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.



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