Answer to Question #339377 in Statistics and Probability for ken

Question

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

Answer 1

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

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

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

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

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