Expected value of random vector-constant matrix-random vector product I was reading through this post on the expected dot product of two random vectors. The answers there are quite interesting but not very conclusive. I was wondering if someone knows if the question is any more clear cut with a few restrictions.
Let $C \in \mathbb R^{n \times n}$ be a PSD correlation matrix (with $-1 \leq C_{ij} \leq 1 \ \ \forall \ i,j$ and $C_{ii} = 1 \ \ \forall \ i$) and $\vec x \in \mathbb \{0,1\}^n$ a binary random vector with $k < n$ ones. Is there anything one can say (upper/lower bounds or even an exact expression) about the expected value of the product?
$$
\mathbb E (\vec x^T  C \ \vec x)
$$
 A: Since $C$ is positive semi-definite, it has a Cholesky decomposition $C=LL^T$ with $L=(l_{ij})$. Thus
$$\newcommand{Var}{\operatorname{Var}}
E(x^TCx)=E((x^TL)(x^TL)^T)=\sum_{j=1}^nE(y_j^2)$$
where the $y_j=\sum_{i=1}^nl_{ij}x_i$ are also random variables. Now, denoting $D=\sum_{i=1}^nl_{ij}^2$ and $T=\sum_{1\le i<k\le n}l_{ij}l_{kj}$ so that $D+2T=\left(\sum_{i=1}^nl_{ij}\right)^2$,
$$E(y_j^2)=E(y_j)^2+\Var(y_j)$$
$$=\left(\frac kn\sum_{i=1}^nl_{ij}\right)^2+\sum_{i=1}^nl_{ij}^2\Var(x_i)+2\sum_{1\le i<k\le n}l_{ij}l_{kj}\operatorname{cov}(x_i,x_k)$$
$$=\frac{k^2}{n^2}(D+2T)+\frac{k(n-k)}{n^2}D-\frac kn\cdot\frac{2(n-k)}{n(n-1)}T$$
$$=\frac kn\left(\frac kn(D+2T)+\frac{n-k} nD-\frac{2(n-k)}{n(n-1)}T\right)$$
$$=\frac kn\left(D+2\frac{k-1}{n-1}T\right)$$
Since only $l_{jj}$ and below can be non-zero,
$$E(y_j^2)=\frac kn\left(\sum_{i=j}^nl_{ij}^2+2\frac{k-1}{n-1}\sum_{j\le i<k\le n}l_{ij}l_{kj}\right)$$
$$E(x^TCx)=\frac kn\sum_{j=1}^n\left(\sum_{i=j}^nl_{ij}^2+2\frac{k-1}{n-1}\sum_{j\le i<k\le n}l_{ij}l_{kj}\right)$$
