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Say $X_{ij}$ are i.i.d. random variables with known moments $\mathbb{E}\left[X_{ij}^n\right] = \mu_n$. Given a random matrix $A = \{X_{ij}\} _{n\times n}$, what is the expected value

$$V_m = \mathbb{E}\left[(\det A)^m\right]$$

equal to? (Where $m$ being a positive integer) By the antisymmetry of a determinant, it is trivial that $V_m$ is $0$ for all odd values of $m$. I was courious whether a general formula can be found for $m\geq 4$, since by an ingenious argument (see this or this mathoverflow article), we are able to find the exact result for $m=2$:

$$V_2 = n! (\mu_2-\mu_1^2)^{n-1}(\mu_2+\mu_1^2(n-1))$$

A special interest is for $X_{ij}$ being exponentialy distributed, i.e. with $\mu_n = n!$

Machinato
  • 2,883
  • @runway44 How specifically can be avoided the presence of integrals of type exp(X12+something)/(X12*(something)-something), which will pop up in the second order for example in the case n=5 (there are 5^2 = 25 variables to be integrated out, and when you do the integration in succesion, it strongly becames unfeasible). I am very interested in the mentioned "prety elementary" properties. – Machinato Feb 01 '21 at 08:20
  • I was talking about $\Phi_{A+B}=\Phi_A\Phi_B$ with the Leibniz formula for determinants, but I made the basic error of assuming ${X_{1\sigma(1)}\cdots X_{n\sigma(n)}\mid \sigma\in S_n}$ were pairwise independent, which is false, sorry. – anon Feb 02 '21 at 06:06

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