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 $d \to \infty$.


2 Answers 2


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 $$s_{nm}\approx \frac{\pi}{d} 2^{\frac{1}{2} (m+n-2)} \left(\frac{4 t}{\Gamma \left(-\frac{m}{2}\right) \Gamma \left(-\frac{n}{2}\right)}-\frac{-2 d+m+n}{\Gamma \left(\frac{1}{2}-\frac{m}{2}\right) \Gamma \left(\frac{1}{2}-\frac{n}{2}\right)}\right),\;\;d\gg 1.$$

  • $\begingroup$ This is quite instructive. Thanks! $\endgroup$
    – dohmatob
    Commented 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
    Commented Oct 11, 2021 at 14:31
  • $\begingroup$ no, I would think that the error term is of order $1/d^2$. $\endgroup$ Commented 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
    Commented Oct 11, 2021 at 14:47

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 $n \ge 0$.


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.


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