# The two variates $x$ and $y$ are uncorrelated and have standard deviation $\sigma_x$ and $\sigma_y$ respectively, the correlation coefficient between $x+y\: and \: x-y$ is

$\begin {array} {1 1} (A)\;\large\frac{\sigma_x \sigma_y}{\sigma_x^2+\sigma_y^2} & \quad (B)\;\large\frac{1}{2}\bigg( \large\frac{1}{\sigma_x}+\large\frac{1}{\sigma_y} \bigg) \\ (C)\;\large\frac{\sigma_x^2-\sigma_y^2}{\sigma_x^2+\sigma_y^2} & \quad (D)\;None \: of \: these \end {array}$

Let $u = x+y\: \: \: v = x-y$
$\therefore \overline u = \overline x+ \overline y\: \: \: \: \: \: \: \: \: \overline v= \overline x- \overline y$
$cov (u,v) = E\{(u-\overline u)\: (v-\overline v)\}$
$= E \{ (x-\overline x)+(y-\overline y) \}.\{ (x-\overline x)-(y-\overline y) \}$
$= E \{ ( x-\overline x)^2-(y-\overline y)^2 \}$
$= \sigma_x^2- \sigma_y^2$
var $(u) = E(u-\overline u)^2$
$= E \{( x-\overline x)+(y-\overline y) \}^2$
$\sigma_x^2+\sigma_y^2$
(Since $x\: and \: y$ are uncorrelated and so $E ( x-\overline x)\: (y-\overline y)=0$)
Similarly var $(v) = \sigma_x^2+\sigma_y^2$
Thus Correlation coefficient=$r(u,v) = \large\frac{cov(u,v)}{\sigma_u \sigma_v}$
$= \large\frac{\sigma_x^2-\sigma_y^2}{\sigma_x^2+\sigma_y^2}$
Hence Ans : (C)
edited Mar 26, 2014