Abstract
Let $X_{ij}$ be independent and identically distributed random variables, from which we construct matrices $A = (X_{ij})_{n \times n}$ and $U = (X_{ij})_{n \times p}$. We denote moments of their entries $X_{ij}$ as $m_r = \mathbb{E} X_{ij}^r$ and their central moments as $\mu_r = \mathbb{E}\,(X_{ij} - m_1)^r$. Is there a way how we can express the even moments of determinants $\det A$ and $(\det U^\top U)^{1/2}$ in an exact form? That is, the objective is to find $f_k(n) = \mathbb{E} \, (\det{A})^k$ and $f_k(n,p) = \mathbb{E} \, (\det U^\top U)^{k/2}$ as a function of $m_r$ (or $\mu_r$). Equivalently, one could first try to find the generating functions $ F_k(t) = \sum_{n=0}^\infty\frac{t^n}{(n!)^2} f_k(n)$ and $F_k(t, \omega) = \sum_{n=0}^\infty \sum_{p=0}^n\frac{(n-p)! }{n!p!} t^p \omega^{n-p} f_k(n,p).$
The exact expression for $F_2(t)$ and $F_2(t,\omega)$ can be easily derived using recurrences for any distribution of $X_{ij}$. For higher moments, it is not that simple. In the case of fourth moment, Nyquist, Rice and Riordan found the expression for $F_4(t)$ when $m_1 =0$. Later, Dembo \cite{Dembo} derived $F_4(t,\omega)$ when $m_1 =0$. The general case for both $F_4(t)$ and $F_4(t,\omega)$ when $m_1 \neq 0$ remained unsolved. However, as shown in recent arXiv preprint \cite{Beck}, we obtained
\begin{align*}
& {\textstyle F_4(t) = \frac{e^{t(\mu_4\!-\!3 \mu _2^2)}}{\left(1-\mu _2^2 t\right){}^3} \!\left(\!\left(1\!+\!m_1 \mu _3 t\right){}^4\!\!+\!6m_1^2\mu _2 t\frac{\left(1\!+\!m_1 \mu_3 t\right){}^2}{1-\mu _2^2 t}\!+\!m_1^4 t\frac{1\!+\!7 \mu _2^2 t\!+\!4 \mu _2^4 t^2}{\left(1-\mu _2^2 t\right){}^2}\right),} \\
& {\textstyle F_4(t,\omega) \!=\! \frac{e^{t(\mu_4 - 3 \mu_2^2)}}{\left(1\!-\! \mu_2^2 t\right) {}^2 \left(1\!-\!\omega\!-\!\mu_2^2 t\right)} \! \!\left[\left(1\! +\! m_1 \mu _3 t\right)^4 \!\! +\! \frac{6 m_1^2 \mu _2 t \left(1\!+\!m_1 \mu_3 t\right){}^2}{1-\mu_2^2 t} \! +\! \frac{m_1^4 t \left(1+7 \mu _2^2 t+4 \mu _2^4 t^2\right)}{\left(1-\mu _2^2 t\right){}^2} \right.} \\
& {\textstyle \left. +\frac{\omega m_1^2 t}{1-\omega -\mu_2^2 t} \left(\frac{2 \mu_2 \left(1+m_1 \mu _3 t\right){}^2}{1-\mu_2^2 t}+\frac{m_1^2 \left(1+5 t \mu_2^2+2 t^2 \mu_2^4\right)}{\left(1-\mu_2^2 t\right){}^2}\right)+\frac{2 t^2 \omega ^2 m_1^4 \mu_2^2}{\left(1-\omega -\mu_2^2 t\right){}^2 \left(1-\mu_2^2 t\right){}^2}\right].}
\end{align*}
One can easily deduce the moments $f_4(n)$ and $f_4(n,p)$ via Taylor expansion.