$$R := A \Lambda^{-1} A^H + \frac{1}{\gamma} I_n$$

where $A$ is a given $n \times m$ matrix (where $m \gg n$),

$$\Lambda := \mbox{diag} \big( \lambda_1, \lambda_2, \dots, \lambda_m \big)$$

and $I_n$ is the $n \times n$ identity matrix. $\lambda_1, \lambda_2, \dots, \lambda_m$ and $\gamma$ are independent random variables with a gamma distribution. I would like to compute the expected value $\mathbb E(R^{-1})$.

Any idea on how to approximate this? I am not able to find any analytical solution for this. Thanks.

  • $\begingroup$ All the random variables are independent, but not identically distributed. $\endgroup$ – Christo Jul 30 '18 at 7:35
  • $\begingroup$ $AA^H \neq I_n$ $\endgroup$ – Christo Jul 30 '18 at 7:36

Let me assume that the gamma distribution of each $\lambda_i$ has shape parameter $k_i>2$, so that $1/\lambda_i$ has a finite average and variance, $${\mathbb E}(1/\lambda_i)=\frac{\Gamma(k_i-1)}{\theta_i\Gamma(k_i)}\equiv \mu_i,\;\; {\rm Var}\,(1/\lambda_i)=\frac{1}{(k_i-2) (k_i-1)^2 \theta_i^2}\equiv\sigma_i^2$$ where $\theta_i$ is the scale parameter of the gamma distribution. The $n\times n$ matrix $R$ has elements $$R_{ij}=\sum_{l=1}^m \frac{1}{\lambda_l} A^{\vphantom{\ast}}_{il}A^\ast_{jl}+\delta_{ij}/\gamma$$ $$\Rightarrow {\mathbb{E}}(R_{ij})=\sum_{l=1}^m \mu_l A^{\vphantom{\ast}}_{il}A^\ast_{jl}+\delta_{ij}/\gamma,$$ where $\ast$ denotes the complex conjugate.
Up to corrections of order $1/m$ we may replace the average of the inverse matrix by the inverse of the average matrix, $$\mathbb{E}(R^{-1})=\bigl(\mathbb{E}(R)\bigr)^{-1}\bigl(1+{\cal O}(m^{-1})\bigr).$$ In particular, for $n=1$ one has $$\mathbb{E}(R^{-1})=\frac{\gamma}{1+\gamma\sum_{l=1}^m \mu_l |A_{1l}|^2}\bigl(1+{\cal O}(m^{-1})\bigr).$$

An alternative approach, that avoids having to invert a matrix, is possible if $\gamma\ll 1$. A Taylor expansion in powers of $\gamma$ gives $$R^{-1}=\gamma I_n+\gamma\sum_{p=1}^\infty(-\gamma)^p(A\Lambda^{-1}A^H)^p.$$ The expectation value to fourth order is $${\mathbb{E}}\bigl((R^{-1})_{ij}\bigr)=\gamma \delta_{ij}-\gamma^2 \sum_{l=1}^m \mu_l A^{\vphantom{\ast}}_{il}A_{jl}^\ast$$ $$+\,\gamma^3\sum_{k,l,l'=1}^m (\mu_l \mu_{l'}+\delta_{ll'}\sigma^2_l)A_{il}^{\vphantom{\ast}}A_{kl}^\ast A_{kl'}^{\vphantom{\ast}}A_{jl'}^\ast+{\cal O}(\gamma^4).$$

  • $\begingroup$ Thanks for the answer. For me, matrix $A$ is such that $n,m$ are very high or can be considered to be $\rightarrow \infty$. So I want to avoid doing a matrix inverse here. $\endgroup$ – Christo Jul 30 '18 at 15:15
  • $\begingroup$ I may be misunderstanding you here, but I don't see how you can avoid doing a matrix inverse; imagine that the $\lambda$'s do not fluctuate at all, so $\mathbb{E}(R^{-1})=R^{-1}$; you have some large $n\times n$ matrix $R$ and you want to invert it; how can you avoid doing a matrix inverse? Of course, if the matrix $A$ has some special structure, you might have a sparse $R$ and that would simplify things, but the question as posed does not provide any special structure for $A$. $\endgroup$ – Carlo Beenakker Jul 30 '18 at 15:24
  • $\begingroup$ Thanks again. Could you explain the second answer? (how it is derived?) $\endgroup$ – Christo Aug 1 '18 at 8:13
  • $\begingroup$ it should be the negative powers of $(A\Lambda^{-1}A^H)$ .. rt? $\endgroup$ – Christo Aug 2 '18 at 8:13
  • $\begingroup$ no, certainly not: it is the Taylor series of $(1+x)^{-1}=1+\sum_{p=1}^\infty (-x)^p$, which has positive powers of $x$ (in this case $x$ is the matrix product $A\Lambda^{-1} A^H$). $\endgroup$ – Carlo Beenakker Aug 2 '18 at 8:17

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