Index: The Book of Statistical Proofs ▷ General Theorems ▷ Estimation theory ▷ Point estimates ▷ Finite-population correction for the variance of a sample mean

Theorem: Let $y_1,\ldots,y_N$ be fixed real values for $N \geq 2$ labeled population members. The labels distinguish members, even if some values are equal. Define

\[\label{eq:srs-meanvar-population} \bar{y}_N = \frac{1}{N} \sum_{j=1}^N y_j, \qquad S_N^2 = \frac{1}{N-1} \sum_{j=1}^N (y_j-\bar{y}_N)^2 \; .\]

Take a simple random sample without replacement of size $1 \leq n \leq N$, i.e. $s \subset \lbrace 1,\ldots,N \rbrace$ with $\lvert s \rvert = n$. Let $\bar{Y}_s$ be the sample mean of the $n$ data points included in this sample. Then,

\[\label{eq:srs-meanvar-result} \mathrm{E}(\bar{Y}_s) = \bar{y}_N,\qquad \mathrm{Var}(\bar{Y}_s) = \left(1-\frac{n}{N}\right)\frac{S_N^2}{n} \; .\]

Proof: Generate a uniformly random permutation $(J_1,\ldots,J_N)$ of all labels and define $Y_i=y_{J_i}$. Any specified subset of $n$ labels occupies the first $n$ positions in $n!(N-n)!$ of the $N!$ permutations, so it has probability $1/\binom{N}{n}$. Thus, the first $n$ labels form a simple random sample, and $\bar{Y}_s = 1/n \sum_{i=1}^n Y_i$.

Each $J_i$ is uniform on $\lbrace 1,\ldots,N \rbrace$. Consequently,

\[\label{eq:srs-meanvar-one-draw} \mathrm{E}(Y_i) = \bar{y}_N, \qquad \mathrm{Var}(Y_i) = \frac{1}{N} \sum_{j=1}^N (y_j-\bar{y}_N)^2 =\frac{N-1}{N}S_N^2 \; .\]

For any two different positions $i$ and $\ell$ in the permutation, the ordered pair $(J_i,J_{\ell})$ is uniform over the $N(N-1)$ pairs of different labels. Put $z_j=y_j-\bar{y}_N$, so $\sum_j z_j=0$. Then,

\[\label{eq:srs-meanvar-covariance} \begin{split} \mathrm{Cov}(Y_i,Y_\ell) &= \frac{1}{N(N-1)}\sum_{j\neq k}z_jz_k\\ &= \frac{(\sum_jz_j)^2-\sum_jz_j^2}{N(N-1)}\\ &=-\frac{(N-1)S_N^2}{N(N-1)} =-\frac{S_N^2}{N} \; . \end{split}\]

This negative covariance is the effect of sampling without replacement. Using the linearity of the expected value gives $\mathrm{E}(\bar{Y}_s)=\bar{y}_N$. Using the variance of a sum gives:

\[\label{eq:srs-meanvar-expand} \begin{split} \mathrm{Var}(\bar{Y}_s) &= \frac{1}{n^2} \left[\sum_{i=1}^n \mathrm{Var}(Y_i) + \sum_{i\neq\ell} \mathrm{Cov}(Y_i,Y_\ell) \right] \\ &= \frac{1}{n^2} \left[n \frac{N-1}{N} S_N^2 - n(n-1) \frac{S_N^2}{N} \right] \\ &= \frac{N-n}{Nn} S_N^2 = \left(1-\frac{n}{N}\right) \frac{S_N^2}{n} \; . \end{split}\]

The off-diagonal sum has $n(n-1)$ ordered terms. Counting unordered pairs instead would require multiplying the covariance sum by two. This proves \eqref{eq:srs-meanvar-result}.

For a census with $n=N$, the formula gives zero variance, because the observed mean is always the fixed population mean. For $n=1$, it gives $(N-1)S_N^2/N$, the variance of one uniform draw. If all population values are equal, $S_N^2=0$ and every sample mean is constant.

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Sources:
  • John A. Rice (2007): "Simple random sampling"; in: Mathematical Statistics and Data Analysis, Third edition, Section 7.3.1, pp. 203-208.

Metadata: ID: P563 | shortcut: meansamp-varcorr | author: kjpetrykowski | date: 2026-10-19, 22:00.