Browse Questions

# If $\sigma_x= \sigma_y$ and $x,y$ are related by $u=x+y\: v=x-y$ then $cov (u,v)$ is

$\begin {array} {1 1} (A)\;-1 & \quad (B)\;0 \\ (C)\;1 & \quad (D)\;cov\: (x,y) \end {array}$

Can you answer this question?

Given
$u = x+y$
$v = x-y$
$\Rightarrow \overline u = \overline x+\overline y$
$\overline v=\overline x-\overline y$
$u - \overline u=(x-\overline x)+(y-\overline y)$
$v-\overline v=(x-\overline x)-(y-\overline y)$
$(u-\overline u)(v-\overline v)=(x-\overline x)^2-(y-\overline y)^2$
$\Rightarrow \large\frac{1}{n} \Sigma ( u-\overline u)(v-\overline v) = \large\frac{1}{n} \Sigma (x-\overline x)^2-\large\frac{1}{x} \Sigma (y-\overline y)^2$
$= \sigma_x^2-\sigma_y^2=0$
$\Rightarrow cov (u,v)=0$
Ans : (B)
answered Jan 31, 2014