2
$\begingroup$

This question is related to: https://math.stackexchange.com/q/4270522/168758


Let $H_n(x) \in \mathbb R[x]$ be the probabilist's $n$th Hermite polynomial. This an $n$th degree polynomial given by the following equivalent formulae (which ever helps)

$$ \begin{split} H_n(x) &= n!\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k}{2^kk!(n-2k)!}x^{n-2k}\\ H_n(x) &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty (x+iy)^n e^{-y^2/2}dy\\ H_n(x) &= e^{-D^2/2}x^n, \end{split} $$ where $D^2$ is the second-derivative-w.r.t-$x$ differential operator $\dfrac{d^2}{dx^2}$, and $e^{-D^2/2}$ should be seen as a power series in $D^2$.

Let $d$ be a large positive integer, $a$ and $b$ be fixed vectors on the unit $(d-1)$-dimensional sphere $S_{d-1}$, and $X$ be uniformly distributed on $S_{d-1}$. For fixed nonnegative integers $n$ and $m$, define

$$ s_{n,m} = \mathbb E[H_n(X^\top a)H_m(X^\top b)]. $$ Due to rotational-invarfiance of $X$, it is clear that $s_{n,m}$ is a polynomial in $t:=a^\top b$. Let $c_{n,m,k}$ be the coefficient of $t^k$ in $s_{n,m}$.

Question. For $k \ge 1$, what is a good Big-O upper-bound for $c_{n,m,k}$ in the limit $d \to \infty$.

$\endgroup$

2 Answers 2

2
$\begingroup$

To find the dependence of $s_{nm}$ on $t=a\cdot b$, we take $a=(t,\sqrt{1-t^2},0,0,\ldots 0)$, $b=(1,0,0,0,\ldots 0)$, so that $$s_{nm} = \mathbb E[H_n(X^\top a)H_m(X^\top b)]=\mathbb E[H_n(X_1 t+X_2\sqrt{1-t^2})H_m(X_1)].$$ The marginal distribution $P(X_1,X_2)$ of two elements from a vector that is uniformly distributed on the $d$-dimensional unit sphere is given by (see, for example, this calculation) $$P(X_1,X_2)=\frac{d-2}{2\pi}(1-X_1^2-X_2^2)^{d/2-2},\;\;X_1^2+X_2^2<1,\;\;d\geq 3.$$ Hence we have for $s_{nm}$ the integral expression $$s_{nm}=\frac{d-2}{2\pi}\int_{0}^{1}rdr\int_0^{2\pi}d\phi\, (1-r^2)^{d/2-2}H_n\left(rt\cos\phi+r\sqrt{1-t^2}\sin\phi\right)H_m(r\cos\phi).$$

For large $d$ the Hermite polynomials can be expanded around $r=0$, which gives $$s_{nm}\approx \frac{\pi}{d} 2^{\frac{1}{2} (m+n-2)} \left(\frac{4 t}{\Gamma \left(-\frac{m}{2}\right) \Gamma \left(-\frac{n}{2}\right)}-\frac{-2 d+m+n}{\Gamma \left(\frac{1}{2}-\frac{m}{2}\right) \Gamma \left(\frac{1}{2}-\frac{n}{2}\right)}\right),\;\;d\gg 1.$$

$\endgroup$
4
  • $\begingroup$ This is quite instructive. Thanks! $\endgroup$
    – dohmatob
    Oct 10, 2021 at 18:16
  • $\begingroup$ Do we have a rough idea how small the error term ignored in the above approximation of $s_{nm}$ is ? For example, is it by any chance of order $\mathcal O(1/d^{2 + \varepsilon})$ for some $\varepsilon>0$ ? $\endgroup$
    – dohmatob
    Oct 11, 2021 at 14:31
  • $\begingroup$ no, I would think that the error term is of order $1/d^2$. $\endgroup$ Oct 11, 2021 at 14:42
  • $\begingroup$ Ah, yes that makes sense on second thought. Thanks. Any idea what the coefficient of $t^k$ (for $k \ge 2$; especially the case $k=2$) would be ? I'm suspecting it would be something like $\mathcal O(1/d^k)$. To get this kind of information, one would have to use a higher-order expansion of the Hermite polynomial expressions (in the integral) around $r=0$, right ? $\endgroup$
    – dohmatob
    Oct 11, 2021 at 14:47
0
$\begingroup$

Disclaimer. This post is just to further simplify @Carlo Beenakker's answer and highlight some potential benefits. It would be a very long comment, so I decided to post it here instead.


