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dohmatob
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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 $$ \begin{split} \mathbb E_z[z^\top B^{-1} A B^{-1} z] &= \mbox{trace}(B^{-1} A B^{-1}/d)\\ &= \frac{1}{d}\sum_{i=1}^n \frac{\lambda_i + a}{(\lambda_i + b)^2}\\ & \overset{a.s}{\to} \int_0^\infty \frac{\lambda + a}{(\lambda + b)^2}\,\mu(\mathrm{d}\lambda)\\ &= m_\mu(-b) - (a-b) m_\mu'(-b), \end{split} $$$$ \begin{split} \mathbb E_z[z^\top B^{-1} A B^{-1} z/d] &= \mbox{trace}(B^{-1} A B^{-1}/d)\\ &= \frac{1}{d}\sum_{i=1}^n \frac{\lambda_i + a}{(\lambda_i + b)^2}\\ & \overset{a.s}{\to} \int_0^\infty \frac{\lambda + a}{(\lambda + b)^2}\,\mu(\mathrm{d}\lambda)\\ &= m_\mu(-b) - (a-b) m_\mu'(-b), \end{split} $$ 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 ?

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 $$ \begin{split} \mathbb E_z[z^\top B^{-1} A B^{-1} z] &= \mbox{trace}(B^{-1} A B^{-1}/d)\\ &= \frac{1}{d}\sum_{i=1}^n \frac{\lambda_i + a}{(\lambda_i + b)^2}\\ & \overset{a.s}{\to} \int_0^\infty \frac{\lambda + a}{(\lambda + b)^2}\,\mu(\mathrm{d}\lambda)\\ &= m_\mu(-b) - (a-b) m_\mu'(-b), \end{split} $$ 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 ?

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 ?

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dohmatob
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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 $$ \begin{split} \mathbb E_z[z^\top B^{-1} A B^{-1} z] &= \mbox{trace}(B^{-1} A B^{-1}/d)\\ &= \frac{1}{d}\sum_{i=1}^n \frac{\lambda_i + a}{(\lambda_i + b)^2}\\ & \overset{a.s}{\to} \int_0^\infty \frac{\lambda + a}{(\lambda + b)^2}\,\mu(\mathrm{d}\lambda)\\ &= \alpha m_\mu(-b) + \beta m_\mu'(-b), \end{split} $$$$ \begin{split} \mathbb E_z[z^\top B^{-1} A B^{-1} z] &= \mbox{trace}(B^{-1} A B^{-1}/d)\\ &= \frac{1}{d}\sum_{i=1}^n \frac{\lambda_i + a}{(\lambda_i + b)^2}\\ & \overset{a.s}{\to} \int_0^\infty \frac{\lambda + a}{(\lambda + b)^2}\,\mu(\mathrm{d}\lambda)\\ &= m_\mu(-b) - (a-b) m_\mu'(-b), \end{split} $$ where $\alpha := 1/c^2$, $\beta := (b-a)/b^2$, $\mu$ 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 ?

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 $$ \begin{split} \mathbb E_z[z^\top B^{-1} A B^{-1} z] &= \mbox{trace}(B^{-1} A B^{-1}/d)\\ &= \frac{1}{d}\sum_{i=1}^n \frac{\lambda_i + a}{(\lambda_i + b)^2}\\ & \overset{a.s}{\to} \int_0^\infty \frac{\lambda + a}{(\lambda + b)^2}\,\mu(\mathrm{d}\lambda)\\ &= \alpha m_\mu(-b) + \beta m_\mu'(-b), \end{split} $$ where $\alpha := 1/c^2$, $\beta := (b-a)/b^2$, $\mu$ 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 ?

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 $$ \begin{split} \mathbb E_z[z^\top B^{-1} A B^{-1} z] &= \mbox{trace}(B^{-1} A B^{-1}/d)\\ &= \frac{1}{d}\sum_{i=1}^n \frac{\lambda_i + a}{(\lambda_i + b)^2}\\ & \overset{a.s}{\to} \int_0^\infty \frac{\lambda + a}{(\lambda + b)^2}\,\mu(\mathrm{d}\lambda)\\ &= m_\mu(-b) - (a-b) m_\mu'(-b), \end{split} $$ 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 ?

