proof of independence of X-Y and X+Y when X,Y come from bivariate normal

Question

I have a bivariate normal distribution:

$$\begin{pmatrix} X\\Y\end{pmatrix} \sim \mathcal{N}(\mu, \Sigma)$$ where: $$\mu = \begin{pmatrix}\mu_X \\ \mu_Y\end{pmatrix}$$ $$\Sigma = \begin{bmatrix} \sigma_X^2 & \rho\sigma_X\sigma_Y \\ \rho\sigma_X\sigma_Y & \sigma_Y^2\end{bmatrix}$$

Now I'm supposed to show that even though $X$ and $Y$ are dependent the sum $X+Y$ and the difference $X-Y$ are independent to each other no matter what $\rho$ is. However, I can't seem to show this.

My working so far:

Let the transformation matrix $A=\begin{bmatrix} 1 & 1 \\ 1 & -1\end{bmatrix}$, then in general:

$$A\begin{pmatrix} X\\Y\end{pmatrix} = \begin{pmatrix} X+Y\\X-Y\end{pmatrix} \sim \mathcal{N}(A\mu, A\Sigma A^T) $$

But the covariance terms are not zero and thus there is no independence: $$A\Sigma A^T = \begin{bmatrix} \sigma_X^2+2\rho \sigma_X\sigma_Y+\sigma_Y^2 & \sigma_X^2 - \sigma_Y^2\\ \sigma_X^2 - \sigma_Y^2 & \sigma_X^2-2\rho \sigma_X\sigma_Y+\sigma_Y^2 \end{bmatrix}$$

Where am I making a mistake?

Answer 1

They're independent when their variances are equal actually; and your result seems OK (except the variance of $X-Y$ as @DilipSarwate states), since off-diagonal entries are zero when $\sigma_X=\sigma_Y$. A simpler way to prove this is going towards covariance, since when covariance is $0$, then the variables are independent (this is specific to joint normal distribution, which is preserved under linear transformation that you've shown).

$cov(X-Y,X+Y)=cov(X,X)+cov(X,Y)-cov(X,Y)-cov(Y,Y)=var(X)-var(Y)$

If this is $0$, then $var(X)=var(Y)$.