# Conditional expectation of product of sums of bernoulli random variables

let's say we have $$X_1,..,X_n$$ i.i.d. Bernoulli random variables. For $$l, we want to calculate: $$E \left[ \sum_{i=1}^{l}X_i \sum_{j=1}^{m}X_j \mid \sum_{k=1}^{n}X_k \right]$$ Does this property apply in this situation? $$E \left[ X Y \mid Z \right] = E \left[X\mid Z \right] E\left[Y\mid Z \right]$$

If $$X_1,\ldots,X_n$$ are i.i.d $$\mathsf{Bernoulli}(p)$$ and $$S_n=\sum\limits_{i=1}^n X_i$$, the conditional expectation $$E\left[S_lS_m\mid S_n\right]$$ is just the unbiased estimator of $$E\left[S_lS_m\right]$$ based on $$S_n$$ by Lehmann-Scheffé theorem. In other words, it is the UMVUE of $$E\left[S_lS_m\right]$$.

Keeping in mind that $$l,

\begin{align} E\left[S_lS_m\right]&=E\left[\sum_{i=1}^l\sum_{j=1}^m X_iX_j\right] \\&=\mathop{\sum_{i=1}^l\sum_{j=1}^m}_{i\ne j} E\left[X_iX_j\right]+\sum_{i=1}^l E\left[X_i^2\right] \\&=l(m-1)E[X_i]E[X_j]+\sum_{i=1}^l \left(\operatorname{Var}(X_i)+(E[X_i])^2\right) \\&=l(m-1)p^2+l(p(1-p)+p^2)=\cdots \end{align}

Using $$S_n\sim \mathsf{Bin}(n,p)$$, we already have

$$E\left[\frac{S_n}{n}\right]=p$$

and $$E\left[\frac{S_n(S_n-1)}{n(n-1)}\right]=p^2$$

No it wouldnt apply. The statement E[XY]=E[X]E[Y] is only valid when X and Y are independent, which is not the case in your setting since the summations run over the same X_i variable, making the sums dependent.

The answer here may be useful: https://math.stackexchange.com/questions/1476906/expectation-of-the-product-of-two-dependent-binomial-random-variable