Consider a random matrix $\mathbf{X} \in \mathbb{C}^{T\times M}$ with independent columns $\mathbf{x}_1,\dots,\mathbf{x}_M$ distributed on the unit sphere $\mathcal{S} = \{\mathbf{x} \in \mathbb{c}^T: \|\mathbf{x}\| = 1\}$. That is, $\|\mathbf{x}_i\| = 1$, for $i = 1,\dots,M$. The joint probability density function of the columns is given by
\begin{align} f(\mathbf{X}) &= \frac{1}{Q} \frac{\exp(tr\{\mathbf{Y}^H(\sigma^2\mathbf{I}_T + \mathbf{X}\mathbf{X}^H)^{-1}\mathbf{Y}\}+tr\{\mathbf{Z}^H\mathbf{X}\})}{\det(\sigma^2I_T + \mathbf{X}\mathbf{X}^H)}, \end{align} where $\mathbf{Y} \in \mathbb{C}^{T\times N}$ and $\mathbf{Z} = [\mathbf{z}_1 \dots \mathbf{z}_M] \in \mathbb{C}^{T\times M}$ are deterministic matrices, $\sigma^2 < 1$, $$Q = \int_{\mathcal{S}^M}\frac{\exp(tr\{\mathbf{Y}^H(\sigma^2\mathbf{I}_T + \mathbf{X}\mathbf{X}^H)^{-1}\mathbf{Y}\}+tr\{\mathbf{Z}^H\mathbf{X}\})}{\det(\sigma^2I_T + \mathbf{X}\mathbf{X}^H)}d\mathbf{X}$$ is the normalization factor. In other words, \begin{align} f(\mathbf{X}) = \frac{1}{Q}\frac{\exp(tr\{\mathbf{Y}^H(\sigma^2\mathbf{I}_T + \mathbf{X}\mathbf{X}^H)^{-1}\mathbf{Y}\})}{\det(\sigma^2I_T + \mathbf{X}\mathbf{X}^H)}\prod_{i = 1}^{M} \exp(\mathbf{z}_i^H\mathbf{x}_i). \end{align}
I want to derive the marginal expectation $\mathbb{E}[\mathbf{x}_i] = \int_{\mathcal{S}^M} f(\mathbf{X})\mathbf{x}_i d\mathbf{X}$, $i = 1,\dots,M$.
This can be solved by factorizing $f(\mathbf{X})$ as a product of functions of each column of $\mathbf{X}$. However, the term $\frac{\exp(tr\{\mathbf{Y}^H(\sigma^2\mathbf{I}_T + \mathbf{X}\mathbf{X}^H)^{-1}\mathbf{Y}\})}{\det(\sigma^2I_T + \mathbf{X}\mathbf{X}^H)}$ makes it difficult and I understand that it is not hopeful for an exact expession of $\mathbb{E}[\mathbf{x}_i]$. So, I am looking for a reasonable approximation of $f(\mathbf{X})$ which can be marginalized more easily.
A straightforward approximation is to impose that the columns of $\mathbf{X}$ are orthogonal, then $\mathbf{X}^H\mathbf{X} = \mathbf{I}_M$ and we get rid of $\frac{\exp(tr\{\mathbf{Y}^H(\sigma^2\mathbf{I}_T + \mathbf{X}\mathbf{X}^H)^{-1}\mathbf{Y}\})}{\det(\sigma^2I_T + \mathbf{X}\mathbf{X}^H)}$. However, this approximation is very loose.
Do you have any idea for a sharper marginalizable approximation?
Thank you.