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Is there an analytic formula for fractional moments of multivariate normal distribution? $E(\prod_{i=1}^k x_i^{\nu_i})={?}$ where $X=(x_1,\ldots,x_k) \sim N_k(\mu, \Sigma)$, $\nu_i\in \mathcal{R}$ and $\nu_i>0$. I know there is one for univariate normal distributions, but can't find one for the multivariate case.

The formula for the univariate case I know of is $E[|x-\mu|^{\nu}]=\sigma^{\nu}\frac{2^{\nu/2}\Gamma(\frac{\nu+1}{2})}{\sqrt{\pi}}$. I understand this is about the central absulate moment, but I am more interested in $E(\prod_{i=1}^k x_i^{\nu_i})$ or $E(\prod_{i=1}^k |x_i|^{\nu_i})$.

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    $\begingroup$ Perhaps you mean $|x_i|^{\nu_i}$ instead? It would be worth writing out the analytic formula you like for the univariate case, to show what approximations and what special functions you're comfortable with. $\endgroup$
    – user44143
    Commented Apr 15, 2020 at 18:42
  • $\begingroup$ Thanks! I have edited my question. $\endgroup$ Commented Apr 15, 2020 at 22:36
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    $\begingroup$ This is now asking for a simultaneous generalization from central to noncentral, from unsigned to signed, and from univariate to multivariate — the combination makes it hopeless to provide a nice formula. $\endgroup$
    – user44143
    Commented Apr 16, 2020 at 7:06
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    $\begingroup$ Thanks. What about the central absolute case $E(\prod_{i=1}^k |x_i-\mu_i|^{\nu_i})$ then? And what if $\nu_i$ are integers? $\endgroup$ Commented Apr 16, 2020 at 14:03

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Mathematica gives some results for the bivariate normal distribution with correlation $\rho$ and standard marginals. If $a$ and $b$ are positive integers then

\begin{align} \text{for odd }a+b,\ \ E[x^a y^b] &=0 \\ \text{for odd }a\text{ and }b,\ \ E[x^a y^b] &= \sqrt{\frac{2^{a+1}}\pi}\ \Gamma\!\left(1+\frac a2\right) f(a,b,\rho) \\ \text{for even }a\text{ and }b,\ \ E[x^a y^b] &= \sqrt{\frac{2^{a}}\pi}\ \Gamma\!\left(1+\frac a2\right) f(a,b,\rho) \end{align} where \begin{align} f(a,1,\rho)&=\rho \\ f(a,3,\rho)&=3\rho+(a-1)\rho^3 \\ f(a,5,\rho)&=15\rho+10(a-1)\rho^3+(a-1)(a-3)\rho^5 \\ f(a,2,\rho)&=1+a\rho^2 \\ f(a,4,\rho)&=3+6a\rho^2+a(a-2)\rho^4 \\ f(a,6,\rho)&=15+45a\rho^2+15a(a-2)\rho^4 + a(a-2)(a-4)\rho^6 \end{align} For odd $a$ and $b$ the coefficients appear to be Ward numbers, and for even $a$ and $b$ they appear to be the exponential Riordan array.

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