Convergence of fraction of expectation values

Question

Let $X_1,...,X_n$ be iid normal random variables.

I am looking for a strategy to establish the following limit for fraction of expectation values

$$\lim_{N \rightarrow \infty} \frac{E(\prod_{1\le i < j\le n} \vert X_i-X_j \vert^{1/n})}{E(\prod_{1\le i < j\le n-1} \vert X_i-X_j \vert^{1/n})}=1.$$

Does anybody have any ideas what to use for this limit?

Answer 1

The Mehta integral is $$M_n(\gamma):=E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma} =\prod_{j=1}^n\frac{\Gamma(1+j\gamma)}{\Gamma(1+\gamma)}.$$ So, your fraction under the limit sign is $$\frac{M_n(1/(2n))}{M_{n-1}(1/(2n)}=\frac{\Gamma(3/2)}{\Gamma(1+1/(2n))}\to\Gamma(3/2)\approx0.886227.$$

Answer 2

The numerator and denominator can also be written as $$E\left[\exp\left(\frac1n\sum\ln|X_i-X_j|\right)\right]$$ where the numerator has $n(n-1)/2$ summands and the denominator has $(n-1)(n-2)/2$ summands.

Let $\mu$ and $\sigma$ be the mean and standard deviation of $\ln|X_i-X_j|$. Since $X_i-X_j$ is a normal distribution with mean $0$ and standard deviation $\sqrt{2}$, $$\mu=E\left[\ln\Big|N(0,\sqrt{2})\Big|\right]=\frac{-\gamma}{2}$$ where $\gamma$ is the Euler-Mascheroni constant.

If the summands were independent, then the fraction in the question would be approximated by a ratio of means of lognormal distributions:

$$ \frac {E\left[LN\left(\frac{n(n-1)}{2n}\mu,\sqrt{\frac{n(n-1)}{2n^2}}\sigma\right)\right]} {E\left[LN\left(\frac{(n-1)(n-2)}{2n}\mu,\sqrt{\frac{(n-1)(n-2)}{2n^2}}\sigma\right)\right]} = \frac {\exp\left(\frac{n(n-1)}{2n}\mu+\frac{n(n-1)}{2n^2}\frac{\sigma^2}{2}\right)} {\exp\left(\frac{(n-1)(n-2)}{2n}\mu+\frac{(n-1)(n-2)}{2n^2}\frac{\sigma^2}{2}\right)} $$

In the limit this is $$ \exp\left(\frac{2(n-1)}{2n}\mu+\frac{2(n-1)}{2n^2}\frac{\sigma^2}{2}\right) \rightarrow \exp(\mu)=\exp(-\gamma/2) \sim 0.749.$$

Empirically the number seems to be between that and $1$.

With some numerical integration we could compute $\sigma$ and the correlation $\rho$ between $\ln|X_i-X_j|$ and $\ln|X_i-X_k|$; with some combinatorics we could compute how often those $\rho$'s show up in the standard deviations of the two lognormals; combining those results we could get a more accurate value for the overall expectation.