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Suppose $X_1, \ldots, X_n$ are independent Bernoulli variables with probability $p_1, \ldots,p_n$ and $Y_1,\ldots,Y_n$ is another set of independent Bernoulli variables with probability $q_1,\ldots,q_n$. ($X_1, \ldots, X_n$ and $Y_1,\ldots,Y_n$ are also independent)

Now, let $\bar{X} = \frac{\sum_{i = 1}^n X_i}{n}$ and $\bar{Y} = \frac{\sum_{i = 1}^{n}Y_i}{n}$

What I want to know is how to calculate $Var(\bar{X}\cdot\bar{Y})$

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Not sure about how much we can simplify, just try to group the terms in a systematic way:

$$ \begin{align} Var[\bar{X}\bar{Y}] &= Cov[\bar{X}\bar{Y},\bar{X}\bar{Y}] \\ &= Cov\left[\left(\frac {1} {n} \sum_{i=1}^n X_i\right) \left(\frac {1} {n} \sum_{j=1}^n Y_j\right), \left(\frac {1} {n} \sum_{k=1}^n X_k\right) \left(\frac {1} {n} \sum_{l=1}^n Y_l\right) \right] \\ &= \frac {1} {n^4} \sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n\sum_{l=1}^n Cov[X_iY_j, X_kY_l] \end{align}$$

For $i \neq k, j \neq l$, by the mutual independence, $$Cov[X_iY_j, X_kY_l] = 0$$

For $i = k, j \neq l$, $$ \begin{align} Cov[X_iY_j, X_kY_l] &= Cov[X_iY_j, X_iY_l] \\ &= E[(X_iY_j)(X_iY_l)] - E[X_iY_j]E[X_iY_l] \\ &= E[X_i^2Y_jY_l] - E[X_i]E[Y_j]E[X_i]E[Y_l] \\ &= E[X_i]E[Y_j]E[Y_l] - p_i^2q_jq_l \\ &= p_i(1 - p_i)q_jq_l \end{align} $$

Similarly, for $i \neq k, j = l$

$$ Cov[X_iY_j, X_kY_l] = p_ip_kq_j(1 - q_j) $$

and for $i = k, j = l$

$$ Cov[X_iY_j, X_kY_l] = p_iq_j(1 - p_iq_j) $$

So putting all these together,

$$ \begin{align} Var[\bar{X}\bar{Y}] &= \frac {1} {n^4} \Bigg[ 2\sum_{i=1}^n \sum_{j=1}^{n-1} \sum_{l=j+1}^n p_i(1 - p_i)q_jq_l \\ &~~~~~~~~~~~~~ + 2\sum_{i=1}^{n-1} \sum_{k=i+1}^n \sum_{j=1}^n p_ip_kq_j(1 - q_j) \\ &~~~~~~~~~~~~~ + \sum_{i=1}^n \sum_{j=1}^n p_iq_j(1 - p_iq_j) \Bigg] \end{align}$$

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  • $\begingroup$ It helps, thank you for your help $\endgroup$
    – kevin
    Nov 4, 2020 at 17:48
  • $\begingroup$ Corrected error in the last term $\endgroup$
    – BGM
    Nov 5, 2020 at 15:56

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