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{\bf Question}

Suppose that $X$ and $Y$ is
$N_2(\mu_x=0,\mu_y=0,\sigma_x=1,\sigma_y=1,\rho)$.  Show that the
sum $X+Y$ and the difference $X-Y$ are independent random
variables. (Hint: Use the transformation technique.)


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{\bf Answer}

Let $U=X+Y,\ V=X-Y$

Find the joint pdf of $U$ and $V$ where the joint pdf of $X$ and
$Y$ is
$$f_{X,Y}(x,y)=\frac{1}{2\pi\sqrt{1-\rho^2}}e^{-\frac{1}{2(1-\rho)^2}\{x^2-2\rho
xy+y^2\}},$$ $$ (-\infty<x<\infty,\ -\infty<y<\infty)$$

$x=\frac{u+v}{2};\ \ \ y=\frac{u-v}{2}$

$J=\left|\begin{array} {cc} \frac{\partial x}{\partial u} &
\frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} &
\frac{\partial y}{\partial
v}\end{array}\right|=\left|\begin{array} {cc} \frac{1}{2} &
\frac{1}{2}\\ \frac{1}{2} & -\frac{1}{2}\end{array}\right| =
-\frac{1}{2}$

Now \begin{eqnarray*} & & x^2-2\rho xy+y^2\\ & = &
\left(\frac{u+v}{2}\right)^2+\left(\frac{u-v}{2}\right)^2-2\rho
\frac{u+v}{2} \frac{u-v}{2}\\ & = &
\frac{u^2}{2}(1-\rho)+\frac{v^2}{2}(1+\rho)\ (\rm{after\ algebra})
\end{eqnarray*}

Therefore
\begin{eqnarray*} f_{U,V}(u,v) & = & \frac{1}{2\pi\sqrt{1-\rho^2}}\, e^{-\frac{1}{2(1-\rho)^2}
\left\{\frac{u^2}{2}(1-\rho)+\frac{v^2}{2}(1-\rho)\right\}}\left|-\frac{1}{2}\right|\\
& = & \frac{1}{4\pi\sqrt{1-\rho^2}}\, e^{-\frac{u^2}{4(1+\rho)}
-\frac{v^2}{4(1+\rho)}}\\ & & -\infty<u<\infty,\ -\infty<v<\infty
\end{eqnarray*}

Easy to see that $U \sim N\{0,\sigma^2=2(1+\rho)\}$ and $V \sim
N\{0,\sigma^2=2(1-\rho)\}$ and $U$ and $V$ are independent.



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