Let $X = \begin{bmatrix} X_1 & X_2 & \dots& \ X_n \end{bmatrix} ^{T} $ be an arbitrary column vector of random variables.

Given an $n \times n$ matrix $A$, how do I determine whether there exists an $X$ such that $K_{XX} = A$ ? I'm asking about both the case where the elements of $X$ are constrained to be independent and the case where the elments of $X$ are not constrained to be independent.

Paraphrasing a bit from Wikipedia, let $X' = X - \text{E}[X]$ and $Y' = Y - \text{E}[Y]$ .

$$ K_{XY} \;\; \stackrel{\text{def}}{=} \;\; \text{E}[X'(Y'^T)] $$

which means:

$$ K_{XX} \;\; \stackrel{\text{def}}{=} \;\; \text{E}[X'(X'^T)] $$

Which I think means that the $ij$th entry looks like this:

$$ (K_{XX})_{ij} \;\;\stackrel{\text{def}}{=}\;\; \text{E}[X'X']_{ij} $$

And this is where I get stuck. Since multiplication is commutative, in order for $A$ to be the covariance matrix of some column vector of random variables, it has to be symmetric.

However, as for actually working backwards and picking individual random variables inside $X'$ I'm stumped.

If I constrain each random variable in $X'$ to be independent, then I'm still picking a càdlàg CDF for each variable, up to the restriction that the mean is zero. That's a huge amount of freedom, so it seems like it should be possible to hit an arbitrary symmetric matrix $A$.

But that's just a vague intuition. I'm not sure how to break the problem down and prove it one way or the other.


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