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Can someone help me understand the following for the variance of a sample proportion.

$$ S^2 = \dfrac{\sum (y_i - p)^2}{N-1} = \dfrac{\sum y_i^2 - 2p\sum yi + Np^2 }{N-1} = \dfrac{N}{N-1}p(1-p) $$

More specifically how do you get to the last step from the second last step.

Thanks!

Will
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    Is $p=\frac1N \sum y_i$ with each $y_i$ zero or one? – Henry Oct 08 '22 at 23:55
  • sorry I wrote it down wrong its fixed now – Will Oct 08 '22 at 23:58
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    If each $y_i$ is zero or one then $y_i^2 = y_i$ so you can show the right hand equality using $\sum y_i^2=\sum y_i = Np$. – Henry Oct 09 '22 at 00:02
  • A more interesting question is the left hand equality, which suggests when $p=\frac12$ a number higher than $\frac14$ (the maximum possible variance of a Bernoulli random variable). This is an issue about the inadmissibility of some unbiased estimators – Henry Oct 09 '22 at 00:03
  • See https://stats.stackexchange.com/questions/52542 – whuber Oct 09 '22 at 13:22
  • https://stats.stackexchange.com/questions/464110 has two different proofs that avoid this specific algebraic manipulation. – whuber Oct 11 '22 at 14:15

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