Given iid $w_1,\dotsc,w_N \sim N(0,1/d)$ iid, find a simple matrix $A$ s.t $\|aa^T-A\|_\text{op}\to0$, where $a_i := E_{G \sim N(0,1)}[f(\|w_i\| G)]$

Question

Let $d$ and $N$ be two large comparable integers, for example assume $$ N,d \to \infty, \quad d/N \to \gamma \in (0,\infty). $$ Let $w_1,\dotsc,w_N$ be iid from $N(0,(1/d)I_d)$ and let $f:\mathbb R \to \mathbb R$ be such that $|f(x)| \le \exp(cx^2/2)$, for all $x \in \mathbb R$ and for some $c < 1$. Note that this implies $f \in L^2(\mathbb R,N(0,1))$. Define an $N \times N$ random matrix $T$ by $$ T_{ij} := \zeta_0(f_i)\zeta_0(f_j), $$ where the function $f_i:\mathbb R \to \mathbb R$ is defined by $f_i(x) := f(\|w_i\|x)$, and $\zeta_k(h)$ is the (probabilist's) $k$th Hermite coefficient of a function $h \in L^2(\mathbb R,N(0,1))$; in particular $\zeta_0(h) := \mathbb E_G[h(G)]$, $\zeta_1(h) := \mathbb E_G[Gh(G)]$, and $\zeta_2(h):=\mathbb E_G[(G^2-1)h(G)]$, for $G \sim N(0,1)$.

I'm interested in a simple approximation of $T$ in terms of simple expressions in the $w_i$'s (e.g polynomial expressions). In this direction, one can obtain the following

(Easy case) If $\zeta_0(f)=0$, then $$ \|T-\mu\mu^\top\|_\text{op} = o_{d,\mathbb P}(1), \tag{1}\label{409440_1} $$ where $\mu \in \mathbb R^N$ is defined by $\mu_i := \zeta_2(f)(\|w_i\|^2-1)/2$.

The idea is to write $T_{ij} = D\mu\mu^\top D$, where $D$ is the diagonal matrix with $$ D_{ii} := \frac{\zeta_0(f_i)}{\zeta_2(f)(\|w_i\|^2-1)/2}. $$ Further, using the assumption that $\zeta_0(f) := \mathbb E_G[f(G)] = 0$, one writes

$$ D_{ii} := \mathbb E_G\left[\frac{f(\|w_i\|G)-f(G)}{\|w_i\|-1}\right]\cdot\frac{1}{\zeta_2(f)(\|w_i\|+1)/2}. $$

One can show that $$ \lim_{t \to 1}\dfrac{f(tG)-f(G)}{t-1} = \zeta_2(f) := \mathbb E_G[(G^2-1)f(G)]. \tag{2}\label{409440_2} $$

This is particularly easy to see if one assumes temporarily that $f$ is $\mathcal C^2$ around $1$ and uses a Taylor expansion combined with the observation that $\zeta_1(f') = \zeta_2(f)$. The general case is slightly more complicated.

Combining \eqref{409440_2} with the fact that $\sup_{1 \le i \le N}||w_i|-1| = o_{d,\mathbb P}(1)$, we obtain that $\|D-I_N\|_{op} = o_{d,\mathbb P}(1)$, from which \eqref{409440_1} follows.

N.B.: Here, $o_{d,\mathbb P}(1)$ is notation for a quantity which converges to zero in probability.

---

Question. Without the condition $\zeta_0(f)=0$, is there a simple approximation of $T_0$ of the kind \eqref{409440_1} ?

Answer 1

It turns our the problem has a simple answer, once the easy case has been solved (see the OP).

---

Indeed, we write $\overline{f} = f + \zeta_0(f)$, so that $\zeta_0(f) = 0$. Now, one has $$ T = \overline{T} + v1^\top+(v1_N^\top)^\top + \zeta_0(f)^2 1_N1_N^\top, $$ where $v = (\zeta_0(\overline{f}_i))_{i \in [N]} = (\zeta_0(f_i)-\zeta_0)_{i \in [N]}$. By the easy case already solved, we know that $$ \|\overline{T}-\mu\mu^\top\|_\text{op} = o_{d,\mathbb P}(1). $$

It remains localize $v1^\top$. To this end, one may write $v1^\top = D\mu 1_N^\top$, where $D$ is the $N \times N$ diagonal matrix with entries $D_{ii} = v_i /\mu_i$. As already seen before, $\|D-I_N\|_\text{op} = o_{d,\mathbb P}(1)$, and so $\|v1^\top - \mu 1_N^\top\|_\text{op} = o_{d,\mathbb P}(1)$. We deduce that

$$ \|T-uu^\top\|_\text{op} = o_{d,\mathbb P}(1), $$ where $u := \mu + \zeta_0(f)1_N$.