Suppose $M$ is a $n\times n$ PSD matrix and $x$ is a $n$ dimensional vector whose elements are drawn iid from $N(0,1/n)$. Assuming that we know the distribution of $x^\top M x,$ including its moments, what can be said of the Empirical Spectral Distribution (ESD) of $M$? I can draw basic conclusions, namely the first and second moment of ESD: Suppose eigen-decomposition of $M$ is given by $M = \sum_{k=1}^n \lambda_k u_k u_k^\top.$ We can write \begin{align} [x^\top M x] &= \sum_i^n \lambda_i x^\top u_i u_i^\top x \\ &= \sum_i^n \lambda_i \langle x,u_i\rangle^2\\ &= \sum_i^n\lambda_i w_i^2, && w_i:=\langle x,u_i\rangle \sim N(0,1/n) \end{align} Thus we have \begin{align} E [x^\top M x] &= \sum_i^n \lambda_iE w_k^2 = \frac1n\sum_i^n \lambda_i\\ E [x^\top M x]^2 &= \sum_i^n \lambda_i^2 E w_i^4 + \sum_{i\neq j} \lambda_i \lambda_jE w_i^2w_j^2\\ &= \frac{3}{n^2} \sum_i^n \lambda_i^2 + \frac{1}{n^2}\sum_{i\neq j} \lambda_i\lambda_j \\ &= \frac{2}{n^2} \sum_i^n \lambda_i^2 + (\frac1n\sum_i^n\lambda_i)^2 \end{align} Thus we have the first and second moments of the ESD can be computed as: $$ \frac1n\sum_i^n \lambda_i =E [x^\top M x], \quad \frac1n\sum_i^n\lambda_i^2 = \frac{n}{2}(E [x^\top M x]^2 - (E [x^\top M x])^2) $$
Questions: are the calculations given above accurate? What more can be said about the ESD of $M$ For example, if we assume that $x^\top M x$ follows a log-normal distribution $\log x^\top M x \sim N(\mu,\sigma^2)$, what can be concluded about ESD of $M$?