With an obvious abuse of notation, let us write $H_n:=H_n(0)$, the $n$th Hermite number. For even $n$, one has $$ \Gamma(1/2-n/2) = \frac{(-4)^{n/2}(n/2)!\sqrt{\pi}}{n!} = \frac{2^{n/2}2^{n/2}(-1)^{n/2}(n/2)!\sqrt{\pi}}{n!} = \frac{2^{n/2}\sqrt{\pi}}{H_n}, $$ and so we deduce that $\dfrac{\pi2^{(m+n)/2}}{\Gamma(1/2-m/2)\Gamma(1/2-n/2)} = H_nH_m$, and $$ \begin{split} \dfrac{\pi2^{(m+n)/2}}{\Gamma(-m/2)\Gamma(-n/2)} &= \dfrac{\pi2^{(m+n)/2}}{\Gamma(1/2-(m+1)/2)\Gamma(1/2-(n+1)/2)}=(1/4)H_{m+1}H_{n+1} \end{split} $$ Thus, we get the following instructive formula $$ s_{nm} \approx \begin{cases}H_nH_m+(1/d)\left(H_nH_{m+1} + H_{n+1}H_m\right),&\mbox{ if }n,m\text{ even},\\ (1/d)H_{n+1}H_{m+1}t,&\mbox{ else,} \end{cases} \tag{1} $$ where we have used the fact that $nH_n = -H_{n+1}$ for every integer $n \ge 0$.


Application

To see the importance of rewriting @Carlo's formula in the form (1), consider the following claim (which settles another question here Approximate the singular values of a certain random dot-product kernel matrix (in the sense of El Karoui, Cheng-Singer, etc.))

Claim. If $g$ is twice continuously-differentiable on $(-1,1)$, then $$ \mathbb E[g'(X^\top a)g'(X^\top b)]=g'(0)^2+\mathcal O(1/d)+\mathcal O(1/d)t. $$

It should be noted that the above estimate has been obtained here https://mathoverflow.net/a/405773/78539, under the much more restrictive condition that $g$ is $\mathcal C^2$ on $(-1,1)$ and $\mathcal C^6$ at $0$.

Proof of Claim. Under the hypothesis, $g'$ has a pointwise convergence Hermite expansion (thanks to this post https://mathoverflow.net/a/145235/78539) $$ g'(x) = \sum_{n \ge 0} b_n(g') H_n(x),\,\forall x \in (-1,1). $$ In particular, $g'(0) = \sum_{n \ge 0\text{ even }}b_n(g') H_n(0)$. Here, $b_n(g') := \mathbb E_{z \sim N(0,1)}[g(z)H_n(z)]$ is the $n$th Hermite coefficient of $g$. Recall the important formula $$ b_n(g') = b_{n+1}(g),\,\forall n \ge 0. \tag{2} $$ Now, one has $$ \begin{split} \mathbb E[g'(X^\top a)g'(X^\top b)] &= \sum_{n \ge 0}\sum_{m \ge 0} b_n(g')b_m(g')\mathbb E[H_n(X^\top a)H_m(X^\top b)]\\ &= \sum_{n}\sum_{m} b_n(g')b_m(g')s_{n,m}\\ &\overset{(1)}{=} \sum_{n,m \ge 0\text{ even }}b_n(g')H_n(0)b_m(g')H_m(0)\\ &\quad+ \mathcal O(1/d)\sum_{n,m}b_n(g')H_nb_m(g')H_{m+1} +b_n(g')H_{n+1}H_m\\ &\quad+ \mathcal O(1/d)t\sum_{n,m \ge 1\text{ odd }}b_n(g')H_{n+1}b_m(g')H_{m+1}\\ &\overset{(2)}{=} \left(\sum_{n \ge 0}b_n(g')H_n\right)^2\\ &\,+ \mathcal O(1/d)\sum_{n,m}b_n(g')H_nb_{m+1}(g)H_{m+1} +b_{n+1}(g)H_{n+1}H_m\\ &\,+ \mathcal O(1/d)t\sum_{n,m \ge 1\text{ odd }}b_{n+1}(g)H_{n+1}b_{m+1}(g)H_{m+1}\\ &= g'(0)^2 + \mathcal O(1/d)g'(0)(g(0)-b_0(g)) + \mathcal O(1/d)t(g(0)-b_0(g))^2\\ &= g'(0)^2 + \mathcal O(1/d) + \mathcal O(1/d) t, \end{split} $$ as claimed.

$\endgroup$

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.