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let's say we have $X_1,..,X_n$ i.i.d. Bernoulli random variables. For $l<m<n$, 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? $$E \left[ X Y \mid Z \right] = E \left[X\mid Z \right] E\left[Y\mid Z \right]$$

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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 $l<m$,

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

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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

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