Does $(X-E(X))^T (cov(X))^{-1}(X-E(X)) \sim \chi^2_p$ imply normality?

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 $$(X-E(X))^T (cov(X))^{-1}(X-E(X)) \sim \chi^2_p$$?

• It is, one example is multivariate skew-$t$ distribution. Details can be found in The Skew-Normal and Related Families Commented Jun 27, 2023 at 20:10

A super simple counter example:

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

• But we speak about (|Y|-E(|Y|))^2 Commented Jun 1, 2023 at 6:41
• 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. Commented Jun 1, 2023 at 6:44
• Is $Y^2$ = $(Y-E(Y))^T (var(Y))^{-1} (Y-E(Y))$? Commented Jun 1, 2023 at 7:53
• Actually, $Y^2$ has a Lévy distribution. Commented Jun 1, 2023 at 7:54

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 $$\chi^2_{q_i}$$]
• What are the covariance matrices for both examples? Commented Jun 3, 2023 at 1:30