I have seen the expectation of a random vector expressed as the solution to the optimization problem:
\begin{equation}
\mathbb{E}[X]=argmin\_{v \in \mathbb{R}^n}\mathbb{E}[\|X-v\|\_{l^2}^2](:= \int\_{\Omega} \|X(\omega)-v\|^2 dP(\omega)).
\end{equation}




---


My question is... can we express the covariance matrix of a random vector as a similar optimization problem?



Covariance matrix as optimization problem solution?
I believe you can do something like this:
Let $X$ be zero mean. Then the covariance matrix is $ \text{cov}(X) = \mathbb{E}[XX^T]$. Hence, by the same argument:
\begin{align\*}
 \text{cov}(X) = \text{arg} \min\_{\Sigma} \; \mathbb{E} \| XX^T - \Sigma\|\_F^2
\end{align\*}
If $X$ not zero mean, let $X'$ be an independent copy. Then, $X - X'$ is zero mean an $\text{cov}(X-X') = 2 \text{cov}(X)$, hence
\begin{align\*}
 \text{cov}(X) = \frac12\text{arg} \min\_{\Sigma} \; \mathbb{E} \| (X-X')(X-X')^T - \Sigma\|\_F^2.
\end{align\*}


