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\newcommand{\ds}{\displaystyle}
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{\bf Question}

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

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

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

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


\vspace{.25in}

{\bf Answer}

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

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

$\Rightarrow 0&lt;x_1&lt;\infty;\ \ \ 0&lt;2z-w&lt;\infty$



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

Therefore the joint pdf of (Z,W) is

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

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

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