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Let $X$ be an $n \times d$ random matrix with iid entries from $N(0, 1/d)$. Let $S:=X^\top X/n$, a $d \times d$ Wishart matrix and let $T = e^{S} := \sum_{k=0}^\infty \dfrac{S^k}{k!}$ be its exponential. We place our self in the asymptotic regime where $n,d \to \infty$ such that $d/n \to \gamma \in (0,\infty)$.

Question. Are there analytic formulae for the limiting values of $\mathbb E\operatorname{trace}(T)$ and $\mathbb E\operatorname{trace}(TS)$, say in terms of the Machenko-Pastur density ?

N.B.: In the absence of analytic formulae, I'm fine with good lower- and upper-bounds.

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A closed-form expression is not forthcoming, but here are the plots of $\mathbb{E}[\text{trace}\, T]$ (left plot) and $\mathbb{E}[\text{trace}\, TS]$ (right plot) as a function of $\gamma$ in the interval $(0,1)$ (obtained by integrating the Marchenko-Pastur distribution). For $\gamma=d/n>1$ there are additionally $d-n$ eigenvalues of $S$ equal to 0, so that case reduces simply to the case $0<\gamma<1$.

If upper and lower bounds are needed, you could just fit a spline to these curves, they increase monotonically between the following end points: $$\mathbb{E}[\text{trace}\, T]=\begin{cases} e&\text{for}\;\;\gamma\rightarrow 0,\\ e^2 I_2(2)&\text{for}\;\;\gamma=1, \end{cases}$$ $$\mathbb{E}[\text{trace}\, TS]=\begin{cases} e&\text{for}\;\;\gamma\rightarrow 0,\\ e^2 I_1(2)&\text{for}\;\;\gamma=1, \end{cases}$$ with $I_p(x)$ a Bessel function.

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  • $\begingroup$ Thanks (upvoted). You mean "addtional $d-n$ eigenvalues equal to $0$ ...", right ? $\endgroup$
    – dohmatob
    Jun 3, 2022 at 16:42
  • $\begingroup$ yes, typo corrected --- thanks. $\endgroup$ Jun 3, 2022 at 17:11
  • $\begingroup$ In math.stackexchange.com/questions/4464861/… I gave the first terms of the expansion. Is it inspiring ? $\endgroup$ Jun 4, 2022 at 14:21

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