I ran the program to calculate \begin{equation*} \mathbb{E}[ \mathrm{tr} (H^{2 p}) ] \end{equation*} for \( p = 1, \ldots, 6 \) (my program is slow for higher values of \(p \) ). The results are \begin{align*} \mathbb{E}[ \mathrm{tr} (H^{2}) ] & = \frac{1}{2} n (n+1)\\ \mathbb{E}[ \mathrm{tr} (H^{4}) ] & =\frac{1}{4} n \left(2 n^2+5 n+5\right)\\ \mathbb{E}[ \mathrm{tr} (H^{6}) ] & =\frac{1}{8} n \left(5 n^3+22 n^2+52 n+41\right)\\ \mathbb{E}[ \mathrm{tr} (H^{8}) ] & =\frac{1}{16} n \left(14 n^4+93 n^3+374 n^2+690 n+509\right)\\ \mathbb{E}[ \mathrm{tr} (H^{10}) ] &= \frac{1}{32} n \left(42 n^5+386 n^4+2290 n^3+7150 n^2+12143 n+8229\right)\\ \mathbb{E}[ \mathrm{tr} (H^{12}) ] & = \frac{1}{64} n \left(132 n^6+1586 n^5+12798 n^4+58760 n^3+167148 n^2+258479 n+166377\right) \end{align*} My results agree with page 3 in the paper ''A recursion formula for the moments of the Gaussian Orthogonal Ensemble'' by Michel Ledoux from 2008. Notice that Ledoux has a different normalization for the GOE, so I needed to insert factors of 2 here and there to compare. Ledoux also proves a recursion relation for these moments, namely if one writes \begin{equation*} b_p = 2^p\ \mathbb{E}[ \mathrm{tr} (H^{2 p}) ] \end{equation*} then \( \forall p \ge 4 \) \begin{align}\label{eq:20151031a} (p+1) b_p &= (4 p -1) (2 n -1) b_{p-1} + (2 p -3) (10 p^2 - 9 p - 8 n^2 + 8 n) b_{p-2}\\ & - 5 (2 p -3)^{\underline{3}}\ (2 n -1) b_{p-3}- 2 (2 p -3)^{\underline{5}}\ b_{p-4}\nonumber \end{align} where I used the notation \begin{equation*} x^{\underline{n}} = x (x-1) (x-2) \cdots (x - n + 1) \end{equation*} It can be seen that the values obtained above indeed satisfy this recursion relation.

__The proof of \eqref{eq:20151031a} in Ledoux's paper is complicated. In a next blog post, I plan to see if a similar recursion relation is easier to prove for other ensembles.__

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