A binomial symmetric sum

Question

Denote \begin{align*} \text{Sym}_k(\textbf{x})=\sum_{i_1<\cdots<i_k}x_{i_1}\cdots x_{i_k} \end{align*} as the $k$th elementary symmetric sum in monomials $\textbf{x} = (x_1, \cdots, x_n)$. If we denote $Y_i \sim \text{Bernoulli}(\pi)$ and $E_i = (Y_i - \pi)/[\pi(1-\pi)]^{1/2}$ (i.e. we standardize $Y_i$), I am hypothesizing that \begin{align*} \mathbb{E}(\text{Sym}_2(\textbf{E}))=\sum_{\mathbb{y}\in\{0, 1\}^n}\pi^{\sum_{i=1}^{n}y_i}(1-\pi)^{n - \sum_{i=1}^{n}y_i}\text{Sym}_2(\textbf{e})= 0 \end{align*} (where $e_i = (y_i - \pi)/[\pi(1-\pi)]^{1/2}$) because the above is related to the expected value of some score equation, which is 0. However, I would like to calculate this directly to give me a better idea of how to tackle more general problems of this type. I'm a bit stuck in the algebra, so any tricks would be much appreciated!

Answer 1

Man, am I embarrassed. I got caught up in the summation form that I totally ignored some basic properties of expectation. Here's my solution: \begin{align*} \text{E}(\text{Sym}_2(\textbf{E})) &= \frac{n(n-1)}{2}\text{E}(E_i E_j) && E_i's \text{ are identically distributed} \\ &= \frac{n(n-1)}{2}\text{E}(E_i) \text{E}(E_j) && E_i's \text{ are independent} \\ &= 0 && E_i' \text{ have mean 0} \end{align*} By this reasoning, $\text{E}(\text{Sym}_k(\textbf{E}))$ are in fact all zero for $k \ge 1$. This is certainly a good start for other problems I am solving, which involves rational functions of these symmetric sums.