Suppose we are given a summable sequence $(c_i)_{i\in\mathbb{N}}$ with $\sum_{i=1}^\infty c_i = C<\infty$ and independent $m$-dimensional, standard Gaussian vectors $\alpha_i\sim\mathcal{N}(0,I_m)$, $i\in\mathbb{N}$. Now we construct the (random) square matrix $$A=\frac{1}{C}\sum_{i=1}^\infty c_i \alpha_i\alpha_i^\top \in\mathbb{R}^{m\times m}. $$ Let $(\lambda_k)_{1\leq k\leq m}$ be the (also random) eigenvalues of $A$ and let $A^{1/2}$ denote a matrix root such that $A^{1/2}(A^{1/2})^\top = A$.
What can we say about $\mathbb{E}[\mathrm{tr}A^{1/2}] = \sum_{k=1}^m\mathbb{E}[\sqrt{\lambda_k}]$?
$A$ has a shape similar to a Wishart distribution ($S=\sum_{i=1}^\nu\alpha_i\alpha_i^\top$ with $\alpha_i\sim\mathcal{N}(0,V)$ has $S\sim W_m(\nu,V)$) but also includes these coefficients $c_i$. Nevertheless every term in the sum is Wishart distributed: $c_i\alpha_i\alpha_i^\top\sim\mathcal{W}_m(1,c_iI_m)$.
We can see easily that $\mathbb{E}[A]=I_m$ and since each term has $\mathbb{E}[\mathrm{tr}\alpha_i\alpha_i^\top]=\sum_{j=1}^m\mathbb{E}[\alpha_{i,j}^2]=m$ we also get $\mathbb{E}[\mathrm{tr}A] = m$. But I don't know how to calculate $\mathbb{E}[\mathrm{tr}A^{1/2}]$.
