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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?

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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*}

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  • $\begingroup$ Perfect! Thank you so much passerby, I'm guessing this would also work for the tensors representing the higher moments?.... $\endgroup$
    – ABIM
    Commented Feb 3, 2016 at 5:34
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    $\begingroup$ No problem... Re tensors, perhaps if you are talking about tensors of central moments. You should be able to use the symmetrization trick. By the way, just double check to see if not wrong! $\endgroup$
    – passerby51
    Commented Feb 3, 2016 at 5:39
  • $\begingroup$ Yes I'll do both those things now, thanks again :) $\endgroup$
    – ABIM
    Commented Feb 3, 2016 at 5:41
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    $\begingroup$ No problem at all :) By the way, it should be argmin, not argmax. $\endgroup$
    – passerby51
    Commented Feb 3, 2016 at 5:42
  • $\begingroup$ Right, my bad; late night typo on my part it seems... $\endgroup$
    – ABIM
    Commented Feb 3, 2016 at 5:50

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