Hypergeometric Variance and the Finite-Population Correction

Why Sampling Without Replacement Changes Variance

When a sample is drawn without replacement from a finite population, each draw affects which objects remain available. Consequently, success indicators for different draws are not independent. In fact, their covariances are negative: finding a success on one draw leaves slightly fewer successes for later draws. This dependence reduces the variability of the total number of successes compared with sampling with replacement.

The variance formula makes that reduction precise. Its derivation also explains the finite-population correction, the factor used to adjust a with-replacement variance when the sample is taken from a finite population. The result is aligned with the SOA VEE Mathematical Statistics syllabus.

Model and notation: A population contains \(N\) distinct objects, of which \(K\) are classified as successes. A simple random sample of size \(n\) is drawn without replacement, where \(N>1\) and \(0\leq n\leq N\). Let \(X\) be the number of successes in the sample, and set \(p=K/N\). Then \(X\) has a hypergeometric distribution. The same \(N\), \(K\), and \(n\) will be used throughout the derivation.

For each draw position \(j\), define an indicator \(I_j\) that equals 1 if the object in position \(j\) is a success and 0 otherwise. The total count is the sum of those indicators. Each draw position is equally likely to contain any one of the \(N\) objects, so \(P(I_j=1)=K/N=p\). Thus each \(I_j\) has a Bernoulli distribution with variance \(p(1-p)\).

X=\sum_{j=1}^{n} I_j, \qquad E[I_j]=p, \qquad \operatorname{Var}(I_j)=p(1-p).

Covariance Between Two Draw Indicators

To find the variance of a sum, the individual variances are not enough: the covariances between distinct terms must also be included. For two different draw positions, \(j\ne k\), the probability that both contain successes is the probability of selecting two successes in order without replacement. There are \(K\) successes available for the first position and \(K-1\) for the second, out of \(N\) and then \(N-1\) possible objects.

P(I_j=1,I_k=1) = \frac{K}{N}\frac{K-1}{N-1} = \frac{K(K-1)}{N(N-1)}.

Covariance is the joint success probability minus the product of the marginal success probabilities. Substituting \(p=K/N\) and simplifying gives a negative value whenever \(0<K<N\). This sign expresses the dependence created by removing each selected object from the population.

\begin{aligned} \operatorname{Cov}(I_j,I_k) &=E[I_jI_k]-E[I_j]E[I_k]\\ &=\frac{K(K-1)}{N(N-1)}-\left(\frac{K}{N}\right)^2\\ &=\frac{K(K-1)N-K^2(N-1)}{N^2(N-1)}\\ &=\frac{K^2N-KN-K^2N+K^2}{N^2(N-1)}\\ &=-\frac{K(N-K)}{N^2(N-1)}\\ &=-\frac{p(1-p)}{N-1}. \end{aligned}

This expression applies to every pair of distinct draw positions. If \(K=0\) or \(K=N\), then \(p(1-p)=0\), so all these variances and covariances are zero, as they should be: the count is then fixed.

Deriving the Hypergeometric Variance

For any finite collection of random variables, the variance of their sum equals the sum of their variances plus twice the sum of the covariances over unordered pairs. Equivalently, one can add the covariance for every ordered pair of distinct indices. There are \(n\) individual variance terms and \(n(n-1)\) ordered pairs.

\begin{aligned} \operatorname{Var}(X) &=\operatorname{Var}\left(\sum_{j=1}^{n} I_j\right)\\ &=\sum_{j=1}^{n}\operatorname{Var}(I_j) +\sum_{\substack{j,k=1\\j\ne k}}^{n}\operatorname{Cov}(I_j,I_k)\\ &=np(1-p)+n(n-1)\left(-\frac{p(1-p)}{N-1}\right)\\ &=np(1-p)\left(1-\frac{n-1}{N-1}\right)\\ &=np(1-p)\frac{N-n}{N-1}. \end{aligned}

Theorem: For a hypergeometric count \(X\) from a population of size \(N\) with \(K\) successes, sampled without replacement in a sample of size \(n\), \(\operatorname{Var}(X)=np(1-p)(N-n)/(N-1)\), where \(p=K/N\). Equivalently, \(\operatorname{Var}(X)=n(K/N)(1-K/N)(N-n)/(N-1)\). This formula assumes \(N>1\); the cases \(n=0\), \(n=N\), \(K=0\), and \(K=N\) give variance zero when applicable.

The expression \(np(1-p)\) is the variance of a binomial count with \(n\) independent trials and success probability \(p\). It is also the variance obtained by drawing with replacement from this population, since each draw then has success probability \(p\), independently of the others. The finite-population correction for the variance is therefore \((N-n)/(N-1)\).

