Question

Reserve Problems Chapter 5 Section 4 Problem 3 Suppose that X and Y are independent continuous random variables. Show that oxy o If X and Y are independant, then fxy (x, y) = – fx (x) - and the range of (X, Y) is rectangular. Therefore, fyy) / xyfx (x) dx E(X) fy(x) [fy(x) dx E(Y) Hence, Oxy = 0 Fan- | | C = 15 If X and Y are independant, then fxx (x, y) = fx (x) +fy (y) and the range of (X, Y) is rectangular. Therefore, E(XY) = xy x (x) + f (y)) dxdy = / xfx (x) dx + yfy (y)dy = E(X) +E(Y) Hence, Oxy = 0 Hence, xy = 0 O f X and Y are independant, then fxx (x,y) = fx (x)-fy(y) and the range of (X, Y) is rectangular. Therefore, E(XY) = //xyx () –fy () dxdy = / xx (x)dx - ſwy()dy = E(X) = E(Y) Hence, oxy = 6 O 18 X and Y are independant, then fxx (x, y) = fx (x)}y () and the range of (X, Y) is rectangular. Therefore, EXY) = S(x850) dxay = a()«r sy (way = E@)EY)

Answer 1

X & Y are independent continuous Random variables. fx y (x,y) = fx (2) fy(Y). shere fx, y (x,y) = Joint pdf of X&Y fx (2) = Marginal belf of x. fy(y) = Marginal pdf of Y. Now, Txy = COV(x,y) = E(XY) – EX) ELY). EXY) = S(xy. fx , y (x, y) dedy = ss xy. fx (x) fy (9) dx dy , as x,y are independent КК = ( x-f(x) dx fy fy lydy - E( E(Y). . Txy = E(XY) - E(X) ELY) = E(X) ELY) – E(X) ECY) -0. - e

Last option is correct. o If x,y are independent, then fxy(x,y)=fxc0f%Cy) and the range of @Y) is rectangular EXY = SJ a4 fx (29fy CD. Szfx Crydx fyfylg) dy = E(X) E (4) Hence, Txy=0