Suppose we have a a set of random matrices in the complex field of the form $a_iv_iv_i^H$ for $i=\{1,\dots,n\}$ where $a_i$ are constant positive real scalars and $v_i$ are **random complex valued matrices** with all elements i.i.d. circular symmetric Gaussian for all $i$. All the $n$ matrices $v_i$ are of same dimension. And $I$ is the identity matrix and $v_i^H$ is the conjugate transpose of $v_i$. Thus an element of the matrices $v_iv_i^H$ for all $i$, is the sum of two squared i.i.d normal random variables which makes it exponential, and $v_iv_i^H$ are symmetric.

Now we want to maximize the following determinant over $a_iv_iv_i^H$ for $i=\{1,\dots,n\}$ $$ \mathbb{E}\det \left( I+\frac{a_iv_iv_i^H}{I+\sum_{j\neq i} a_jv_jv_j^H} \right).$$

Here $\mathbb E$ is the expectation. Essentially we pick one matrix for the numerator and all the rest go in the denominator. Since all matrices are i.i.d, can I claim that the matrix which should go on the numerator is the one with the highest $a_i$ ?

Edit: With regards to first comments, it seems **asymptotic** analysis is the appropriate for such a question. So we consider the solution when the **dimensions of all $v_i$ grow**.

I ran a simulation with the $a_i=i$, where $i\in \{1,\dots,10\}$ and matrices of $20 \times 20$. With randomly generated normals with mean zero and variance 1. The results seems to confirm the claim. The function is maximum when the matrix with $a_{10}=10$ goes on the numerator.

P.S.: Random version of my previous question A determinant problem with symmetric PSD matrices

May be related to Expected determinant of a random NxN matrix.