x = 1 n (x = 1 (x n 1 ι(ι ι) 1 ι x) (x ι(ι ι) 1 ι x) = 1

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1 Estimation and Inference in Econometrics Exercises, January 24, 2003 Solutions 1. a) cov(wy ) = E [(WY E[WY ])(WY E[WY ]) ] = E [W(Y E[Y ])(Y E[Y ]) W ] = W [(Y E[Y ])(Y E[Y ]) ] W = WΣW b) Let Σ be a p p covariance matrix of random vector Y =(Y 1,...,Y n ). Then for any a IR p, Z = a Y is a scalar random variable, and a Σa = a cov[y ]a = a E [(Y E[Y ])(Y E[Y ])] a = E [(ay E[aY ])(ay E[aY ]) ] = E[(Z E[Z])(Z E[Z])] (Z is a scalar rv) = E[(Z E[Z]) 2 ] = var[z] 0 (by definition), i.e., Σ is p.s.d. c) Define ι =(1, 1,...,1) an n-vactor of ones. Then x = 1 n n x i = 1 n ι x, i=1 where x = (x 1,...,x n ). Because n = = ι ι, we can also write x =(ι ι) 1 ι x (c.f. the least squares estimator ˆβ =(X X) 1 X y). In the same manner s 2 x = 1 ni=1 (x n 1 i x) 2 = 1 (x n 1 ι(ι ι) 1 ι x) (x ι(ι ι) 1 ι x) = 1 n 1 x (I ι(ι ι) 1 ι )x, where I is the n n unitmatrix,andwehaveusedtheproperty (I ι(ι ι) 1 ι )(I ι(ι ι) 1 ι )=I ι(ι ι) 1 ι. (Again, c.f. in the regression leas squares solution s 2 = 1 (y n p 1 ŷ) (y ŷ) = 1 n p 1 y (I X(X X) 1 X )y.) 2. The problem is to find the solution for the constrained optimization problem: min ω ω Σω (1)

2 subject to ω µ = µ p and ω ι =1, where ι =(1,...,1) is the vector of ones. The Lagrangian function is L = ω Σω δ 1 (ω µ µ p ) δ 2 (ω ι 1). Partial derivatives L ω = 2Σω δ 1µ δ 2 ι = 0 (2) L δ 1 = ω µ µ p =0 (3) L δ 2 = ω ι 1=0. (4) Solving (2) for ω, and denoting the solution by ω p,weobtain ω p = 1 δ1 Σ 1 µ + δ 2 Σ 1 ι (5) 2 Multiplying (5) from the left first by µ,thenbyι, and taking into account (3) and (4) gives equations or µ p = 1 2 (δ 1µ Σ 1 µ + δ 2 µ Σ 1 ι) (6) 1= 1 2 (δ 1ι Σ 1 µ + δ 2 ι Σ 1 ι) (7) where a = ι Σ 1 µ, b = µσ 1 µ,andc = ι Σ 1 ι. Solving (8) and (9) for δ 1 and δ 2 gives µ p = 1 2 (δ 1b + δ 2 a) (8) 1= 1 2 (δ 1a + δ 2 c), (9) δ 1 = 2(cµ p a) d (10) δ 2 = 2(b aµp) d, (11) 2

3 where d = bc a 2. Using these in (5) and arranging terms, we get ω p = g + hµ p, (12) where g = 1 d (bσ 1 ι aσ 1 µ) (13) h = 1 d (cσ 1 µ aσ 1 ι). (14) 3. Theorem. (Gauss-Markov) Consider the regression model y = X β +, where y is an n 1 vector, X is an n (p +1) matrix, β is a (p +1) 1 parameter vector, and is an n 1 random vector. Then under the (classical) asumptions (1) E[ ] =0, (2) cov[ ] =σ 2 I,whereI is the n n unit matrix, (3) X is nonstochastic matrix with rank equal to p +1, the least squares estimator ˆβ =(X X) 1 X y is the best lienar unbiast estimator (BLUE) of β. Proof. Linearity: By denoting A = (X X) 1 X, we find ˆβ = Ay, i.e., a linear transformationm of y. Unbiasedness: Write ˆβ = β +(X X) 1 X. Then because X is non-stochastic, we get E[ ˆβ] = β +(X X) 1 X E[β] = β, because E[ ] =0. by assumption (1) That is ˆβ is unbiased. Best: Let β be another linear unbiased estimator of β. The task is to show that cov[ β] cov[ ˆβ] is positive semi definite (p.s.d). 3

