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The Mehta integral is the following expression:

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

To simplify the notation, we introduce a measure $$d\mu_{n,\gamma}(t):=\prod_{i=1}^n e^{-t_i^2/2} \prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma} \ dt. $$ It is easy to see that $1$ is orthogonal to $(t^2-1)$ when weighted with a Gaussian measure \begin{equation} \label{eq:ortho} \frac{1}{(2\pi)^{1/2}}\int_{-\infty}^{\infty} e^{-t^2/2} (t^2-1)dt =0. \end{equation}

Now, it seems that again for different scalings of $\gamma$ we find interesting phenomena for the value of

$$\nu:=\lim_{n \rightarrow \infty}\frac{\int (t_1^2-1) \ d\mu_{n,\gamma}(t)}{\sqrt{\int (t_1^2-1)^2 \ d\mu_{n,\gamma}(t)}}$$

where I admit that I use the limit without actually knowing whether it exists.

Case 1:

As one can guess from this thread, if we choose $\gamma=1/n^2$ it seems that the product $F(t)$ does not contribute to the value of the above limit and we have $\nu=0.$

Case 2:

If we choose $\gamma=1/n$ then Carlo Beenakker's answer that treats the case 3 rigorously suggests that we find $\nu=0$ in this case, too.

Case 3:

If we choose $\gamma=1$ then it seems like we get that $\nu$ is of order one, which is confirmed by Carlo Beenakker's answer.

My question is: Consider Case 2 with scaling $\gamma=1/n$, then find the value of $\nu$ for large $n$?

The Mehta integral is the following expression:

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

To simplify the notation, we introduce a measure $$d\mu_{n,\gamma}(t):=\prod_{i=1}^n e^{-t_i^2/2} \prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma} \ dt. $$ It is easy to see that $1$ is orthogonal to $(t^2-1)$ when weighted with a Gaussian measure \begin{equation} \label{eq:ortho} \frac{1}{(2\pi)^{1/2}}\int_{-\infty}^{\infty} e^{-t^2/2} (t^2-1)dt =0. \end{equation}

Now, it seems that again for different scalings of $\gamma$ we find interesting phenomena for the value of

$$\nu:=\lim_{n \rightarrow \infty}\frac{\int (t_1^2-1) \ d\mu_{n,\gamma}(t)}{\sqrt{\int (t_1^2-1)^2 \ d\mu_{n,\gamma}(t)}\sqrt{\int 1 \ d\mu_{n,\gamma}(t)}}$$

where I admit that I use the limit without actually knowing whether it exists.

Case 1:

As one can guess from this thread, if we choose $\gamma=1/n^2$ it seems that the product $F(t)$ does not contribute to the value of the above limit and we have $\nu=0.$

Case 2:

If we choose $\gamma=1/n$ then Carlo Beenakker's answer that treats the case 3 rigorously suggests that we find $\nu=0$ in this case, too.

Case 3:

If we choose $\gamma=1$ then it seems like we get that $\nu$ is of order one, which is confirmed by Carlo Beenakker's answer.

My question is: Consider Case 2 with scaling $\gamma=1/n$, then find the value of $\nu$ for large $n$?

This is a follow-up question on an earlier question that I asked on the Mehta integral, see here [on mathoverflow.][1]

The Mehta integral is the following expression:

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

In the previous thread, linked above, Iosif Pinelis showed that if we choose the scaling $\gamma=1/n^2$ in the Mehta integral, then the value of the integral is essentially $n$-independent.

This is essentially because the product $F(t):=\prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma}$ contains of order $n^2$ many terms.

