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Any vector is assumed to be a column vector by default. Suppose $f(\mathbf{x})$ is the $d$-dimensional standard Gaussian density. I am interested in the following optimization problem: $$ \max_g ~\det \left(\int_{\mathbb{R}^d} \mathbf{x}\mathbf{x}^T f(\mathbf{x}) g(\mathbf{x}) d \mathbf{x}\right), $$ where $\det$ stands for the determinant, and the maximization is taken over all measurable $g$ such that $0\le g\le 1$ and $$ \int_{\mathbb{R}^d} f(\mathbf{x})g(\mathbf{x})d\mathbf{x}=1/2. $$

Remark 1: My conjecture is that the maximizer $g$ is the indicator function of the complement of a ball in $\mathbb{R}^d$ of a suitable radius, but I could not find a proof.

Remark 2: It would be even useful to only prove that this optimization problem has a solution, namely, there exists such a $g$ which attains the maximum.

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    $\begingroup$ The determinant is bounded by the trace divided by $d$ to the $d$-th power and the equality is achieved on the multiples of identity. The maximal trace is trivially attained for $g$ you suggested and the resulting matrix is a multiple of identity. Am I missing some subtlety? $\endgroup$
    – fedja
    Dec 7, 2018 at 21:28
  • $\begingroup$ Never thought about the trace argument. Thanks. If you are willing to put your comment as answer I would check it. $\endgroup$
    – Uchiha
    Dec 7, 2018 at 22:03

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