Variance of the Mean for SRS Derivation (full details)

PUBHBIO 7225

Assume an SRS of size \(n\) from population of size \(N\), with \(\bar{y}= \frac{1}{n} \sum_{i \in \mathcal{S}} y_i\) = sample mean
\(Z_i\) is the selection/inclusion indicator (=1 if unit \(i\) is in the sample, =0 if not)
As shown in Lecture 4.1,
\(E(Z_i) = \pi_i = \frac{n}{N}\)
\(V(Z_i) = \pi_i(1-\pi_i)\)
\(\text{Cov}(Z_i,Z_j) = -\frac{n}{N} \left(\frac{1}{N-1}\right) \left(1-\frac{n}{N}\right)\)

\[\begin{aligned} V(\bar{y}) &= V \left( \sum_{i=1}^N Z_i \frac{y_i}{n} \right) \\ &= \frac{1}{n^2} V \left( \sum_{i=1}^N Z_i y_i \right)\\ &=\frac{1}{n^2} \text{Cov}\left(\sum_{i=1}^N Z_i y_i, \sum_{j=1}^N Z_j y_j \right) \\ &=\frac{1}{n^2} \sum_{i=1}^N \sum_{j=1}^N y_i y_j \text{Cov}(Z_i,Z_j)\\ &= \frac{1}{n^2} \left[\sum_{i=1}^N y_i^2 V(Z_i) + \sum_{i=1}^N \sum_{j \ne i}^N y_i y_j \text{Cov}(Z_i,Z_j) \right] \\ &= \frac{1}{n^2} \left[\frac{n}{N} \left(1-\frac{n}{N}\right)\sum_{i=1}^N y_i^2 + \left[-\frac{n}{N} \left(\frac{1}{N-1}\right) \left(1-\frac{n}{N}\right)\right] \sum_{i=1}^N \sum_{j \ne i}^N y_i y_j \right] \\ &= \frac{1}{n^2} \frac{n}{N} \left(1-\frac{n}{N}\right)\left[ \sum_{i=1}^N y_i^2 - \frac{1}{N-1} \sum_{i=1}^N \sum_{j \ne i}^N y_i y_j \right] \\ &= \frac{1}{n}\frac{1}{N} \left(1-\frac{n}{N}\right)\frac{1}{N-1} \left[ (N-1) \sum_{i=1}^N y_i^2 - \sum_{i=1}^N \sum_{j \ne i}^N y_i y_j \right]\\ &= \frac{1}{n}\frac{1}{N} \left(1-\frac{n}{N}\right)\frac{1}{N-1} \left[ (N-1) \sum_{i=1}^N y_i^2 - \left[ \left(\sum_{i=1}^N y_i\right)^2 - \sum_{i=1}^N y_i^2 \right]\right] \\ &= \frac{1}{n}\frac{1}{N} \left(1-\frac{n}{N}\right)\frac{1}{N-1} \left[N\sum_{i=1}^N y_i^2 - \left(\sum_{i=1}^N y_i\right)^2 \right] \\ &= \frac{1}{n}\frac{1}{N} \left(1-\frac{n}{N}\right)\frac{1}{N-1} \left[N\sum_{i=1}^N y_i^2 - \left(N \bar{y}_U \right)^2 \right] \\ &= \frac{1}{n}\left(1-\frac{n}{N}\right)\frac{1}{N-1} \left[\sum_{i=1}^N y_i^2 - N \bar{y}_U^2 \right] \\ &= \frac{1}{n}\left(1-\frac{n}{N}\right)S^2 \\ &= \left(1-\frac{n}{N}\right)\frac{S^2}{n} \\ \end{aligned}\]