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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?

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    $\begingroup$ Uncorrelated variables can be dependent see stats.stackexchange.com/q/188242/35989 $\endgroup$ – Tim Feb 21 '18 at 6:39
  • $\begingroup$ Another example: if $Z \sim \text{Normal}(0, \sigma^2)$, then $Z^2$ and $Z$ are uncorrelated but dependent. $\endgroup$ – Taylor Feb 27 '18 at 14:09
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Examples where the Pearson correlation coefficient is zero but variables are dependent (from Wikipedia):

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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.

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  • $\begingroup$ You haven't addressed the case in which a density does not exist, a so-called singular Multivariate Gaussian $\endgroup$ – Mark L. Stone May 8 '18 at 17:33
  • $\begingroup$ @MarkL.Stone Do you mean when the covariance matrix is singular? In the bivariate case this would imply perfect correlation between X and Y, no? So it would be outside the scope of the question, which starts with the premise of zero correlation. $\endgroup$ – user20160 May 8 '18 at 19:14
  • $\begingroup$ "In the bivariate case this would imply perfect correlation between X and Y, no?" No. If Y has variance of zero, then it is a constant (with probability one), and therefore independent of X (and not perfectly correlated with X). I am not disagreeing with the statement to be proved, merely with the adequacy of the proof as presented. $\endgroup$ – Mark L. Stone May 8 '18 at 19:29
  • $\begingroup$ True, I wasn't considering zero variance. But, then the correlation would be undefined. I'd be happy to improve the proof if needed, but I'm not seeing a bivariate, singular case with zero correlation as the OP asked for. Am I missing something? $\endgroup$ – user20160 May 11 '18 at 12:57

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