The goal of this exercise is to show the claim on Slide 23 (Topic 5) that the variance of the OLS estimator is higher due to measurement error. Consider the linear regression Y = Bo + BIX* + wi, where E(u;X* ) = 0. We are interested in estimating 1. X* is not observed perfectly but with error, i.e, we observe X; = X# + e;. The measurement error and the regression error are such that E(Xfe;) = 0, E(e) = 0 and u; le; X*. This is a stronger version of the "First Case" framework of Slide 23. We saw that in this case, the OLS estimator is consistent. Write of* = Var(X*), ox = Var(X), ? = Var(e), etc. (a) For vi = ui - Bier, show that the large sample variance of B1 is 1 E((X; - ux)2v?) ox (b) Use the assumptions of the exercise to show that ox = ox. - o?. (c) We assume that the linear regression satisfies homoskedasticity, i.c, E(u? X#) = 02. Show that E((Xi - ux)'u?) = oxo? and E(equi(Xi - ux)?) = 0. (d) Use the previous results to show that if o? > 0, the variance of the OLS estimator is larger than that of the OLS estimator when there are no measurement error, and even larger if B, # 0.Random Coefficients Measurement Errors First Case Y = B'X + u - Bkek = B'X + vx with vK = u - BKek . First case: we assume Cov(XK, ek ) = 0. . In the linear regression above, the OLS estimator is consistent if E(XVK) = 0 (and if no perfect multicollinearity holds) E(VK) . . . Here E(XVK) = IE(XK 1VK) E(XKVK) . Thus, the OLS estimator converges to B ! . What about its variance? You will show in HW5 that if BK * 0, then the OLS estimator has higher variance