Question

1. The joint probability density function (pdf) of X and Y is given by fxy(x, y) = A (1 – xey, 0<x<1,0 < y < 0 (a) Find the c

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Answer #1

1) The given joint PDF is fx.r(x,y) = A(1 - x)e-V; 0<x<1,0 <y <a .

a) The condition for PDF is

fx,y(x,y)dxdy = 1 Ae | (1-X)dxdy = 1 ए|N (1) = 1 |N

b) The marginal PDFs are

fy(y) = (fx,x(x,y)dx fy(Y) = Aev 1 (1 - x)dx fr (Y) = 4ev fy(y) = e-Y;0<y < 0

fx(x) = fx,x(x,y)dy fx(x) = A ( M (1 - x)dy fx (X) = A(1 - x) /e Ydy JO fx(x) = 2 (1-x); 0<x<1

c) The expectations are

E(X) = 5 *(x)dx E(X) = 1*2*(1-x)dx E(X) = 2 [(x – x2) dx E(X) = 2 (x2/2 – x3/3) E(X) = WENNS

E(Y) = 1 °yfr (y)dy E(Y) = ( ye Ydy E(Y) = 1

d) The expected value

0 1 E(XY) = f ( xyfx,x(x,y)dxdy E(XY) = 26 [xy (1-x) e Ydxdy E(XY) = 2 (yev (x(1 - x) dxdy E(XY) = 3 | ºye Ydy E(XY) = Jo

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