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We know that if $X \sim N_p(\mu,\Sigma)$ then $(X-\mu)^T \Sigma^{-1}(X-\mu) \sim \chi^2_p$, does the converse hold? Is it possible for a non-multivariate Gaussian random variable to satisfy $(X-E(X))^T (cov(X))^{-1}(X-E(X)) \sim \chi^2_p$?

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A super simple counter example:

Let $X \sim \mathcal{N}(0, 1)$, but let $Y = |X|$. Well, what's the distribution of $Y^2$?

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    $\begingroup$ But we speak about (|Y|-E(|Y|))^2 $\endgroup$ Commented Jun 1, 2023 at 6:41
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    $\begingroup$ The example may still work when we multiply the values of that half normal $Y$ with -1 if the values are above some level, such that the resulting variable has an expectation value equal to zero. $\endgroup$ Commented Jun 1, 2023 at 6:44
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    $\begingroup$ Is $Y^2$ = $(Y-E(Y))^T (var(Y))^{-1} (Y-E(Y))$? $\endgroup$ Commented Jun 1, 2023 at 7:53
  • $\begingroup$ Actually, $Y^2$ has a Lévy distribution. $\endgroup$ Commented Jun 1, 2023 at 7:54
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There are various artificial solutions to this

  • Let $Y_1$ be any zero-mean variable that is lighter-tailed than Normal and has variance at most 1. Take $Q\sim \chi^2_2$ correlated with $Y_1^2$ so that $Q-Y_1^2$ is always non-negative and then take $Y_2=\sqrt{Q}$ with a random $\pm$ sign (so that its mean is zero).
  • Let $Y_i$ be independent $\sqrt{\chi^2_{q_i}}$ with a random sign (so that the mean is zero) and with $\sum_i q_i=p$. As an extreme case, take $Y_1\sim \sqrt{\chi^2_p}$ with a random sign and the other $Y_i=0$. [if you follow wikipedia in saying $\chi^2$ distributions have to have integer df, then pretend I said $\Gamma(q_i/2)$ rather than $\chi^2_{q_i}$]
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  • $\begingroup$ What are the covariance matrices for both examples? $\endgroup$ Commented Jun 3, 2023 at 1:30

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