Jointly Gaussian uncorrelated random variables are independent 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:

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

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?
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.