Solutions to Odd-Numbered End-of-Chapter Exercises: Chapter 14

Transcription

1 Introduction to Econometrics (3 rd Updated Edition) by James H. Stock and Mark W. Watson Solutions to Odd-Numbered End-of-Chapter Exercises: Chapter 14 (This version July 0, 014) 015 Pearson Education, Inc. Publishing as Addison Wesley

2 Stock/Watson - Introduction to Econometrics - 3 rd Updated Edition - Answers to Exercises: Chapter (a) Since the probability distribution of t is the same as the probability distribution of t 1 (this is the definition of stationarity), the means (and all other moments) are the same. (b) E( t ) 0 1 E( t 1 ) E(u t ), but E(u t ) 0 and E( t ) E( t 1 ). Thus E( t ) 0 1 E( t ), and solving for E( t ) yields the result. 015 Pearson Education, Inc. Publishing as Addison Wesley

3 Stock/Watson - Introduction to Econometrics - 3 rd Updated Edition - Answers to Exercises: Chapter (a) To test for a stochastic trend (unit root) in ln(ip), the ADF statistic is the t- statistic testing the hypothesis that the coefficient on ln(ip t 1 ) is zero versus the alternative hypothesis that the coefficient on ln(ip t 1 ) is less than zero. The calculated t-statistic is t From Table 14.4, the 10% critical value with a time trend is 3.1. Because , the test does not reject the null hypothesis that ln(ip) has a unit autoregressive root at the 10% significance level. That is, the test does not reject the null hypothesis that ln(ip) contains a stochastic trend, against the alternative that it is stationary. (b) The ADF test supports the specification used in Exercise 14.. The use of first differences in Exercise 14. eliminates random walk trend in ln(ip). 015 Pearson Education, Inc. Publishing as Addison Wesley

4 Stock/Watson - Introduction to Econometrics - 3 rd Updated Edition - Answers to Exercises: Chapter (a) E W c E W c [( ) ] {[ W) ( W )] } E[( W ) ] E( W )( c) ( c) ( c). W W W W W W (b) Using the result in part (a), the conditional mean squared error E[( f ),,...] ( f ) t t1 t1 t tt 1 t t1 t1 with the conditional variance tt 1 E t tt 1 [( ) ]. This equation is minimized when the second term equals zero, or when f 1. 1 (An alternative is to use the hint, and notice that the result follows immediately from exercise.7.) (c) Applying Equation (.7), we know the error u t is uncorrelated with u t 1 if E(u t u t 1 ) 0. From Equation (14.14) for the AR(p) process, we have t tt u f(,,..., ), t1 t1 0 1 t t3 p tp1 t1 t tp1 a function of t 1 and its lagged values. The assumption Eu ( t t 1, t,...) 0 means that conditional on t 1 and its lagged values, or any functions of t 1 and its lagged values, u t has mean zero. That is, Eu ( u ) Eu [ f(,,..., )] 0. t t1 t t1 t tp Thus u t and u t 1 are uncorrelated. A similar argument shows that u t and u t j are uncorrelated for all j 1. Thus u t is serially uncorrelated. 015 Pearson Education, Inc. Publishing as Addison Wesley

5 Stock/Watson - Introduction to Econometrics - 3 rd Updated Edition - Answers to Exercises: Chapter (a) From Exercise (14.1) E( t ).5 0.7E( t 1 ) E(u t ), but E( t ) E( t 1 ) (stationarity) and E(u t ) 0, so that E( t ).5/(10.7). Also, because t t 1 u t, var( t ) 0.7 var( t 1 ) var(u t ) 0.7 co1v( t 1, u t ). But cov( t 1, u t ) 0 and var( t ) var( t 1 ) (stationarity), so that var( t ) 9/(1 0.7 ) (b) The 1 st autocovariance is cov( t, t1 ) cov( t1 u t, t1 ) 0.7 var( t1 ) cov(u t, t1 ) The nd autocovariance is cov(, ) cov[(1 0.7) u 0.7 u, ] t t t t t1 t 0.7 var( t) cov( ut 0.7 ut 1, t) 0.7 (c) The 1 st autocorrelation is cov(, ) 0.7 corr (, ) 0.7. t t t1 t1 var( t) var( t 1) The nd autocorrelation is cov(, ) 0.7 corr (, ) t t t t var( t) var( t ) (d) The conditional expectation of T 1 given T is T T 1/ T 015 Pearson Education, Inc. Publishing as Addison Wesley

6 Stock/Watson - Introduction to Econometrics - 3 rd Updated Edition - Answers to Exercises: Chapter (a) E( t ) 0 E(e t ) b 1 E(e t 1 ) b q E(e t q ) 0 [because E(e t ) 0 for all (b) values of t]. where the final equality follows from var(e t ) e for all t and cov(e t, e i ) 0 for i t. (c) t 0 e t b 1 e t 1 b e t b q e t q and t j 0 e t j b 1 e t 1 j b e t j b q e t q j and q q cov( t, t j ) 0 0bb cov( e, e ), where b 0 1. k m k m tk tjm Notice that cov(e t k, e t j m ) 0 for all terms in the sum. (d) t e b1 var( ) 1, and cov(, ) 0 for j 1. cov( t, t j) eb1, t t j 015 Pearson Education, Inc. Publishing as Addison Wesley

7 Stock/Watson - Introduction to Econometrics - 3 rd Updated Edition - Answers to Exercises: Chapter Write the model as t t ( t 1 t ) u t. Rearranging yields t 0 (1 1 ) t 1 1 t u t. 015 Pearson Education, Inc. Publishing as Addison Wesley