Now, I consider a related object:

To simplify the notation, we introduce a measure $$d\mu_{n,\gamma}(t):=\prod_{i=1}^n e^{-t_i^2/2} \prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma} \ dt. $$ It is easy to see that $1$ is orthogonal to $(t^2-1)$ when weighted with a Gaussian measure \begin{equation} \label{eq:ortho} \frac{1}{(2\pi)^{1/2}}\int_{-\infty}^{\infty} e^{-t^2/2} (t^2-1)dt =0. \end{equation}

Now, it seems that again for different scalings of $\gamma$ we find interesting phenomena for the value of

$$\nu:=\lim_{n \rightarrow \infty}\frac{\int (t_1^2-1) \ d\mu_{n,\gamma}(t)}{\sqrt{\int (t_1^2-1)^2 \ d\mu_{n,\gamma}(t)\int 1 \ d\mu_{n,\gamma}(t)}}$$

where I admit that I use the limit without actually knowing whether it exists.

Case 1:

As one can guess from the other thread if we choose $\gamma=1/n^2$ it seems that the product $F(t)$ does not contribute to the value of the above limit and we have $\nu=0.$

Case 2:

If we choose $\gamma=1/n$ then Carlo Beenakker's answer that treats the case 3 rigorously suggests that we find $\nu=0$ in this case, too.

Case 3:

If we choose $\gamma=1$ then it seems like we get that $\nu$ is of order one, which is confirmed by Carlo Beenakker's answer.

My question is: Consider Case 2 with scaling $\gamma=1/n$, then find the value of $\nu$ for large $n$?

The Mehta integral is the following expression:

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

To simplify the notation, we introduce a measure $$d\mu_{n,\gamma}(t):=\prod_{i=1}^n e^{-t_i^2/2} \prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma} \ dt. $$ It is easy to see that $1$ is orthogonal to $(t^2-1)$ when weighted with a Gaussian measure \begin{equation} \label{eq:ortho} \frac{1}{(2\pi)^{1/2}}\int_{-\infty}^{\infty} e^{-t^2/2} (t^2-1)dt =0. \end{equation}

Now, it seems that again for different scalings of $\gamma$ we find interesting phenomena for the value of

$$\nu:=\lim_{n \rightarrow \infty}\frac{\int (t_1^2-1) \ d\mu_{n,\gamma}(t)}{\sqrt{\int (t_1^2-1)^2 \ d\mu_{n,\gamma}(t)}}$$

where I admit that I use the limit without actually knowing whether it exists.

Case 1:

As one can guess from this thread, if we choose $\gamma=1/n^2$ it seems that the product $F(t)$ does not contribute to the value of the above limit and we have $\nu=0.$

Case 2:

If we choose $\gamma=1/n$ then Carlo Beenakker's answer that treats the case 3 rigorously suggests that we find $\nu=0$ in this case, too.

Case 3:

If we choose $\gamma=1$ then it seems like we get that $\nu$ is of order one, which is confirmed by Carlo Beenakker's answer.

My question is: Consider Case 2 with scaling $\gamma=1/n$, then find the value of $\nu$ for large $n$?

This is a follow-up question on an earlier question that I asked on the Mehta integral, see here on mathoverflow.

The Mehta integral is the following expression:

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

In the previous thread, linked above, Iosif Pinelis showed that if we choose the scaling $\gamma=1/n^2$ in the Mehta integral, then the value of the integral is essentially $n$-independent.

This is essentially because the product $F(t):=\prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma}$ contains of order $n^2$ many terms.

Now, I consider a related object:

To simplify the notation, we introduce a measure $$d\mu_{n,\gamma}(t):=\prod_{i=1}^n e^{-t_i^2/2} \prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma} \ dt. $$ It is easy to see that $1$ is orthogonal to $(t^2-1)$ when weighted with a Gaussian measure \begin{equation} \label{eq:ortho} \frac{1}{(2\pi)^{1/2}}\int_{-\infty}^{\infty} e^{-t^2/2} (t^2-1)dt =0. \end{equation}

Now, it seems that again for different scalings of $\gamma$ we find interesting phenomena for the value of

$$\nu:=\lim_{n \rightarrow \infty}\frac{\int (t_1^2-1) \ d\mu_{n,\gamma}(t)}{\sqrt{\int (t_1^2-1)^2 \ d\mu_{n,\gamma}(t)\int 1 \ d\mu_{n,\gamma}(t)}}$$

where I admit that I use the limit without actually knowing whether it exists.

