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?


1 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 $n\to\infty$.

  • 1
    $\begingroup$ interesting, but how did you estimate this product? Also, the dimension of the integral changes, which is perhaps a bit different from the usual DCT. $\endgroup$ Jan 23, 2020 at 1:47
  • $\begingroup$ @SolidStatePhysicist : I have added the requested details. $\endgroup$ Jan 23, 2020 at 1:56
  • $\begingroup$ @SolidStatePhysicist : You were right. I had indeed overlooked that the dimension of the integral, $n$, goes to infinity. This is now fixed. Instead of the integration over $\mathbb R^n$, with the variable $n$, we now use the expectation, which is the integration over a fixed background probability space. Instead of dominated convergence, we now use uniform integrability, which complicates the reasoning just a bit. $\endgroup$ Jan 23, 2020 at 4:00
  • 1
    $\begingroup$ @IosifPinelis I think there is an issue with this answer. The OP claims and asking for a proof that this property holds for $\gamma=1/n^2$. In your answer however, you assume that $\gamma=o(1/n^2).$ This seems to be irrelevant for the uniform integrability, although the property is mentioned there, but it is clearly used to show convergence in probability. I'd be curious to hear what you think? $\endgroup$
    – Xin Wang
    Feb 24, 2020 at 1:39
  • 1
    $\begingroup$ @SolidStatePhysicist : I have added the modification you requested. $\endgroup$ Feb 24, 2020 at 4:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.