Springer Texts in Business and Economics

Independence and Simple Correlation

a. Assume that X and Y are continuous random variables. The proof is similar if X and Y are discrete random variables and is left to the reader. If X and Y are independent, then f(x, y) = f1(x)f2(y) where f1(x) is the marginal probability density function (p. d.f.) of X and f2(y) is the marginal p. d.f. of Y. In this case,

E(XY) = ' xyf(x, y)dxdy = ' xyf1(x)f2(y)dxdy =(/ xf1(x)dx)(/ yf2(y)dy) = E(X)E(Y)

B. H. Baltagi, Solutions Manual for Econometrics, Springer Texts in Business and Economics, DOI 10.1007/978-3-642-54548-1_2, © Springer-Verlag Berlin Heidelberg 2015

butvar(Y) = b2 var(X) from problem 2.1a. Hence, pXY ± 1 depending on the sign of b.

b var(X) v/b2(var(X))2

E(YX) = E(X2.X) = E(X3) = 0

E(X - E(X))(Y - E(Y)) = E(X - 0)(Y - 2) E(XY) - 2E(X) = E(XY) = E(X3) = 0

_ cov(X, Y) _ о

-s/var(X)var(Y) '