0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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 $\sum_{i\in\mathcal{S}}(y_i-\bar{y})=0$. $\endgroup$ Commented 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
    Commented 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
    Commented 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
    Commented Aug 24, 2022 at 14:56

1 Answer 1

3
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.