Key distinction: The variance is multiplied by \((N-n)/(N-1)\). The standard deviation is multiplied by the square root of that factor: \(\sqrt{(N-n)/(N-1)}\). Do not apply the square-root factor to a variance.

For the sample proportion \(\hat p=X/n\), when \(n>0\), variance scales by the square of the constant \(1/n\). Hence its with-replacement variance \(p(1-p)/n\) is reduced by the same variance correction.

\operatorname{Var}(\hat p) = \operatorname{Var}\left(\frac{X}{n}\right) = \frac{p(1-p)}{n}\frac{N-n}{N-1}.

Worked Examples

Worked Example: Deriving a Count Variance From Draw Indicators

A hypothetical inspection sample contains \(n=5\) items drawn uniformly without replacement from a batch of \(N=20\) items. Exactly \(K=8\) items in the batch meet a specified quality standard. Find the variance of the number \(X\) meeting the standard, showing the contribution from the indicator covariances.

Method. This is a hypergeometric count: \(N=20\), \(K=8\), and \(n=5\). The success proportion is \(p=K/N=8/20=2/5=0.4\), so \(1-p=3/5=0.6\).

For each draw indicator, the variance is \(\operatorname{Var}(I_j)=p(1-p)=(2/5)(3/5)=6/25=0.24\). The covariance between any two distinct indicators is \[ \operatorname{Cov}(I_j,I_k)=-\frac{p(1-p)}{N-1} =-\frac{6/25}{19}=-\frac{6}{475}\approx-0.0126316. \] There are five individual variance terms and \(5(5-1)=20\) ordered covariance terms. Therefore, \[ \begin{aligned} \operatorname{Var}(X) &=5\left(\frac{6}{25}\right)+20\left(-\frac{6}{475}\right)\\ &=\frac{6}{5}-\frac{24}{95}\\ &=\frac{114}{95}-\frac{24}{95}\\ &=\frac{90}{95}=\frac{18}{19}\approx0.9474. \end{aligned} \] As a check, the compact formula gives \(np(1-p)(N-n)/(N-1)=5(6/25)(15/19)=18/19\), the same result.

Common mistake avoided: Treating the five draws as independent would give variance \(5(0.24)=1.2\). That omits the 20 negative covariance terms and overstates the variance.

Worked Example: Comparing With- and Without-Replacement Variance

A hypothetical archive has \(N=50\) files, of which \(K=15\) are flagged for review. A random sample of \(n=10\) files is drawn without replacement. Find the variance of the number flagged, the corresponding with-replacement variance, and the variance reduction as a percentage of the with-replacement value.

Method. The sampling model is hypergeometric with \(N=50\), \(K=15\), and \(n=10\). Here \(p=15/50=0.3\) and \(1-p=0.7\). The with-replacement (binomial) variance is \[ np(1-p)=10(0.3)(0.7)=2.1. \] The variance correction is \[ \frac{N-n}{N-1}=\frac{50-10}{50-1}=\frac{40}{49}. \] Thus the without-replacement variance is \[ \operatorname{Var}(X)=2.1\left(\frac{40}{49}\right) =\frac{21}{10}\frac{40}{49} =\frac{84}{49}=\frac{12}{7}\approx1.7143. \] The absolute reduction is \[ 2.1-\frac{12}{7}=\frac{21}{10}-\frac{12}{7} =\frac{147-120}{70}=\frac{27}{70}\approx0.3857. \] As a percentage of the with-replacement variance, the reduction is \[ \frac{27/70}{21/10}\times100\% =\frac{27}{147}\times100\% \approx18.3673\%. \] The same result follows from the correction factor: \(1-40/49=9/49\approx18.3673\%\).

Common mistake avoided: The correction ratio is the without-replacement variance divided by the with-replacement variance, not the percentage reduction. The reduction is \(1-40/49\), not \(40/49\).

Worked Example: Variance and Standard Deviation of a Sample Proportion

A hypothetical population contains \(N=12\) records, exactly \(K=3\) of which have a specified feature. Four records are selected uniformly without replacement. Find the variance and standard deviation of the sample proportion \(\hat p\) with the feature.

