The Best Predictor. Let X and Y be the two random variables considered in problem 16. Now consider predicting Y by a general, possibly non-linear, function of X denoted by h(X).

(a) Show that the best predictor of Y based on X, where best in this case means the minimum mean squared error predictor that minimizes E[Y − h(X)]2 is given by h(X) = E(Y/X).

Hint: Write E[Y − h(X)]2 as E{[Y − E(Y/X)] + [E(Y/X) − h(X)]}2. Expand the square and show that the cross-product term has zero expectation. Conclude that this mean squared error is minimized at h(X) = E(Y/X).

(b) If X and Y are bivariate Normal, show that the best predictor of Y based on X is identical to the best linear predictor of Y based on X.