# Covariance zero for two gaussian variables

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?

• If $(X,Y)$ is jointly gaussian, yes. If $X$ is gaussian and $Y$ is gaussian, then no, not necessarily. Example: $X$ centered gaussian, $Y=BX$ with $B=\pm1$ symmetric Bernoulli independent of $X$, then $Y$ is also centered gaussian but $(X,Y)$ is not jointly gaussian. – Did Oct 13 '16 at 20:38
• @Did what means jointly gaussian? – penguina Oct 13 '16 at 20:42
• That (X,Y) is gaussian as a family. What is your source on normal random variables? – Did Oct 13 '16 at 21:05
• If you mean what I'm understanding by normal random variables, it's from wiki (en.wikipedia.org/wiki/Normal_distribution) – penguina Oct 13 '16 at 21:08
• – Henry Oct 17 '16 at 17:40

## 1 Answer

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 • "For normal random variables, uncorrelated implies independent." Not true, see my comment on main. – Did Oct 13 '16 at 21:03
• Clarification acknowledged. I assumed jointly normal. But the question does say 'gaussian distribution' not 'jointly gaussian distribution.' – BruceET Oct 13 '16 at 21:07