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Flip an unweighted coin $N$ times, where we denote $X_i = 1$ if flip $i$ resulted in heads, and $X_i = 0$ if flip $i$ resulted in tails. Now take the average result $\bar X = \frac{\sum_{i=1}^N X_i}{N}$.

Conduct the same experiment again, this time denoting the $N$ flips $Y_i$. What is the expected value of the square of the difference between the two averages? In other words, $E \left[ (\bar X - \bar Y)^2 \right] = E \left[ \left( \frac{\sum_{i=1}^N X_i - \sum_{i=1}^N Y_i}{N} \right)^2 \right] = ?$

I'm interested in all of the steps leading up to the answer, which I assume should end up being some function of $N$...

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    @ThomasA$n$drews Thanks--I clarified it (I want square of differences).2012-04-23

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first note that $E(\bar X) = E(\bar Y)$ as $\bar X$ and $\bar Y$ have the same distribution, also $E(\bar X^2) = E(\bar Y^2)$. As $\bar X$ and $\bar Y$ are independent, we have $E(\bar X \bar Y) = E(\bar X)E(\bar Y)$. So we have \begin{align*} E\bigl((\bar X - \bar Y)^2\bigr) &= E(\bar X^2 - 2\bar X \bar Y + \bar Y^2)\\\ &= E(\bar X^2) - 2E(\bar X)E(\bar Y) + E(\bar Y^2)\\\ &= 2\bigl(E(\bar X^2) - E(\bar X)^2\bigr)\\\ &= 2V(\bar X). \end{align*} Now as the $X_i$ are independent $V(\bar X) = \frac 1{N^2}\sum_{i=1}^N V(X_i)$. We have $E(X_i^2) = E(X_i) = \frac 12$, so $V(X_i) = \frac 12 - \frac 14 = \frac 14$ and so $V(\bar X) = N \cdot \frac 1{4N^2} = \frac 1{4N}$ and by the calculation above $E[(\bar X - \bar Y)^2] = \frac 1{2N}$.

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    Hallelujah! I was banging my head against the wall, rereading your answer, and I didn't catch it either. Thanks for your help!2012-04-23