Question about correlations and independence My professor was discussing correlation and when it implies independence. It was fairly clear to me that, if $X$ and $Y$ are independent, then their correlation is zero. The reverse direction is less clear. Correlation being zero doesn't imply independence. This makes some sense to me because correlation measures linear association in some sense, but I am having trouble coming up with a concrete example of this. He also stated that, if $X$ and $Y$ are bivariate normal and correlation is zero, then $X$ and $Y$ are independent. Can someone provide a proof/intuition about this?
 A: Examples where the Pearson correlation coefficient is zero but variables are dependent (from Wikipedia):


if X and Y are bivariate normal and correlation is zero then X and Y are independent. Can someone provide a proof/intuition about this?

Consider Gaussian random variables $X, Y$ with means $\mu_X, \mu_Y$, and variances $\sigma_X^2, \sigma_Y^2$. Suppose $X$ and $Y$ are jointly Gaussian, and their correlation coefficient is $\rho$.
The joint distribution is:
$$p(x,y) = $$
$$\frac{1}{2 \pi \sigma_X \sigma_Y \sqrt{1-\rho^2}}
\exp \left [
  -\frac{1}{2 (1-\rho^2)} \left (
    \frac{(x-\mu_X)^2}{\sigma_X^2}
    + \frac{(y-\mu_Y)^2}{\sigma_Y^2}
    - 2 \rho \frac{(x-\mu_X)(y-\mu_Y)}{\sigma_X \sigma_Y}
  \right )
\right ]$$
In the case of zero correlation ($\rho=0$) this reduces to:
$$p(x,y) = \frac{1}{2 \pi \sigma_X \sigma_Y }
\exp \left [
  -\frac{1}{2} \left (
    \frac{(x-\mu_X)^2}{\sigma_X^2}
    + \frac{(y-\mu_Y)^2}{\sigma_Y^2}
  \right )
\right ]$$
This expression can be factored as follows:
$$p(x,y) = \frac{1}{\sigma_X \sqrt{2 \pi}}
\exp \left [
 -\frac{(x-\mu_X)^2}{2 \sigma_X^2}
\right ]
\frac{1}{\sigma_Y \sqrt{2 \pi}}
\exp \left [
 -\frac{(y-\mu_Y)^2}{2 \sigma_Y^2}
\right ]$$
We can see that this is simply the product of the Gaussian marginal distributions of $X$ and $Y$. The joint distribution being equal to the product of the marginal distributions implies independence.
