Scaling in Mehta's integral

Question

The following expression is known as Mehta's integral and deeply connected to random matrix theory:

$$\frac{1}{(2\pi)^{n/2}}\int_{-\infty}^{\infty} \cdots \int_{-\infty}^{\infty} \prod_{i=1}^n e^{-t_i^2/2} \prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma} dt_1 \cdots dt_n =\prod_{j=1}^n\frac{\Gamma(1+j\gamma)}{\Gamma(1+\gamma)}.$$

An interesting question is what happens if one assumes $\gamma$ to be a function of $n.$ For example by choosing $\gamma=1/n$ one finds that as $n$ tends to infinity, the value of the integral tends to zero whereas for $\gamma=1/n^2$ the value of the integral approaches a positive constant value as $n$ tends to infinity.

These properties one can deduce from the asymptotics of the product of gamma functions. I would like to ask:

It is not too surprising that for some suitable scaling $\gamma=1/n^{\alpha}$ one approaches a constant value, as $\vert t_i-t_j \vert^{1/n} \xrightarrow 1$ for fixed $t_i,t_j$ and

$$\frac{1}{(2\pi)^{n/2}}\int_{-\infty}^{\infty} \cdots \int_{-\infty}^{\infty} \prod_{i=1}^n e^{-t_i^2/2} dt_1 \cdots dt_n =1.$$

Can one also conclude these two properties from the integral directly without evaluating it?

Answer 1

Yes, this follows by the de la Vallée-Poussin necessary and sufficient condition for the uniform integrability. Indeed, suppose that \begin{equation} \gamma n^2\to a \end{equation} (as $n\to\infty$) for some real $a\ge0$. Your integral is $$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma},$$ where the $X_i$'s are independent standard normal random variables. Introducing $N:=n(n-1)/2$, $X:=(X_1,\dots,X_n)$, and $\|X\|:=\sqrt{\sum_1^n X_i^2}$, and then using the arithmetic-geometric-mean inequality, we have $$\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma} \le\Big(\frac1N\,\sum_{1\le i<j\le n}|X_i-X_j|^2\Big)^{N\gamma} \\ =O\Big(\frac{\|X\|^2}n\Big)^{N\gamma}=O\Big(1+\frac{\|X\|^2}n\Big)^C $$ for $C:=a/2+1$ and all large enough $n$.

Note also that $\|X\|^2$ has the gamma distribution with parameters $n/2$ and $2$ and hence $E\|X\|^{2C}=O(n^C)$. So, $$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}=O(1)$$ and, similarly, $$E\Big[\Big(\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}\Big)^2\Big]=O(1).$$ Also, obviously, $t^2/t\to\infty$ as $t\to\infty$. So, we have the uniform integrability.

So, we only need to establish the convergence of \begin{equation*} {2\gamma}\sum_{1\le i<j\le n}\ln|X_i-X_j|=2\gamma NU_n \end{equation*} in probability, where \begin{equation*} U_n:=\frac1N\,\sum_{1\le i<j\le n}h(X_i,X_j) \end{equation*} is a so-called U-statistic with kernel $h(X_i,X_j):=\ln|X_i-X_j|$, and still $N=\binom n2=n(n-1)/2$. It is easy to see (cf. e.g. page 20) that $Var\,U_n=O(1/n)=o(1)$, whereas $$EU_n=m:=E\ln|X_1-X_2|.$$ So, $U_n\to m$ in probability, whence \begin{equation*} 2\gamma NU_n\to am \end{equation*} in probability and thus, by the uniform integrability,
$$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}\to e^{am}=\exp\{a\,E\ln|X_1-X_2|\}$$ as $n\to\infty$.