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Let $X=(X_1,\ldots,X_d)$ be uniformly distributed on the sphere of radius $\sqrt{d}$ in $\mathbb R^d$. Fix a "sufficiently integrable" function $h:\mathbb R \to \mathbb R$, and define another function $f:[-1,1] \to \mathbb R$ by $$ f(q)=E[h(X_1)h(qX_1 + (1-q^2)^{1/2} X_2)]. $$

Examples

  • If $h(z) \equiv \mathrm{sign}(z)$, then one can show that $f(q) \equiv \arcsin(q)/pi$.
  • If $h(z) \equiv 1[z \ge 0]$, then $f(q) \equiv \arccos(-q)/(2\pi)$
  • If $h(z) \equiv z$, then one can easily show that $f(q) \equiv q/2$. More generally, if $h(z) \equiv z^k$ (for an integer $z \ge 0$), then $$ f(q) \equiv k!2^k (-i\cos\alpha)^k P_k(-i\tan\alpha), $$ where $\alpha = \arcsin(q)$, $i = \sqrt{-1}$ is the imaginary unit, and $P_k$ is the $k$th Legendre polynomial.
  • ...

The problem

Now, suppose $f$ is smooth around $0$ and define the coefficients $$ \alpha_d := f(0),\,\beta_d := f'(0),\,\gamma_d := f(1), $$ where we've emphasized the fact that these coefficients depend on the dimension $d$.

Goal. I'm interested in the limit of these coefficients when $d \to \infty$.

To this end, further suppose $h$ is smooth on $\mathbb R$ and define more coefficients $$ \begin{split} a &:= (E[h(G)])^2 =\zeta_0(h)^2,\text{ with }G \sim N(0,1),\\ b &= (E[h'(G)])^2=(E[Gh(G)])^2 = \zeta_1(h)^2, \\ c &:= E[h(G)^2]=\zeta_0(h^2), \end{split} $$ where $\zeta_k(h)$ is the $k$th Hermite coefficient of $h$.

Question. In the limit $d \to \infty$, is it true that $\alpha_d \to a$, $\beta_d \to b$, and $\gamma_d \to c$ for any "sufficiently integrable" function $h$ ?

Notes

  • Of course, it might well be that in order to answer the above question in the affirmative, one needs some additional assumptions on the function $h$. I'm fine with this, as long as the additional conditions are as general as possible (for example, should cover Lipschitz-continuous functions, polynomials, etc. because these special cases can be established via "direct computation").

  • The motivation of this question is that, after all, the marginal distribution of each coordinate of $X$ convergence to $N(0,1)$ (in both Wasserstein distance, say).


Update

See partial solution below (here https://mathoverflow.net/a/403398/78539), under additional Lipschitz continuity assumoptions.

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  • $\begingroup$ The sign and indicator functions given are example are very special as they are homogeneous of order 0. Since $(X_1,X_2)$ is equal in distribution to $(G_1, G_2)/\sqrt{G_1^2+G_2^2 + \chi^2_{d-2}}$ for independent $G_1,G_2,\chi^2_d$, so that the case of homogeneous $h$ of order 0 reduces to the Gaussian case. $\endgroup$
    – jlewk
    Commented Sep 15, 2021 at 0:49

1 Answer 1

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Solution with added restriction that $h$ is Lipschitz continuous

Below, I do some computations which seem to suggest the result is true under some additional smoothness constraints on $h$. I'm not 100% sure of my calculations. I'd be grateful if someone else would double-check. Thanks in advance!

So, suppose $h$ satisfies

  • (1) $h$ is Lipschitz-continuous.
  • (2) $h$ is square-integrable w.r.t $N(0,1)$.

Claim 1. Under the above assumptions, the following hold in the limit $d \to \infty$, $$ \begin{split} |\alpha_d-a| &= \mathcal O(1/d),\\ |\beta_d-b| &= \mathcal O(1/d). \end{split} $$

Under the additional assumptions

  • (3) $h$ is smooth with Lipschitz-continuous gradient $h'$,
  • (4) $h'$ is square-integrable w.r.t $N(0,1)$,

we have

Claim 2. $|\gamma_d-c| = \mathcal O(1/d)$ in the limit $d \to \infty$.

Proof of Claim 1. For any $p \in [1,\infty)$, define (when it exists!) the $L_p$ norm of a (real-valued) random variable $R$ by $$ \|R\|_p := (E[R^p])^{1/p}. $$

Note. If $R_1$ and $R_2$ are random variables, $\|R_1-R_2\|_p$ will always be understood w.r.t to an underlying [coupling][1] $(R_1,R_2)$ of $R_1$ and $R_2$.

