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

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?

• Your calculation of the variance of $X-Y$ is incorrect but you are not making any errors when you conclude that $X+Y$ and $X-Y$ have covariance $\sigma_X^2-\sigma_Y^2$ and thus are not independent regardless of the value of $\rho$ unless you also furher assume that $X$ and $Y$ have equal variance. Get back to your instructor and inform him/her that the problem asks for a proof of an incorrect statement. – Dilip Sarwate Mar 20 at 14:51
• @DilipSarwate Thanks for pointing out the, I believe, sign error and thanks for the comment – Dahlai Mar 20 at 14:59

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 $$var(X)=var(Y)$$.