Skip to main content
limit
Source Link
Carlo Beenakker
  • 188.1k
  • 18
  • 448
  • 651

let $\lambda_{1},\lambda_2,\ldots\lambda_N$ be the eigenvalues of $M^{-1}XX^{T}$; including for convenience a factor $1/N$, the quantity you seek is

$$N^{-1}E[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=\int d\lambda \rho(\lambda)(\lambda\gamma+1)^{-1}$$

where $\rho(\lambda)=E[N^{-1}\sum_n\delta(\lambda-\lambda_n)]$ is the eigenvalue density of Wishart matrices; this quantity is known in closed form for $N,M\rightarrow\infty$ at finite ratio $N/M=r\in(0,1]$ (Marchenko-Pastur distribution). I find in this way the answer

$$\lim_{N,M\rightarrow\infty}N^{-1}E[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=(2r\gamma)^{-1}\left(-1-\gamma\sqrt{ab}+\sqrt{(1+a\gamma)(1+b\gamma)}\right)$$

with $a=(1+\sqrt r)^2$ and $b=(1-\sqrt r)^2$. As a check, you can take the limit $\gamma\rightarrow 0$ of this expression and obtain $1$, as it should.

let $\lambda_{1},\lambda_2,\ldots\lambda_N$ be the eigenvalues of $M^{-1}XX^{T}$; including for convenience a factor $1/N$, the quantity you seek is

$$N^{-1}E[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=\int d\lambda \rho(\lambda)(\lambda\gamma+1)^{-1}$$

where $\rho(\lambda)=E[N^{-1}\sum_n\delta(\lambda-\lambda_n)]$ is the eigenvalue density of Wishart matrices; this quantity is known in closed form for $N,M\rightarrow\infty$ at finite ratio $N/M=r\in(0,1]$ (Marchenko-Pastur distribution). I find in this way the answer

$$\lim_{N,M\rightarrow\infty}N^{-1}E[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=(2r\gamma)^{-1}\left(-1-\gamma\sqrt{ab}+\sqrt{(1+a\gamma)(1+b\gamma)}\right)$$

with $a=(1+\sqrt r)^2$ and $b=(1-\sqrt r)^2$.

let $\lambda_{1},\lambda_2,\ldots\lambda_N$ be the eigenvalues of $M^{-1}XX^{T}$; including for convenience a factor $1/N$, the quantity you seek is

$$N^{-1}E[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=\int d\lambda \rho(\lambda)(\lambda\gamma+1)^{-1}$$

where $\rho(\lambda)=E[N^{-1}\sum_n\delta(\lambda-\lambda_n)]$ is the eigenvalue density of Wishart matrices; this quantity is known in closed form for $N,M\rightarrow\infty$ at finite ratio $N/M=r\in(0,1]$ (Marchenko-Pastur distribution). I find in this way the answer

$$\lim_{N,M\rightarrow\infty}N^{-1}E[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=(2r\gamma)^{-1}\left(-1-\gamma\sqrt{ab}+\sqrt{(1+a\gamma)(1+b\gamma)}\right)$$

with $a=(1+\sqrt r)^2$ and $b=(1-\sqrt r)^2$. As a check, you can take the limit $\gamma\rightarrow 0$ of this expression and obtain $1$, as it should.

typo
Source Link
Carlo Beenakker
  • 188.1k
  • 18
  • 448
  • 651

let $\lambda_{1},\lambda_2,\ldots\lambda_N$ be the eigenvalues of $M^{-1}XX^{T}$; including for convenience a factor $1/N$, the quantity you seek is

$${\rm E}[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=N\int d\lambda \rho(\lambda)(\lambda\gamma+1)^{-1}$$$$N^{-1}E[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=\int d\lambda \rho(\lambda)(\lambda\gamma+1)^{-1}$$

where $\rho(\lambda)=E[N^{-1}\sum_n\delta(\lambda-\lambda_n)]$ is the eigenvalue density of Wishart matrices; this quantity is known in closed form for $N,M\rightarrow\infty$ at finite ratio $N/M=r$$N/M=r\in(0,1]$ (Marchenko-Pastur distribution). I find in this way the answer

$$\lim_{N,M\rightarrow\infty}N^{-1}E[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=(2r\gamma)^{-1}\left(-1-\gamma\sqrt{ab}+\sqrt{(1+a\gamma)(1+b\gamma)}\right)$$

with $a=(1+\sqrt r)^2$ and $b=(1-\sqrt r)^2$.

