Question #339377

The joint probability density of X, Y is

f(x, y) = e −(x + y) xi > 0 i = 1, 2

0 otherwise

Using the change of variable technique, determine the joint distribution of Z = X and W = X + Y


1
Expert's answer
2022-05-12T08:56:55-0400

At first, we check that f(x,y)f(x,y) is a valid joint probability density. We have to prove that 0+0+e(x+y)dxdy=1\int_0^{+\infty}\int_0^{+\infty}e^{-(x+y)}dxdy=1. Compute the integral at the left side: 0+0+e(x+y)dxdy=0+e(x+y)0+ dy=0+eydy=ey0+=1\int_0^{+\infty}\int_0^{+\infty}e^{-(x+y)}dxdy=\int_{0}^{+\infty}-e^{-(x+y)}|_0^{+\ \infty}dy=\int_{0}^{+\infty}e^{-y}dy=-e^{-y}|_0^{+\infty}=1.

Thus, it is the valid distribution. In order to find the joint distribution, consider the probability: P{X<z,X+Y<w}P\{X<z,X+Y<w\}. In case z<0z<0 or w<0w<0, P{X<z,X+Y<w}=0P\{X<z,X+Y<w\}=0. In case z>0,w>0z>0,w>0, we get: P{Z<z,X+Y<w}=x<z,x+y<wf(x,y)dxdy=0min(z,w)(0wxe(x+y)dy)dx=0min(z,w)e(x+y)0wxdx=0min(z,w)(exew)dx=ex0min(z,w)min(z,w)ew=1emin(z,w)min(z,w)ew.P\{Z<z,X+Y<w\}=\iint_{x<z,x+y<w}f(x,y)dxdy=\int_0^{min(z,w)}(\int_0^{w-x}e^{-(x+y)}dy)dx=\int_0^{min(z,w)}-e^{-(x+y)}|_0^{w-x}dx=\int_0^{min(z,w)}(e^{-x}-e^{-w})dx=-e^{-x}|_0^{min(z,w)}-min(z,w)e^{-w}=1-e^{-min(z,w)}-min(z,w)e^{-w}.

Thus, in case z<wz<w, we get: P{Z<z,X+Y<w}=1ezzewP\{Z<z,X+Y<w\}=1-e^{-z}-ze^{-w}. For z>wz>w we have: P{Z<z,X+Y<w}=1ewwewP\{Z<z,X+Y<w\}=1-e^{-w}-we^{-w}. Denote F(z,w)=P{Z<z,X+Y<w}F(z,w)=P\{Z<z,X+Y<w\}

The joint probability density is: p(z,w)=2F(z,w)zw=ewp(z,w)=\frac{\partial^2F(z,w)}{\partial z\partial w}=e^{-w}, z<wz<w, z>0,w>0z>0,w>0.

Answer: The joint probability density function is: p(z,w)=ewp(z,w)=e^{-w}, z<wz<w, z>0,w>0.z>0,w>0.


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