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 $W \sim N(0,\Sigma_d)$

Question

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 ?

Answer 1

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.