# Jointly Gaussian uncorrelated random variables are independent [closed]

Let $X,Y$ be jointly normally distributed and uncorrelated. Why are they independent?

In short, they are independent because the bivariate normal density, in case they are uncorrelated, i.e. $\rho =0$, reduces to a product of two normal densities the support of each one ranges from $(-\infty, \infty)$. If the joint distribution can be written as a product of nonnegative functions, we know that the RVs are independent. Moreover, we know, and can show, that each marginal density is normal on its own.

That is easy to see in the bivariate density below:

$$f(x,y)= \frac{1}{2 \pi \sigma_1 \sigma_2 \left( 1-\rho^2 \right)^{1/2}} \exp\{-q/2 \}, \quad -\infty<x<\infty,\quad -\infty<y<\infty$$

where $$q= \frac{1}{1-\rho^2} \left[ \left( \frac{x-\mu_1}{\sigma_1} \right)^2-2\rho \left(\frac{x-\mu_1}{\sigma_1} \right) \left(\frac{y-\mu_2}{\sigma_2} \right)+\left(\frac{y-\mu_2}{\sigma_2} \right)^2 \right]$$

Put $\rho=0$. There is also a nice proof involving mfgs. Is that what you were looking for?

• Thank you! I knew this formula for the joint density, but forgot that the $\rho$ is actually the correlation coefficient. It makes perfect sense now! Commented Dec 17, 2013 at 14:22
• What if the covariance matrix is singular so there is no density? How can you show independence then? Commented Oct 25, 2023 at 3:28
• @ashpool A joint normal distribution is entirely determined by the mean $\mu$ and the covariance matrix $\Sigma$; this is easy to show using the joint MGF, since it depends only on these quantities. Now imagine we were to sample two independent random variables $X', Y'$ with the same marginal distributions as $X, Y$. Then the joint distribution of $X', Y'$ would have the same mean and covariance matrix as $X, Y$, so this tells us the joint distributions of $X, Y$ and $X', Y'$ are the same. Thus $X$ and $Y$ are independent.
– lily
Commented Mar 18 at 18:32