$Cov(X, Y)=0$ if and only if $X$ and $Y$ are independent

Question

When $X$ and $Y$ are jointly normally distributed,

$Cov(X, Y)=0$ if and only if $X$ and $Y$ are independent

I don't get it. How can I show this?

Answer 1

Assume without loss of generality that $\mathbb{E}X=\mathbb{E}Y=0$.

Since $X,\,Y$ are jointly normally distributed, their joint pdf is $\frac{1}{\sqrt{\det(2\pi\Sigma)}}\exp-\frac{1}{2}r^T\Sigma^{-1}r$.

If $\Sigma_{XY}=\operatorname{Cov}(X,\,Y)=0$, $\Sigma$ is diagonal, so the pdf is $\frac{1}{\sigma_X\sqrt{2\pi}}\exp-\frac{x^2}{2\sigma_X^2}\frac{1}{\sigma_Y\sqrt{2\pi}}\exp-\frac{y^2}{2\sigma_Y^2}$.

Hence $X$ and $Y$ are independent.