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.


  • 1
    $\begingroup$ Is $p=\frac1N \sum y_i$ with each $y_i$ zero or one? $\endgroup$
    – Henry
    Oct 8, 2022 at 23:55
  • $\begingroup$ sorry I wrote it down wrong its fixed now $\endgroup$
    – Will
    Oct 8, 2022 at 23:58
  • 1
    $\begingroup$ If each $y_i$ is zero or one then $y_i^2 = y_i$ so you can show the right hand equality using $\sum y_i^2=\sum y_i = Np$. $\endgroup$
    – Henry
    Oct 9, 2022 at 0:02
  • $\begingroup$ 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 $\endgroup$
    – Henry
    Oct 9, 2022 at 0:03
  • $\begingroup$ See stats.stackexchange.com/questions/52542 $\endgroup$
    – whuber
    Oct 9, 2022 at 13:22


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