  • Now, let $(G_1,G_2) \sim N(0,I_2)$, independent of $X$. Let $W_p(X_1,G_1)$ be the Wasserstein distance of order $p$, between $X_1$ and $G_1$. For $W_2$-optimal couplings $(X_1,G_1)$ and $(X_2,G_2)$, one computes $$ \begin{split} &|\alpha_d-a| = |E[h(X_1)h(X_2) - h(G_1)h(G_2)]|\\ &\quad= |E[h(X_1)(h(X_2)-h(G_2))] + E[h(G_2)(h(X_1)-h(G_1))]|\\ &\quad\le |E[h(X_1)(h(X_2)-h(G_2))]| + |E[h(G_2)(h(X_1)-h(G_1))]|\\ &\quad\le \|h(X_1)\|_2\|h(X_2)-h(G_2)\|_2 + \|h(G_2)\|_2\|h(X_1)-h(G_1)\|_2\\ &\quad\lesssim \|X_2-G_2\|_2 + \|X_1-G_1\|_2\\ &\quad\lesssim W_2(X_2,G_2) + W_2(X_1,G_1)\\ &\quad\lesssim W_2(X_1,G_1). \end{split} \tag{1} $$ The first inequality is the triangle inequality, the second is Cauchy-Schwarz, and the third is thanks to the Lipschitz continuity of $h$. Recall the following fact (established here https://mathoverflow.net/a/401444/78539)

Fact 1. $W_p(X_1,G_1) = \mathcal O(1/d)$ for all $ p \ge 1$.

We deduce from (1) that $|\alpha_d-a| = \mathcal O(1/d)$ as claimed.

  • Next, we prove that $|\gamma_d-c| = \mathcal O(1/d)$. Indeed, $$ \begin{split} |\gamma_d-c| &= |E[h(X_1)^2 - h(G_1)^2| = |E[(h(X_1)+h(G_1))(h(X_1)-h(G_1)]|\\ &\le \|h(X_1)+h(G_1)\|_2\|h(X_1)-h(G_1)\|_2\\ &\le (\|h(X_1)\|_2+\|h(G_1)\|_2)\|X_1-G_1\|_2\\ & \lesssim \|X_1-G_1\|_2 = W_2(X_1,G_1) = \mathcal O(1/d). \end{split} $$ This completes the proof of Claim 1. $\quad\quad\Box$

Proof of Claim 2. Consider the function $r:\mathbb R \to \mathbb R$ defined by $r(x) := xh(x)$. Let $(X_2,G_2)$ be an $W_2$-optimal coupling of $X_2$ and $G_2$ and let $(X_1,G_1)$ be a $W_4$-optimal coupling for $X_1$ and $G_1$. One computes $$ \begin{split} |\beta_d-b| &= |E[r(X_1)h'(X_2) - r(G_1)h'(G_2)]|\\ &= |E[r(X_1)(h'(X_2)-h'(G_2)) + h'(G_2)(r(X_1)-r(G_1))]|\\ &\le \|r(X_1)\|_2\|h'(X_2)-h'(G_2)\|_2 + \|h'(G_2)\|_2\|r(X_1)-r(G_1)\|_2\\ &\lesssim \|X_2-G_2\|_2 + \|r(X_1)-r(G_1)\|_2\\ &= W_2(X_2,G_2) + \|r(X_1)-r(G_1)\|_2\\ &= \|r(X_1)-r(G_1)\|_2+\mathcal O(1/d), \end{split} $$ where the first inequality is thanks to the Lipschitzness of $h'$. It only remains to bound the term $\|r(X_1)-r(G_1)\|_2$. To this end, note that $$ \begin{split} \|r(X_1)-r(G_1)\|_2 &= |X_1h(X_1)-G_1h(G_1)|\\ &= \|X_1(h(X_1)-h(G_1))+h(G_1)(X_1-G_1)\|_2\\ &\le \|X_1\|_4\|h(X_1)-h(G_1)\|_4+\|h(G_1)\|_4\|X_1-G_1\|_4\\ &\lesssim \|X_1-G_1\|_4 = W_4(X_1,G_1) = \mathcal O(1/d). \end{split} $$ This completes the proof of Claim 2. $\quad\quad\quad\Box$

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