CHAPTER 6 LARGE SAMPLE INFERENCE AND PREDICTION 1 Chapter 6 Large Sample Inference and Prediction 6.1 Large sample inference Consider the null hypothesis H0 : Rβ = r. where R is a J ×K matrix. Wald test for this null hypothesis is defined as W = (Rb?r)′ bracketleftBig ?σ2R(X′X)?1 R′ bracketrightBig?1 (Rb?r) as n →∞, W d→χ2 (J). This result does NOT require a normality assumption. This result follows from 1. √n(Rb?r) d→ N (0,σ2RQ?1R′). 2. ?σ2 p→σ2. 3. parenleftBigX′X n parenrightBig?1 p →Q?1. Writing W = √n(Rb?r)′ ? ??σ2R parenleftBiggX′X n parenrightBigg?1 R′ ? ? ?1√ n(Rb?r) and applying these, we have W d→N parenleftBig 0,σ2RQ?1R′ parenrightBigbracketleftBig σ2RQ?1R′ bracketrightBig?1 N parenleftBig 0,σ2RQ?1R′ parenrightBig = N (0,IJ)′ N (0,IJ) = χ2 (J) Consider H0 : βk = β0k. For this null hypothesis, we have t = bk ?β 0 kradicalBig s2 (X′X)?1kk . Writing t = √nparenleftBigb k ?β0k parenrightBig radicalBig s2 (X′X)?1kk , we find that t d→ N parenleftBig 0,σ2Q?1kk parenrightBig radicalBig σ2Q?1kk = N (0,1) CHAPTER 6 LARGE SAMPLE INFERENCE AND PREDICTION 2 6.2 Testing nonlinear restrictions H0 : c(β) = q. Since c parenleftBig? β parenrightBig ? c(β) + parenleftBigg?c(β) ?β parenrightBigg′ parenleftBig ?β ?βparenrightBig, Var parenleftBig c parenleftBig? β parenrightBigparenrightBig ? parenleftBigg?c(β) ?β parenrightBigg′ Var parenleftBig? β parenrightBigparenleftBigg?c(β) ?β parenrightBigg , the test we use is Z = c parenleftBig? β parenrightBig ?q parenleftBig?c(β) ?β parenrightBig′ |β=?β hatwidestVarparenleftBig?βparenrightBigparenleftBig?c(β) ?β parenrightBig |β=?β . As n →∞, Z d→N (0,1). Example 1 A Long—Run MPC. hatwidestlnCt s.e. = =a0.003142 (0.01055) + =b0.07495 (0.02873) lnYt+ =c0.09246 (0.02859) lnCt?1 R2 = 0.999712 s = 0.00874 Estimated asymptotic Cov[b,c] = ?0.0003298. H0 : δ = β1?γ = 1. d = b1?c = 0.074951?0.9246 = 0.99403 gb = ?d?b = 11?c = 13.2626 gc = ?d?c = b(1?c)2 = 13.1834. The estimated asymptotic variance of d is g2bEst.Asy.Var[b] +g2cEst.Asy.Var[c] + 2gbgcEst.Asy.Cov[b,c] = 13.26262 ×0.028732 + 13.18342 ×0.028592 + 2(13.2626)(13.1834)(?0.0003298) = 0.17192. Thus, Z = 0.99403?1√0.17192 = ?0.0144. CHAPTER 6 LARGE SAMPLE INFERENCE AND PREDICTION 3 6.3 Prediction Suppose that we want to predict y0 = X0′β +ε0. The minimum variance linear unbiased estimator of E(y0|X0) = X0′β is ?y0 = X0′b. The forecast error is e0 = y0 ? ?y0 = (β ?b)′ X0 +ε0. The prediction variance is Var parenleftBig e0|X,X0 parenrightBig = σ2 +Var bracketleftBig (β ?b)′ X0|X,X0 bracketrightBig = σ2 +X0′ bracketleftBig σ2 (X′X)?1 bracketrightBig X0. Theprediction variance can be estimatedbyusing s2 in place of σ2. Aconfidence interval for y0 would be formed using ?y0 ±tλ/2 ·se parenleftBig e0 parenrightBig . This formula is based on the assumption of normality for the regression errors. Measures for assessing the predictive accuracy of forecasting models are 1. Root mean squared error RMSE = radicaltpradicalvertex radicalvertexradicalbt 1 n0 summationdisplay i (yi ? ?yi)2 n0 : # of periods being forecasted 2. Mean absolute error MAE = 1n0 summationdisplay i |yi ? ?yi| 3. Theil U?statistic U = radicaltpradicalvertex radicalvertexradicalvertex radicalbt parenleftBig 1 n0 parenrightBigsummationtext i (yi ? ?yi) 2 parenleftBig 1 n0 parenrightBigsummationtext i y2i U? = radicaltpradicalvertex radicalvertexradicalvertex radicalbt parenleftBig 1 n0 parenrightBigsummationtext i (?yi ???yi) 2 parenleftBig 1 n0 parenrightBigsummationtext i (?yi) 2