4 By the linearity β must be of the form β = Dy = DXβ + D, where D is some (p + 1) n non-stochastic matrix. Furthermore, because of unbiasedness and E[D ] =D E[ ] =0, wemusthave β = E[ β] =DXβ, which implies that DX = I. Let C = D (X X) 1 X.ThenCX = 0, and cov[ β] = cov[c +(X X) 1 X y] = (C +(X X) 1 X )cov[y(c +(X X) 1 X ) = σ 2 (C +(X X) 1 X )I(C +(X X) 1 X ) = σ 2 CC + σ 2 CX(X X) 1 +σ 2 (X X) 1 X C + σ 2 (X X) 1 X X(X X) 1 = σ 2 CC + σ 2 (X X) 1 = σ 2 CC +cov[ˆβ]. Thus, finally cov( β) cov( ˆβ) =σ 2 CC which is p.s.d. This completes the proof. 4. Now C t = β 0 + β 1 Y t + β 2 C t 1 + t. The long run propensity to consume is by taking expectations, and considering obtained by setting C t = C t 1 = C, and ignoring the error term (i.e., consider the expected relationship) so that (1 β 2 )C = β 0 + β 1 Y or C = β0 + β1y, where β0 = β 0 1 β 2 and β1 = β 1. 1 β 2 a) Below are time series and scatter plots of the consumption and income series. 4

5 2800 US Consumtion and Income 1950 to USD bn CS Y Year Y CS 5

6 The estimation result using EViews are as follows Dependent Variable: CS Method: Least Squares Date: 01/31/03 Time: 16:00 Sample(adjusted): Included observations: 35 after adjusting endpoints Variable Coefficient Std. Error t-statistic Prob. C Y CS(-1) R-squared Mean dependent var Adjusted R-squared S.D. dependent var S.E. of regression Akaike info criterion Sum squared resid Schwarz criterion Log likelihood F-statistic Durbin-Watson stat Prob(F-statistic) The R-square is pretty high. The Durbin-Watson test statistic indicates autocorrelation in the residuals. Nevertheless the statistic is not fully useful here because of lagged endogenous variables in the equation. Nevertheless, the residual graph and correlogram below strongly support the idea that the residuals are correlated Residual Actual Fitted 6

7 Correlogram of Residuals Date: 01/31/03 Time: 16:24 Sample: Included observations: 35 Autocorrelation Partial Correlation AC PAC Q-Stat Prob Let us next re-estimate the model with first order autocorrelation in resuiduals, so that where ν NID(0, σ 2 ν). Estimating the model gives, C t = β 0 + β 1 Y t + β 2 C t 1 + t t = φ t 1 + ν t, Dependent Variable: CS Method: Least Squares Date: 02/02/03 Time: 23:46 Sample(adjusted): Included observations: 34 after adjusting endpoints Convergence achieved after 8 iterations Variable Coefficient Std. Error t-statistic Prob. C Y CS(-1) AR(1) R-squared Mean dependent var Adjusted R-squared S.D. dependent var S.E. of regression Akaike info criterion Sum squared resid Schwarz criterion Log likelihood F-statistic Durbin-Watson stat Prob(F-statistic) Inverted AR Roots.54 Now, as seen from the correlogram, there is no empirical evidence for additional autocorrelation remined. 7

8 Correlogram of Residuals Date: 02/02/03 Time: 23:51 Sample: Included observations: 34 Q-statistic probabilities adjusted for 1 ARMA term(s) Autocorrelation Partial Correlation AC PAC Q-Stat Prob Nevertheless, there is left behind some large deviations, especially in early 1970 s and in the end of the sample period, as is seen in the graph below. We, however, do not refine the model further in this case Residual Actual Fitted b) On the basis of these estimation results the the estimate of the short run MPC is ˆβ 1 = and the long run MPC ˆβ 1 = ˆβ 1 /(1 ˆβ 2 )= /( ) c) To test the null hypothesis H 0 : β1 = 1 can be accomplished in several ways. The null hypothesis implies β 1 /(1 β 2 ) = 1, i.e., β 1 =1 β 2,orβ 1 + β 2 = 1. If there is a ready made program to work out linear restrictions on the parmaeters it can be used right away. If not the we have to estimate the model under the null hypothesis separately and calculate the F -test. Under the 8

9 null hypothesis the model becomes C t = β 0 +(1 β 2 )Y t + β 2 C t 1 + t = β 0 + Y t + β 2 (C t 1 Y t )+ t. So technically we estimate the following model under the null hypothesis C t Y t = β 0 + β 2 (C t 1 Y t )+ t. Dependent Variable: CS-Y Method: Least Squares Date: 02/03/03 Time: 00:23 Sample(adjusted): Included observations: 34 after adjusting endpoints Convergence achieved after 12 iterations Variable Coefficient Std. Error t-statistic Prob. C CS(-1)-Y AR(1) R-squared Mean dependent var Adjusted R-squared S.D. dependent var S.E. of regression Akaike info criterion Sum squared resid Schwarz criterion Log likelihood F-statistic Durbin-Watson stat Prob(F-statistic) Inverted AR Roots.90 From these and the unrestricted results we get the F -statistic F = SSE R SSE U SSE U df U df R df U = From the F -table F.01 (1, 30) = 7.56 < 7.97 = F. So the null hypothesis is rejected, and we infer that the long run MPCs is different from one (less than one). 9