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Let $X~N_d(\mu,\Sigma)$ be a multivariate Gaussian random vector. Is there a convenient formula for each of $$ \mu_p\triangleq \mathbb{E}\left[\sum_{i=1}^d |X_i|^p\right], $$ in terms of $\mu$ and of $\Sigma$?

Reasoning This is true for the univariate case; see this table. For example, when $d=1$ and $p=3$, then $\mu_p= \mu^3 + 3\mu\sigma^2$.

This is to be distinguished from this question, which consideres the second higher-order moment but defining it via the outer and not inner product.

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    $\begingroup$ "or example, when d=1 and p=, then" .... did you intend to up a 3 in there after "p="? $\endgroup$
    – Glen_b
    Mar 13, 2019 at 3:59

1 Answer 1

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Since $$\mu_p\triangleq \mathbb{E}_{\mu,\sigma}\left[\sum_{i=1}^d |X_i|^p\right]=\sum_{i=1}^d\mathbb{E}_{\mu,\sigma}\left[|X_i|^p\right]$$ the multivariate nature of $X=(X_1,\ldots,X_n)$ and in particular the correlations between the $X_i$'s have no relevance. For a Normal distribution $\mathcal{N}(\mu,\sigma^2)$, the moments are connected by the recurrence relation (Patel & Read, 1986)$$\xi_p\triangleq\mathbb{E}_{\mu,\sigma}[X^p]=\mu\xi_{p-1}+(p-1)\sigma^2\xi_{p-2}$$which provides $\mu_{2q}=\xi_{2q}$. As stated by Wikipedia]2, the generic absolute moment of order $p>0$ is $$\mathbb{E}_{\mu,\sigma}\left[|X|^p \right] =\sigma^p 2^{p/2} \frac {\Gamma\left(\frac{1+p} 2\right)}{\sqrt\pi} {}_1F_1\left( -\frac{p}{2}, \frac{1}{2}, -\frac{1}{2} \left( \frac \mu \sigma \right)^2 \right)$$

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  • $\begingroup$ Amazing, this is very helpful... but do you have a reference also? I would need that. Merci Xi'an $\endgroup$
    – AIM
    Mar 13, 2019 at 9:20
  • $\begingroup$ Nono, for the recurrence relation. $\endgroup$
    – AIM
    Mar 13, 2019 at 9:49

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