the total marks. Answer ONE question. 1. ANS 2. ANS c LSE 2012/Ec402 Page 6

7 Section C This section carries 20% of the total marks. Answer ONE question. 1. ANS 2. Consider the linear model: y t = X tβ+u t X t n ½ β ¼ true parameter valueº Suppose E(X t s u t) = ¼ at s = ¼ but is otherwise unrestricted: In particular, E(X t s u t) for s < ¼ is likely different from 0. Assume that ÔÐ ÑT ½ T T X t X t = Q X full rank, finite and T ½/¾ T X t u t L N(¼ Ω)º (a) Provide conditions under which the OLS estimator β T is consistent and asymptotically normal. (b) Since in general Ω Q X σ ¾, OLS software packages will typically report incorrect estimated standard errors. Describe how correct standard errors can be calculated for OLS. Instead of using OLS as above, a researcher attempts GLS (there might also be an unspoken suspicion this will make the resulting estimator efficient). Stacking the data, the researcher constructs y ½. y T = X ½ X T u ½. β+. u T c LSE 2012/Ec402 Page 7

8 or, in vector notation, y = Xβ+uº Suppose some economic theory specifies the value of the T T matrix V = Euu. Suppose, moreover, that we know a T T matrix Λ satisfying Λ Λ = V ½. (c) Using Λ, obtain the GLS estimator for β. (d) Show that under the conditions given in this question, GLS is typically inconsistent. (e) Which assumptions above might be strengthened to make GLS consistent? Describe economic circumstances under which this strengthening would be unreasonable, i.e., so that the researcher should just stick with OLS and apply the corrected standard errors. ANS (a) The following conditions will make OLS consistent and asymptotically normal (repeating some statements already given in the question itself): i. E[X t u t ] = ¼, i.e., E[X t (y t X t β)] = ¼ at β = β ¼; ii. ÔÐ Ñ T T ½ T X tx t = Q X full rank, finite; iii. T ½/¾ T X L tu t N(¼ Ω), Ω full rank, finite. (b) To correct the standard errors from OLS, use the HAC estimator for Ω, i.e., define Z t = X t u t (β) at β = β T and the HAC covariance matrix estimator Ω T =T ½ [ T Z t Z t+ k j=½ ( ½ j ) T ] [Z t Z t j k+½ +Z t jz t] t=j+½ c LSE 2012/Ec402 Page 8

9 for Ω. (c) The GLS estimator obtains from first transforming the vector equation by premultiplying with Λ to obtain: noticing that Λy = ΛXβ+Λu = ỹ = Xβ+ũ Eũũ = ΛEuu Λ = ΛVΛ = ΛΛ ½ (Λ ) ½ Λ = Iº Therefore, GLS is b T = ( X X) ½ ( X ỹ)º (d) That GLS is typically inconsistent follows from continuing the calculation above: b T = ( X X) ½ ( X ỹ) = β ¼ +( X X) ½ ( X ũ)º If we denote the matrix Λ to have (s t) entry λ st then writing out that last term X ũ gives: [ T ( T T ) X ũ = st X t)( ] λ st u t º s=½ λ Notice that each of the constituent terms ( T λ st X t)( T λ st u t ) involves taking the product of u with all X s, past, current and future. For this cross-product normalized by T ½ to converge in probability to zero, it is not enough that EX t s u t = ¼ at s = ¼; instead, we would require EX t s u t = ¼ for all s. c LSE 2012/Ec402 Page 9

10 (e) Following the reasoning just given, the condition could be strengthened to EX t s u t = ¼ at s = ¼ EX t s u t = ¼ for all sº This, however, might be unreasonable: for instance, the model might interpret u t as forecast errors, whereupon they will likely be correlated with future X t s (s < ¼). c LSE 2012/Ec402 Page 10