# Expectation of log determinant of "weighted" Wishart matrix

Let $$g_1,\dots,g_d$$ be $$n$$ dimensional Gaussian iid vectors drawn from $$N(0,I_n),$$ and $$p_1,\dots, p_d \in [0,1]$$ be a discrete probability vector $$\sum_i^d p_i=1,$$ and define weighted Wishart matrix $$W = \sum_i^d p_i g_i g_i^\top.$$ I am wondering if we can compute the log-determinant of this matrix? $$E \log\det(W) = ? \quad W = \sum_i^d p_i g_i g_i^\top, \qquad g_1,\dots,g_d\sim N(0,I_n)$$ Note that the probability vector is uniform $$p_1=\dots = p_d = 1/d$$ this falls back to the classic Wishart matrix, which is why I called this particular matrix "weighted Wishart" to emphasis its non-uniformity. and we can rely on known results, such those here or here, which would roughly be $$-n/2d$$ when $$n$$ is sufficiently smaller than $$d$$.

It seems intuitively clear to me that when $$p_i$$'s deviate from uniform distribution, the effective $$d$$ in this expectation will be lower, because we can view non-uniform distribution as seeing some samples many more times than others, and effectively reducing $$d$$. But I am not sure how to formalise this notion, and come to a more accurate, or even approximately true formula that will hold for general $$p_i$$'s.