problem involving a bivariate gaussian

Question

I need some help with this exercise, I tried to do this but my calculations seem to go nowhere, any help or hint can be very useful

Answer 1

$\newcommand{\var}{\operatorname{var}} \newcommand{\cov}{\operatorname{cov}}$

To be independent they must be uncorrelated, i.e. the covariance must be $0$. I'm going to use capital $X,Y,U,V$ to refer to the random variables. We have $$ \var(X) = \sigma_X^2, \quad\var(Y) = \sigma_Y^2, \quad \cov(X,Y) = \rho\sigma_X\sigma_Y. $$ Then $$ \begin{align} & \cov(U,V) \\[8pt] ={} & \cov(\ (\cos\theta) X -(\sin\theta)Y,\ (\sin\theta)X+(\cos\theta)Y\ ) \\[8pt] = {} & (\cos\theta)\cov(\ X,\ (\sin\theta)X+(\cos\theta)Y\ ) - (\sin\theta)\cov(\ Y,\ (\sin\theta)X + (\cos\theta)Y\ ) \\[8pt] = {} & (\cos\theta\sin\theta)\var(X) +(\cos^2\theta)\cov(X,Y)-(\sin^2\theta)\cov(Y,X) - (\sin\theta\cos\theta)\var(Y) \\[8pt] = & (\cos\theta\sin\theta)(\sigma^2_X-\sigma^2_Y) + (\cos^2\theta-\sin^2\theta)\rho\sigma_X\sigma_Y. \end{align} $$ What must $\theta$ be in order that this last quantity be $0$?

Recall that $2\sin\theta\cos\theta=\sin(2\theta)$ and $\cos^2\theta-\sin^2\theta=\cos(2\theta)$. So we need $$ \frac 1 2 \sin(2\theta) (\sigma^2_X-\sigma^2_Y) + (\cos(2\theta)) \rho\sigma_X \sigma_Y = 0. $$ $$ \sin(2\theta) (\sigma^2_X-\sigma^2_Y) = -2 (\cos(2\theta)) \rho\sigma_X \sigma_Y $$ $$ \tan(2\theta) = \frac{2\sigma_X \sigma_Y}{\sigma_Y^2 - \sigma_X^2} $$

$$ \frac{2\tan\theta}{1-\tan^2\theta} =\tan(2\theta) = \frac{2\sigma_X \sigma_Y}{\sigma_Y^2 - \sigma_X^2} = \frac{2\left(\frac{\sigma_X}{\sigma_Y}\right)}{1 - \left(\frac{\sigma_X}{\sigma_Y}\right)^2} $$

So we need $\tan\theta=\dfrac{\sigma_X}{\sigma_Y}$, so $\theta=\arctan\dfrac{\sigma_X}{\sigma_Y}$.