[Math] Approximating the expectation of a matrix inverse

expectationmatricesmatrix inversepr.probabilityrandom matrices

Let

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

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.

Best Answer

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 $${\mathbb{E}}\bigl((R^{-1})_{ij}\bigr)=\gamma \delta_{ij}-\gamma^2 \sum_{l=1}^m \mu_l A^{\vphantom{\ast}}_{il}A_{jl}^\ast$$ $$+\,\gamma^3\sum_{k,l,l'=1}^m (\mu_l \mu_{l'}+\delta_{ll'}\sigma^2_l)A_{il}^{\vphantom{\ast}}A_{kl}^\ast A_{kl'}^{\vphantom{\ast}}A_{jl'}^\ast+{\cal O}(\gamma^4).$$

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