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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?

\begin{equation} 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) \end{equation}

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!

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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 $P(X=Y)=P(X=-Y)=\frac{1}{2}$.

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  • $\begingroup$ Hi Herb. I apologize for the late response but I would like to thank you so much for this answer. It is very helpful. Take care :) $\endgroup$ Apr 15, 2021 at 17:56

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