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Oct 20, 2021 at 3:42 comment added jlewk The last part of the abstract of arxiv.org/abs/1912.10754 suggests the paper will have some answers to these issues, and possibly answers to your question.
Oct 20, 2021 at 3:38 comment added jlewk Absolute continuity is not suficient. If $d=1$ and $n\in\{1,2\}$ then $E[(\chi^2_n)^{-1}]=+\infty$.
Oct 20, 2021 at 2:23 history edited Drew Brady CC BY-SA 4.0
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Oct 20, 2021 at 2:17 history edited Drew Brady CC BY-SA 4.0
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Oct 20, 2021 at 2:15 comment added Drew Brady But even somewhat mild conditions would prevent that from being true (e.g. $x_j$ has law absolutely continuous w.r.t. Lebesgue measure). This question makes sense, provided we assume that $\mathbb{E}_x[\mathrm{Tr}(\Sigma^{-1})]$ exists and is finite?
Oct 20, 2021 at 1:27 comment added jlewk What is the distribution of the $x_j$? If $P(x_j=0)>0$ then $E_x[trace\Sigma^{-1}]$ is $+\infty$.
Oct 16, 2021 at 7:59 history asked Drew Brady CC BY-SA 4.0