let $\lambda_{1},\lambda_2,\ldots\lambda_N$ be the eigenvalues of $M^{-1}XX^{T}$; the quantity you seek is

$${\rm E}[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=N\int d\lambda \rho(\lambda)(\lambda\gamma+1)^{-1}$$

where $\rho(\lambda)=E[N^{-1}\sum_n\delta(\lambda-\lambda_n)]$ is the eigenvalue density of Wishart matrices; this quantity is known in closed form for $N,M\rightarrow\infty$ at finite ratio $N/M=r$ (Marchenko-Pastur distribution).

let $\lambda_{1},\lambda_2,\ldots\lambda_N$ be the eigenvalues of $M^{-1}XX^{T}$; including for convenience a factor $1/N$, the quantity you seek is

$$N^{-1}E[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=\int d\lambda \rho(\lambda)(\lambda\gamma+1)^{-1}$$

where $\rho(\lambda)=E[N^{-1}\sum_n\delta(\lambda-\lambda_n)]$ is the eigenvalue density of Wishart matrices; this quantity is known in closed form for $N,M\rightarrow\infty$ at finite ratio $N/M=r\in(0,1]$ (Marchenko-Pastur distribution). I find in this way the answer

$$\lim_{N,M\rightarrow\infty}N^{-1}E[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=(2r\gamma)^{-1}\left(-1-\gamma\sqrt{ab}+\sqrt{(1+a\gamma)(1+b\gamma)}\right)$$

with $a=(1+\sqrt r)^2$ and $b=(1-\sqrt r)^2$.

ratio
Source Link
Carlo Beenakker
  • 188.1k
  • 18
  • 448
  • 651

let $\lambda_{1},\lambda_2,\ldots\lambda_N$ be the eigenvalues of $M^{-1}XX^{T}$; the quantity you seek is

$${\rm E}[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=N\int d\lambda \rho(\lambda)(\lambda\gamma+1)^{-1}$$

where $\rho(\lambda)=E[N^{-1}\sum_n\delta(\lambda-\lambda_n)]$ is the eigenvalue density of Wishart matrices; this quantity is known in closed form for large $N$$N,M\rightarrow\infty$ at finite ratio $N/M=r$ (Marchenko-Pastur distribution).

let $\lambda_{1},\lambda_2,\ldots\lambda_N$ be the eigenvalues of $M^{-1}XX^{T}$; the quantity you seek is

$${\rm E}[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=N\int d\lambda \rho(\lambda)(\lambda\gamma+1)^{-1}$$

where $\rho(\lambda)=E[N^{-1}\sum_n\delta(\lambda-\lambda_n)]$ is the eigenvalue density of Wishart matrices; this quantity is known in closed form for large $N$ (Marchenko-Pastur distribution).

let $\lambda_{1},\lambda_2,\ldots\lambda_N$ be the eigenvalues of $M^{-1}XX^{T}$; the quantity you seek is

$${\rm E}[{\rm Tr}(XX^{T}\gamma/M)+I)^{-1}]=N\int d\lambda \rho(\lambda)(\lambda\gamma+1)^{-1}$$

where $\rho(\lambda)=E[N^{-1}\sum_n\delta(\lambda-\lambda_n)]$ is the eigenvalue density of Wishart matrices; this quantity is known in closed form for $N,M\rightarrow\infty$ at finite ratio $N/M=r$ (Marchenko-Pastur distribution).

Source Link
Carlo Beenakker
  • 188.1k
  • 18
  • 448
  • 651
Loading