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Going through Sharon L. Lohr's Sampling design book (2nd Edition), I have no issues with the content all the way until it goes into the proof in chapter 2 on SRSWOR that $E[s^2] = S^2$, where $S^2$ is defined as: $$ S^2 = \frac{1}{N-1}\sum_{i=1}^{N} (y_i - \bar{y}_{U})^2 $$ And the sample variance estimator is: $$ s^2 = \frac{1}{n-1}\sum_{i\in\mathcal{S}}(y_i-\bar{y})^2 $$ where $U$ is the index set of the finite population: $$ U = \{1,2,\dotsc,N\} $$ And $\mathcal{S}$ is the particular sample chosen, a subset consisting of $n$ of the units in $U$.

It says:

and then find the multiplicative constant that will give the unbiasedness: $$ \begin{align} E\left[\sum_{i\in\mathcal{S}}(y_i-\bar{y})^2\right] & = E\left[\sum_{i\in\mathcal{S}}((y_i-\bar{y}_U) - (\bar{y}-\bar{y}_U))^2\right]\\ & = E\left[\sum_{i\in\mathcal{S}}(y_i-\bar{y}_U)^2 - n(\bar{y}-\bar{y}_U)^2\right]\\ & = E\left[\sum_{i=1}^NZ_i(y_i-\bar{y}_U)^2\right] - n\textrm{Var}(\bar{y})\\ & = \frac{n}{N}\sum_{i=1}^N(y_i-\bar{y}_U)^2-\left(1-\frac{n}{N}\right)S^2\\ & = \frac{n(N-1)}{N}S^2 - \frac{N-n}{N}S^2\\ & = (n-1)S^2 \end{align} $$ Thus, $$ E\left[\frac{1}{n-1}\sum_{i\in\mathcal{S}}(y_i-\bar{y})^2\right] = E[s^2] = S^2 $$

I have no issues with most of the derivation, except for how the first line turns into the second line. I assume there must be the intermediary step: $$ \begin{align} E\left[\sum_{i\in\mathcal{S}}(y_i-\bar{y})^2\right] & = E\left[\sum_{i\in\mathcal{S}}((y_i-\bar{y}_U) - (\bar{y}-\bar{y}_U))^2\right]\\ & = E\left[\sum_{i\in\mathcal{S}}(y_i-\bar{y}_U)^2 - \sum_{i\in\mathcal{S}}(\bar{y}-\bar{y}_U)^2\right]\\ & = E\left[\sum_{i\in\mathcal{S}}(y_i-\bar{y}_U)^2 - n(\bar{y}-\bar{y}_U)^2\right]\\ \end{align} $$ But I still can't get from the first line to the second. My guesses are that either I am getting confused about what the terms mean, the notation, the summation, or I have made an obvious mistake.

Either way, any help would be greatly appreciated.

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  • $\begingroup$ FOIL (sry for my american lingo, lmk if youre not familiar with this acronym) the terms in the parens $((y_i-\bar{y}_U))-(\bar{y}_U-\bar{y}))$ and note that $\sum_{i\in\mathcal{S}}(y_i-\bar{y})=0$. $\endgroup$ Aug 23, 2022 at 16:33
  • $\begingroup$ Does this answer your question? How exactly did statisticians agree to using (n-1) as the unbiased estimator for population variance without simulation? $\endgroup$
    – Xi'an
    Aug 23, 2022 at 18:16
  • $\begingroup$ I don't think it does? My issue was simply getting from the first line of this specific proof to the second, which was more of an algebra issue on my part. $\endgroup$
    – philiptomk
    Aug 24, 2022 at 2:10
  • $\begingroup$ This is an algebra FAQ (making it difficult to search for). It has been asked here on CV several dozen times and always has the same answer: expand the sum and simplify using the fact that the sum of the residuals is zero. $\endgroup$
    – whuber
    Aug 24, 2022 at 14:56

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Following the intermediary step:

\begin{align} \mathbb E\left[\sum_{i\in\mathcal{S}}(y_i-\bar{y})^2\right] & = \mathbb E\left[\sum_{i\in\mathcal{S}}((y_i-\bar{y}_U) - (\bar{y}-\bar{y}_U))^2\right]\\ &= \mathbb E\left[\sum_{i\in\mathcal{S}}(y_i-\bar{y}_U)^2 +n(\bar{y}-\bar{y}_U)^2-2(\bar{y}-\bar{y}_U)\underbrace{\sum_{i\in\mathcal{S}}(y_i-\bar{y}_U)}_{= n(\bar{y}-\bar{y}_U)}\right] . \end{align}

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