Question

Professor E.Z. Stuff has decided that the least squares estimator is too much trouble. Noting that two points determine a line, Dr. Stuff chooses two points from a sample of size N and draws a line between them, calling the slope of this line the EZ estimator of B2 in the simple regression model. Algebraically, if the two points are (xı,y) and (x2,y2), the EZ estimation rule is 2.8 y2-y1 EZ Assuming that all the assumptions of the simple regression model hold: (a) Show that bez is a "linear" estimator. (b) Show that bgz is an unbiased estimator. c) Find the variance of bEz ( - (d) Find the probability distribution of bEz-

All listed parts please.

Answer 1

B_Ez = y_2 - y_1/x_2 - x_1 b_Ex = beta_2(x_2 - x_1) + (y_2 - y_1)/(x_2 - x_1) b_Ez = beta_2 + u_2 - u_1/x_2 - x_1 where x_2 - x_1 rightarrow is non-stochastic so, it is a linear y_1 = beta_1 + beta_2x_i + ui y_2 = beta_2 + beta_2 x_2 + u_2 y_1 = beta_1 + beta_2 x_1 + u_1 y_2 - y_1 = beta_2(x_2 - x_1) + (u_2 - u_1) now b) E(b_Ez) = E[y_2 - y_1/x_2 - x_1] = beta_2 = (x_2 - x_1)/x_2 - x_1 + E(y_2 - y_1)/x_2 - x_1 since E(u_2 - u_1) = 0 so, E(b_Ez) = beta_2 so, b elementof z is an unbiased estimator of beta_2 c) variance of be_z = var(b_Ez) = var(beta_2 + u_2 - u_1/x_2 - x_1) var(b_EZ) = sigma^2/sigma(x_2 - x_1)^2 since var constant = 0 b_EZ is normally distributed z = b_EZ - beta_2 /squareroot var(b_EZ) tilde N(0, 1)