Solution
= Solution
Expanding $\sum(X_i-\bar X)(Y_i-\bar Y)=\sum X_iY_i-n\bar X\bar Y$ and taking expectations gives $(n-1)(E[XY]-EX,EY)$. Division by $n-1$ proves unbiasedness.
= Solution
Expanding $\sum(X_i-\bar X)(Y_i-\bar Y)=\sum X_iY_i-n\bar X\bar Y$ and taking expectations gives $(n-1)(E[XY]-EX,EY)$. Division by $n-1$ proves unbiasedness.