Question

PART B: Application 5. Suppose that you observe a random variable X. and then, on the basis of the observed value. you attempt to predict the value of a second random variable Y. Let Y denote the predictor or an estimator of Y ; that is, if X is observed to equal , then Y is your prediction for the value of Y, and your goal is to choose Y so that it tends to be close to Y First, let's consider the situation where there are no data available to use, i.e. X is not available. We measure the distance between a random variable Y and Y by (Y-Y)? The closer Y is to y, the smaller this quantity is. However, if you search for a value of Y that minimizes (Y-Y). then Y would depend on Y, and making Y a useless predictor of Y. One possible approach is to minimize the mean square error (MSE), i.e. Therefore, determine the value of Y that minimizes the MSE will provide a good predictor of (a) Show that E (Y - »)") - E[(y - E(Y))?] + (E(Y) - Y). (b) Argue that choosing Y = E(Y) will minimize the MSE. The result from part (1) suggests that if we want to predict the value of Y when there are no data available to use, a possible prediction (with respect to MSE) is a mean of Y, i.e. Y = E(Y). Now, suppose that X is observed to be , then the prediction problem remains exactly as no-data case, but we now use conditional for all probabilities and expectations, such as Y = E(YX). The estimation error between Y and Y is defined as Y = Ý - Y. Note that Y is completely determined by X while Y is a random variable. (c) Prove that E( X) = 0 and E(Y) = 0. It means that the expectation E(Y|X = ?) is identically zero for all values of x. (d) Show that Var(Y) =Var(Ỹ) + Var(*).

Answer 1

(a) E(Y - Y)2 = E(Y – E(Y) + E(Y) - Y)2 = E(Y - E(Y))2 + (E(Y) - Y)2 +2(E(Y) - Y) E(Y - E(Y)) = E(Y - E(Y))2 + (E(Y) - Y)2 +2(E(Y) - Y(E(Y) – E(Y)) = E(Y – E(Y))² + (E(Y) - Y)? (6) MSE = E(Y - Y)2 = E(Y – E(Y))2 + (E(Y) - Y)? > E(Y – E(Y))? since (E(Y) –Y)?> 0 Hence MSE is minimum i.e. E(Y – E(Y))2 when E(Y) - Y = 0 or Y = E(Y). (c) E(Ỹ|X) = E (Ý –Y)|X) =Ý - EY|X) =Ý – Ý = 0; E(Y) = E(E(Y|X)) = 0 (d) We know, Var(Y) = Var(E(Y|X)) + E(Var(Y|X)) = Var(*) + E(E(Y – EY|X))) =Var(*) + E(E(Y – Ý)?|X) Var(*) + E(E(Y –Y)?|x) = ar(Ý) + E(E(Ỹ)? X) = Var(Y) + E(Y)2 = Var(Y) +Var(Y) (since E(Y) = 0)