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dohmatob
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Limiting value of $\dfrac{1_m^\top1_n^\top B^{-1} A B^{-1} 1_m1_n}{d}$, where $A=XX^\top$A=WW^\top + a I_m$I_n$, $B = XX^\topWW^\top + b I_m$I_n$, and $W \sim N(0,\Sigma\Sigma_d)$

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{tr}(\Sigma_d) = 1$$\mbox{trace}(\Sigma_d) = 1$.
  • $\|\Sigma_d\|_{op} = \mathcal O(1/d)$.
  • SpectrumThe 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 $$ \begin{split} A &= WW^\top + a I_m,\\ B &= WW^\top + b I_m. \end{split} $$$$ \begin{split} A &= WW^\top + a I_n,\\ B &= WW^\top + b I_n. \end{split} $$

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_m$$1_n$ by $z$, where $z \sim N(0,I_n)$ is independent of $W$, then $$ \begin{split} \mathbb E_z[z^\top B^{-1} A B^{-1} z] = \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)\\ & = \alpha m_\mu(-b)-\beta m_\mu'(-b), \end{split} $$$$ \begin{split} \mathbb E_z[z^\top B^{-1} A B^{-1} z] &= \mbox{trace}(B^{-1} A B^{-1}/d)\\ &= \frac{1}{d}\sum_{i=1}^n \frac{\lambda_i + a}{(\lambda_i + b)^2}\\ & \overset{a.s}{\to} \int_0^\infty \frac{\lambda + a}{(\lambda + b)^2}\,\mu(\mathrm{d}\lambda)\\ &= \alpha m_\mu(-b) + \beta m_\mu'(-b), \end{split} $$ where $\alpha := 1/c^2$, $\beta := (b-a)/b^2$, $\mu$ 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 ?

Limiting value of $\dfrac{1_m^\top B^{-1} A B^{-1} 1_m}{d}$, where $A=XX^\top + a I_m$, $B = XX^\top + b I_m$, and $W \sim N(0,\Sigma)$

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

  • $\mbox{tr}(\Sigma_d) = 1$.
  • $\|\Sigma_d\|_{op} = \mathcal O(1/d)$.
  • Spectrum 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 $$ \begin{split} A &= WW^\top + a I_m,\\ B &= WW^\top + b I_m. \end{split} $$

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 $1_m$ by $z$, where $z \sim N(0,I_n)$ is independent of $W$, then $$ \begin{split} \mathbb E_z[z^\top B^{-1} A B^{-1} z] = \mbox{trace}(B^{-1} A B^{-1}/d) &= \frac{1}{d}\sum_{i=1}^n \frac{\lambda_i + a}{(\lambda_i + b)^2} \overset{a.s}{\to} \int_0^\infty \frac{\lambda + a}{(\lambda + b)^2}\,\mu(\mathrm{d}\lambda)\\ & = \alpha m_\mu(-b)-\beta m_\mu'(-b), \end{split} $$ where $\alpha := 1/c^2$, $\beta := (b-a)/b^2$, $\mu$ 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 ?

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 $$ \begin{split} \mathbb E_z[z^\top B^{-1} A B^{-1} z] &= \mbox{trace}(B^{-1} A B^{-1}/d)\\ &= \frac{1}{d}\sum_{i=1}^n \frac{\lambda_i + a}{(\lambda_i + b)^2}\\ & \overset{a.s}{\to} \int_0^\infty \frac{\lambda + a}{(\lambda + b)^2}\,\mu(\mathrm{d}\lambda)\\ &= \alpha m_\mu(-b) + \beta m_\mu'(-b), \end{split} $$ where $\alpha := 1/c^2$, $\beta := (b-a)/b^2$, $\mu$ 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 ?

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dohmatob
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