Case 1:

As one can guess from the other thread if we choose $\gamma=1/n^2$ it seems that the product $F(t)$ does not contribute to the value of the above limit and we have $\nu=0.$

Case 2:

If we choose $\gamma=1/n$ then it seems like the value of $\nu$ converges neither to zero. However, Carlo Beenakker's answer that treats the case 3 rigorously suggests that we find $\nu=0$ in this case, too.

Case 3:

If we choose $\gamma=1$ then it seems like we get that $\nu$ is of order one, which is confirmed by Carlo Beenakker's answer.

Now, I must admit I was unable to evaluate these high-dimensional integrals numerically and so I did something much less reliable (see also Carlo Beenakker's response for some constructivite criticism regarding this approach) to obtain a conjecture about the behaviour of the integral for $n$ large:

I chose a set $t_2,...,t_n$ of random numbers sampled from a Gaussian distribution and then computed

$$\frac{\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} (t^2-1) e^{-t^2/2} \prod_{j=2}^n \vert t-t_j \vert^{2\gamma} \ dt}{\sqrt{\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} (t^2-1)^2 e^{-t^2/2} \prod_{j=2}^n \vert t-t_j \vert^{2\gamma} \ dt \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-t^2/2} \prod_{j=2}^n \vert t-t_j \vert^{2\gamma} \ dt}}.$$

My question is: Can we say analytically what happens in Case 2 with scaling $\gamma=1/n$ to the value of $\nu$ for large $n$?

Please let me know if you have any questions.

As functions of $n$, I then obtained the following plots for cases 1-3 respectively:

This is a follow-up question on an earlier question that I asked on the Mehta integral, see here [on mathoverflow.][1]

The Mehta integral is the following expression:

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

In the previous thread, linked above, Iosif Pinelis showed that if we choose the scaling $\gamma=1/n^2$ in the Mehta integral, then the value of the integral is essentially $n$-independent.

This is essentially because the product $F(t):=\prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma}$ contains of order $n^2$ many terms.

Now, I consider a related object:

To simplify the notation, we introduce a measure $$d\mu_{n,\gamma}(t):=\prod_{i=1}^n e^{-t_i^2/2} \prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma} \ dt. $$ It is easy to see that $1$ is orthogonal to $(t^2-1)$ when weighted with a Gaussian measure \begin{equation} \label{eq:ortho} \frac{1}{(2\pi)^{1/2}}\int_{-\infty}^{\infty} e^{-t^2/2} (t^2-1)dt =0. \end{equation}

Now, it seems that again for different scalings of $\gamma$ we find interesting phenomena for the value of

$$\nu:=\lim_{n \rightarrow \infty}\frac{\int (t_1^2-1) \ d\mu_{n,\gamma}(t)}{\sqrt{\int (t_1^2-1)^2 \ d\mu_{n,\gamma}(t)\int 1 \ d\mu_{n,\gamma}(t)}}$$

where I admit that I use the limit without actually knowing whether it exists.

Case 1:

As one can guess from the other thread if we choose $\gamma=1/n^2$ it seems that the product $F(t)$ does not contribute to the value of the above limit and we have $\nu=0.$

Case 2:

If we choose $\gamma=1/n$ then Carlo Beenakker's answer that treats the case 3 rigorously suggests that we find $\nu=0$ in this case, too.

Case 3:

If we choose $\gamma=1$ then it seems like we get that $\nu$ is of order one, which is confirmed by Carlo Beenakker's answer.

My question is: Consider Case 2 with scaling $\gamma=1/n$, then find the value of $\nu$ for large $n$?