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{\bf Question}

Suppose that $X$ and $Y$ have the joint pdf

$$f(x,y)=\left\{\begin{array}{ll} e^{-(x+y)} & {\rm if}\ x > 0,\ y
> 0\\ 0 & {\rm otherwise}
\end{array} \right.$$

By using a suitable transformation, show that the pdf of
$U=\frac{(X+Y)}{2}$ is given by

$$f_U(u)=4ue^{-2u},\ u>0.$$


\vspace{.25in}

{\bf Answer}

Let $Z=\ds\frac{X_1 + X_2}{2}$ and $W=X_2$

$\Rightarrow x_1=2z+w;\ \ \ x_2=w$

$\Rightarrow 0<x_1<\infty;\ \ \ 0<2z-w<\infty$



$\left|\begin{array} {cc} \frac{\partial x_1}{\partial z} &
\frac{\partial x_1}{\partial w}\\ \frac{\partial w_2}{\partial z}
& \frac{\partial x_2}{\partial w}\end{array}\right| =
\left|\begin{array}{cr} 2 & -1\\ 0 & 1 \end{array}\right| = 2$

Therefore the joint pdf of (Z,W) is

$f_{Z,W}(z,w)=e^{-2z} . 2,\ \ \ 0<2z-w<\infty$

Therefore $f_Z(z)=2\ds\int_0^{2z} e^{-2z} \,dw = 4ze^{-2z},\ \ \
z>0$

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