TITLE: Covariance zero for two gaussian variables
QUESTION [5 upvotes]: Say we have two random variables $X$ and $Y$ and both of them have a gaussian distribution. Further, we know that $cov(X,Y) = 0$, where $cov(X,Y)$ is the covariance of two variables (i.e $cov(X,Y) = E[(X-E[X])(Y-E[Y])]$, where $E[X]$ is the mean (expectation) of variable $X$.
Can we say that $X$ and $Y$ are independent variables? I know that, in general, $cov(X,Y)= 0 $ does not imply that $X$ and $Y$ are independent, but what about the case when $X$ and $Y$ have a gaussian(normal) distribution? Can we take this as a theorem?

REPLY [2 votes]: For jointly (per @Did) normal random variables, uncorrelated implies independent.
In particular, it is easy to see that the joint density function factors,
giving the product of the two marginal density functions.
Also, for normal data, the sample mean $\bar X$ and sample SD $S$ are
independent. (Proof via linear algebra or moment generating functions.) But $\bar X$ and $S$ are not independent except for normal
data. 
In the left panel below $S$ is plotted against $\bar X$ for 30,000
randomly generated standard normal datasets of size $n = 5.$ As a 'naturally occurring' instance where zero correlation and dependence coexist: in the right
panel the same is done for 30,000 samples of size $n = 5$ from $Beta(.5, .5).$
For these beta data $\bar X$ and $S$ are uncorrelated, but not independent.
m = 30000;  n = 5
x = rnorm(m*n); NRM = matrix(x, nrow=m)
ax = rowMeans(NRM); sx = apply(NRM, 1, sd)
cor(ax, sx)
## -0.001177232  # consistent with uncorrelated
y = rbeta(m*n, .5, .5);  BTA = matrix(y, nrow=m)
ay = rowMeans(BTA);  sy = apply(BTA, 1, sd)
cor(ay, sy)
## -0.001677063  # consistent with uncorrelated