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

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closed as off-topic by user1337, Shuchang, Eric Naslund, Did, user66733 Dec 17 '13 at 1:23

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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?

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  • $\begingroup$ 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! $\endgroup$ – Balerion_the_black Dec 17 '13 at 14:22
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Not true http://en.wikipedia.org/wiki/Normally_distributed_and_uncorrelated_does_not_imply_independent http://probability.ca/jeff/teaching/uncornor.html

Consider a simple example. Let X have a standard normal distribution. Let Z be independent of X, with Z equally likely to be +1 or -1 (i.e., Pr[Z=+1] = Pr[Z=-1] = 1/2). Let Y = X Z (the product of X and Z). Then it is easy to see that Y also has a standard normal distribution, and that Cov(X,Y) = 0. On the other hand, clearly X and Y are not independent: indeed, it always holds that |X| = |Y|. The point is that, just because each of X and Y has a normal distribution, that does not mean that the pair (X,Y) has a bivariate normal distribution, nor even that (X,Y) is jointly absolutely continuous, nor does it mean that zero covariance implies independence.

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    $\begingroup$ I said "jointly normally distributed"! Under this assumption the claim is true. $\endgroup$ – Balerion_the_black Dec 16 '13 at 23:50
  • $\begingroup$ What you are saying is nice to know, but Balerion is right. Action is needed here. If you do not want to delete your answer (as I said, it is nice to know) then at least edit and make clear in it that it is not really an answer to the question of the OP. $\endgroup$ – drhab Dec 17 '13 at 10:15

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