# Is multinormality required for zero covariance to imply independence?

I have two questions question which are bugging me regarding the multivariate normal/ multinormal framework. I can not find a satisfactory answer anywhere I have looked so far, perhaps it is because I don't know what to look for.

Firstly: Can we simply assemble a multinormal distribution from two univariate normals and a covariance?

i.e can we take two univariate normals, and then express a multinormal using a covariance between them?

$$$$X \sim N(0,1) Y \sim N(0,1) \operatorname{Cov}(X, Y)=0.5 \Leftrightarrow\left(\begin{array}{l} X \\ Y \end{array}\right) \sim N_{2}\left(\left(\begin{array}{l} 0 \\ 0 \end{array}\right),\left(\begin{array}{cc} 1 & 0.5 \\ 0.5 & 1 \end{array}\right)\right)$$$$

Also, regarding independence, I have been showed many times the perennial example of $$Y=X^2$$ where $$X \sim N(0,1)$$ as an example of two random variables which have zero covariance, yet a clear form of dependence between them. However, it occured to me, that Y is of course chi-squared, rather than normal, so of course we are not in a "multinormal setting".

Therefore, my second question is as follows: If we are in a truly multi-normal framework, is it fair to say that if we have zero covariance, than the two variables are truly independent? Can there exist a form of dependence expressed in a multinormal setting which is not captured by covariance?

Thanks for the insight!

The answer to your question is: $$0$$ covariance does not imply independence unless the variables are jointly normal. Simple example: let $$X$$ be standard normal ($$N(0,1)$$) and let $$Y=cX$$, where $$c$$ (independent of $$X$$) is standard Bernoulli $$(P(c=1)=P(c=-1)=\frac{1}{2})$$. Then $$Y$$ is also standard normal and $$E(XY)=0$$. However they are not independent since $$P(X=Y)=P(X=-Y)=\frac{1}{2}$$.