# Limiting value of $\dfrac{1_n^\top B^{-1} A B^{-1} 1_n}{d}$, where $A=WW^\top + a I_n$, $B = WW^\top + b I_n$, and $W \sim N(0,\Sigma_d)$

Let $$n$$ and $$d$$ be positive integers with $$n,d \to \infty, \quad n/d \to \rho \in (0,\infty).$$ Let $$\Sigma_d$$ be a psd matrix such that

• $$\mbox{trace}(\Sigma_d) = 1$$.
• $$\|\Sigma_d\|_{op} = \mathcal O(1/d)$$.
• The empirical eigenvalue distribution of $$d \cdot \Sigma_d$$ converges weakly to some distribution $$D$$ on $$\mathbb R$$.

Let $$W$$ be a random $$n \times d$$ matrix with iid rows from $$N(0,\Sigma_d)$$. Finally, let $$a,b \ge 0$$ be fixed constants, and define random $$A$$ and $$B$$ by $$\begin{split} A &= WW^\top + a I_n,\\ B &= WW^\top + b I_n. \end{split}$$

Question. What is an analytic formula for the limiting value of the (random) scalar $$\dfrac{1_n^\top B^{-1}AB^{-1} 1_n}{d}$$, where $$1_n := (1,1,\ldots,1) \in \mathbb R^n$$ ?

## Observations

If we replace $$1_n$$ by $$z$$, where $$z \sim N(0,I_n)$$ is independent of $$W$$, then $$\begin{split} \mathbb E_z[z^\top B^{-1} A B^{-1} z/d] &= \mbox{trace}(B^{-1} A B^{-1}/d)\\ &= \frac{1}{d}\sum_{i=1}^n \frac{\lambda_i + a}{(\lambda_i + b)^2}\\ & \overset{a.s}{\to} \int_0^\infty \frac{\lambda + a}{(\lambda + b)^2}\,\mu(\mathrm{d}\lambda)\\ &= m_\mu(-b) - (a-b) m_\mu'(-b), \end{split}$$ where $$\mu$$ is the LSD of $$WW^\top$$, and $$m_\mu$$ is its Stieltjes transform (which can be evaluated via $$D$$ and Silverstein's fixed-point equation).

Maybe this computation is somehow linked to the an answer to my main question ?

• $E[z^TMz]=trace[ME[zz^T]] = trace[M]$ for $z\sim N(0,I_n)$. Is the $1/d$ in your $trace[B^{-1}AB^{-1}]$ a typo? Commented Dec 11, 2021 at 19:43
• Yes, thats a typo. Fixed. Commented Dec 11, 2021 at 19:51

The distribution of $$v^TB^{-1}AB^{-1}v$$ is the same for every vector $$v$$ in the unit sphere either deterministic or independent of $$W$$. Once this is established, you are allowed to take $$v=z/\|z\|$$ independent of $$W$$; you can then apply the argument given at the end of the question together with concentration of the quadratic form in $$z$$ (Hanson Wright inequality) and $$\|z\|^2/n\to^{as}1$$.
The fact that the distribution of $$v^TB^{-1}AB^{-1}v$$ is the same for all $$v$$ in the unit sphere can be seen as follows. Let $$u=R^Tv$$ for a rotation $$R\in O(n)$$. Then $$u^TB^{-1}AB^{-1}u = v^T R B^{-1} R^T R A R^T R B^{-1} R^T v = v^T \tilde B^{-1} \tilde A \tilde B^{-1} v$$ where $$\tilde A = RAR^T = \tilde W\tilde W^T + aI$$ where $$\tilde W=RW$$ and $$\tilde B = RBR^T = \tilde W\tilde W^T + bI$$. But $$W=^d\tilde W$$ by rotational invariance of the Gaussian distribution.