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Consider a random matrix $A \in \mathbb{R}^{m\times n}$ with i.i.d. entries, with mean zero and variance 1 and $m <n $. Has anyone studied this expectation in asymptotics $$E_{A}(\mathrm{Tr}( (A^T A + \lambda \mathrm{Id})^{-1} A^T A))?$$

Any papers/resources would be helpful, ideal fidnings would be $m,n \mapsto \infty$ as $\frac{m}{n} \mapsto 0$, but anything would be great.

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The answer is similar to that of your earlier question:

For $m,n\gg 1$, and $m/n\equiv r\in (0,1)$ fixed, an integration over the Marchenko–Pastur distribution gives (with $x_\pm=(1\pm\sqrt{r})^2$) $$\lim_{m,n\rightarrow\infty}\mathbb{E}\bigl[m^{-1}\mathrm{Tr}\,\bigl((n^{-1}AA^\top + \lambda I)^{-1}n^{-1}AA^\top\bigr)\bigr]=\int_{x_-}^{x_+} \frac{x}{x+\lambda}\frac{\sqrt{\left(x_+-x\right) \left(x-x_-\right)}}{2 \pi r x}\,dx$$ $$\qquad=\frac{1}{2r}\bigl(\lambda+r+1-\sqrt{\lambda^2+2 \lambda (r+1)+(r-1)^2}\bigr).$$ The rescaling of $AA^\top$ by a factor $1/n$ is needed for a $\lambda$-dependent answer in the large $n$ limit.

In the limit $r\rightarrow 0$ this tends to $1/(1+\lambda)$.

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  • $\begingroup$ Could you please give integration bounds to show its similar to the other answer? $\endgroup$
    – goku
    Commented Jun 20, 2022 at 20:59

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