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?

  • 2
    $\begingroup$ 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. $\endgroup$ – Dilip Sarwate Mar 20 '19 at 14:51
  • $\begingroup$ @DilipSarwate Thanks for pointing out the, I believe, sign error and thanks for the comment $\endgroup$ – Dahlai Mar 20 '19 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).


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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.