Expectation of a combination of random variables. Assume that $\xi_i, i=1, \dots, n$ are independent identically distributed discrete random variables. What is an expectation of $$\frac{\sum_{i=1}^{n} a_i \xi_i^2}{\left( \sum_{i=1}^n \xi_i \right)^2}$$ Note: It is possible to show that $E\frac{\sum_i a_i \xi_i}{\sum_i \xi_i} = \frac{\sum_i a_i}{n}$.

The expectation of $\eta_n^{-1}\sum\limits_{i=1}^na_i\xi_i$ is $\theta_1\sum\limits_{i=1}^na_i$, where $\theta_1=E[\eta_n^{-1}\xi_1]$ and $\eta_n=\sum\limits_{i=1}^n\xi_i$. Note that, by symmetry, $n\theta_1=E[\eta_n^{-1}\xi_1]+E[\eta_n^{-1}\xi_2]+\cdots+E[\eta_n^{-1}\xi_n]=1$ hence $\theta_1=n^{-1}$, irrespectively of the distribution of $\xi_1$. This yields the result recalled in the question.

Likewise, the answer here is $\theta_2\sum\limits_{i=1}^na_i$, where $\theta_2=E[\eta_n^{-2}\xi_1^2]$. Note that $n^2\theta_2\geqslant1$ by Cauchy-Schwarz inequality and $n\theta_2\lt1$ for every $n\geqslant2$ since $n\theta_2+n(n-1)\varphi=1$, where $\varphi=E[\eta_n^{-2}\xi_1\xi_2]$. However, there is no reason to expect that $\theta_2$ is independent of the distribution of $\xi_1$, nor that it has a simple form in general.