Method. Here \(n=4\), \(p=3/12=1/4\), and \(1-p=3/4\). First calculate the variance of the count: \[ \begin{aligned} \operatorname{Var}(X) &=4\left(\frac14\right)\left(\frac34\right)\frac{12-4}{12-1}\\ &=\frac34\frac{8}{11}\\ &=\frac{6}{11}\approx0.5454545. \end{aligned} \] Since \(\hat p=X/4\), its variance is the count variance divided by \(4^2=16\): \[ \operatorname{Var}(\hat p)=\frac{6/11}{16} =\frac{3}{88}\approx0.0340909. \] The standard deviation is the square root of this variance: \[ \operatorname{SD}(\hat p)=\sqrt{\frac{3}{88}}\approx0.184637. \] Rounded to four decimal places, \(\operatorname{Var}(\hat p)=0.0341\) and \(\operatorname{SD}(\hat p)=0.1846\). As a check, the with-replacement proportion variance would be \((1/4)(3/4)/4=3/64=0.046875\); multiplying by \(8/11\) gives \(3/88\).

Common mistake avoided: The correction factor \(8/11\) multiplies the variance, while its square root multiplies the standard deviation. Also, converting from \(X\) to \(X/4\) divides variance by 16, not by 4.

Interpreting the Finite-Population Correction

The factor \((N-n)/(N-1)\) measures how much the finite population constrains the sample total. When \(n=1\), it equals 1, so the count variance agrees with the one-draw variance \(p(1-p)\). When \(n=N\), it is zero: observing the entire population makes the count exactly \(K\), with no sampling variability. For intermediate sample sizes, it lies between zero and one.

For a small sampling fraction \(n/N\), the factor is often close to 1, so sampling with replacement may provide a useful variance approximation. For a sample that is a substantial fraction of the population, the reduction can be material. The exact formula uses \(N-1\), not \(N\), in the denominator because the covariance between two distinct draw indicators is \(-p(1-p)/(N-1)\).

Identify the finite population.
Record \(N\), the number \(K\) of successes, and the sample size \(n\), and verify that sampling is without replacement.

Find the population success proportion.
Calculate \(p=K/N\), then \(p(1-p)\).

Apply the correction to the right quantity.
For the count, begin with \(np(1-p)\) and multiply by \((N-n)/(N-1)\). For the sample proportion, use \(p(1-p)/n\) and multiply by the same factor.

Check the result.
The corrected variance cannot exceed the corresponding with-replacement variance; it is zero for a census or a population with no variation in success status.

Common Exam Traps

Assuming independence. Draw indicators are not independent under sampling without replacement. Leaving out their negative covariances gives the with-replacement variance instead.
Using \(N\) instead of \(N-1\). The correction is \((N-n)/(N-1)\), which follows from the covariance of two distinct draw indicators.
Confusing variance and standard deviation corrections. Multiply variance by \((N-n)/(N-1)\); multiply standard deviation by its square root.
Forgetting the sample-proportion scaling. Since \(\hat p=X/n\), its variance is \(\operatorname{Var}(X)/n^2\), not \(\operatorname{Var}(X)/n\).
Using the correction for the wrong model. The factor here addresses sampling without replacement from a finite population. It is not a general adjustment for independent observations or for an arbitrary probability model.

Key takeaway: The hypergeometric count is a sum of dependent indicators. Each indicator has variance \(p(1-p)\), and each distinct pair has covariance \(-p(1-p)/(N-1)\). Adding these terms yields \(np(1-p)(N-n)/(N-1)\): the with-replacement variance multiplied by the finite-population correction.

Check Your Understanding

For each problem, identify the finite-population model before applying a variance formula.

A sample of \(n=6\) objects is drawn without replacement from \(N=30\), where \(K=9\) are successes. Find the variance of the success count. (Answer: \(30.24/29\approx1.0428\))
A population has \(N=40\) members, \(K=10\) of whom have a certain attribute. A sample of \(n=8\) is taken without replacement. Find the count variance and the variance it would have under sampling with replacement. (Answer: Without replacement, \(16/13\approx1.2308\); with replacement, \(1.5\))
From \(N=60\) items, \(K=18\) have a defect. A sample of \(n=15\) is drawn without replacement. Find the variance of the sample proportion of defective items. (Answer: \(0.63/59\approx0.0107\))
A population contains \(N=24\) objects, \(K=6\) of them successes, and a sample of \(n=9\) is taken without replacement. Find the count variance and the absolute reduction from the with-replacement variance. (Answer: Count variance \(405/368\approx1.1005\); absolute reduction \(27/46\approx0.5870\))
For \(N=100\), \(K=40\), and \(n=20\), calculate the covariance between two distinct draw indicators and the variance of the total count. (Answer: Indicator covariance \(-2/825\approx-0.002424\); count variance \(128/33\approx3.8788\))

What You'll Learn