The Hypergeometric Mean from Draw Indicators

From Draws to a Count

A hypergeometric random variable counts how many successes appear in a sample drawn without replacement from a finite population. Its probability mass function can be obtained by counting samples, but there is a shorter route to its mean: describe the count as a sum of indicators, one for each draw position. The method also illustrates a general principle used throughout probability theory: expectations add even when the random variables being added are dependent.

Let a population contain \(N\) objects, of which \(K\) are designated successes. Draw \(n\) objects uniformly without replacement, where \(N\geq 1\) and \(K,n\) are integers satisfying \(0\leq K\leq N\) and \(0\leq n\leq N\). Let \(X\) be the number of successes in the sample. We can view the sample as an ordered sequence of \(n\) draws for the purpose of defining indicators; the count \(X\) does not depend on the order.

Definition: For draw position \(j\), where \(1\leq j\leq n\), define \(I_j=1\) if the object at that position is a success and \(I_j=0\) otherwise. Then \(I_j\) is an indicator random variable, and the total number of successes is \(X=\sum_{j=1}^{n} I_j\). When \(n=0\), the sum is empty and \(X=0\).

An indicator has a particularly simple expectation: if \(I\) equals 1 on an event \(A\) and 0 otherwise, then \(E[I]=P(A)\). Indeed, its expectation is \(1\cdot P(A)+0\cdot P(A^{c})=P(A)\). Thus finding \(E[X]\) reduces to finding the probability that each draw position contains a success.

The Mean by Linearity of Expectation

For any specified position \(j\), every one of the \(N\) population objects is equally likely to occupy that position. Since exactly \(K\) objects are successes, the probability of a success there is \(K/N\). This is true for every \(j\), even though the draws are made without replacement.

Theorem: If \(X\) is the number of successes in a uniform sample of size \(n\) drawn without replacement from a population of size \(N\) containing \(K\) successes, with \(N\geq1\), \(0\leq K\leq N\), and \(0\leq n\leq N\), then \[ E[X]=\frac{nK}{N}. \]

Proof. For each draw position \(j\), the indicator \(I_j\) has expectation \(E[I_j]=P(I_j=1)=K/N\). Since \(X=\sum_{j=1}^{n} I_j\), linearity of expectation gives

E[X] =E\!\left[\sum_{j=1}^{n}I_j\right] =\sum_{j=1}^{n}E[I_j] =\sum_{j=1}^{n}\frac{K}{N} =\frac{nK}{N}.

If \(n=0\), the sum has no terms, so the same result gives \(E[X]=0\). The argument uses only the marginal probability for each draw position and linearity of expectation; it does not require the indicators to be independent. \(\square\)

The formula has a direct interpretation. The fraction of the population that consists of successes is \(K/N\). In a sample of \(n\) draws, the expected number of successes is \(n\) times that fraction. An expected count need not be an integer: it is an average over repetitions of the sampling process, not necessarily a count that any one sample can attain.

Worked Applications

Worked Example: A Small Sample From a Larger Population

A box contains \(20\) labeled components, of which \(7\) are defective. Five components are selected uniformly without replacement. Find the expected number of defective components in the sample.

Let \(I_j\) indicate that the component selected at draw position \(j\) is defective, for \(j=1,\ldots,5\). At any one specified position, each of the \(20\) components is equally likely to appear, so

E[I_j]=P(I_j=1)=\frac{7}{20}.

The number \(X\) of defective components is \(I_1+I_2+I_3+I_4+I_5\). Therefore,

E[X] =\sum_{j=1}^{5}E[I_j] =5\left(\frac{7}{20}\right) =\frac{35}{20} =1.75.

Thus the expected number of defective components is \(1.75\). Each actual sample contains a whole number of defective components; \(1.75\) describes the long-run average across repeated samples of five.

Worked Example: A Mean That Is an Integer

A collection contains \(12\) specimens, \(3\) of which have a specified marker. A sample of \(8\) is drawn without replacement. Find the expected number with the marker, and explain what the result does and does not say.

For each of the eight positions, let \(I_j=1\) if the specimen has the marker and \(I_j=0\) otherwise. Each position has success probability \(3/12=1/4\), so \(E[I_j]=1/4\). Summing the expectations gives

E[X] =\sum_{j=1}^{8}E[I_j] =8\left(\frac{1}{4}\right) =2.

The expected number is \(2\). This does not say that every sample has exactly two marked specimens. It says that the average count over repeated samples of size \(8\) is \(2\). The calculation is about an expectation; identifying the probabilities of the possible counts is a separate question.

Worked Example: Checking a Boundary Case

A population has \(18\) objects, \(11\) of them successes. Suppose all \(18\) objects are drawn, so \(n=18\). Use indicators to find the mean and check it against what must happen in every sample.

For each position \(j\), \(E[I_j]=11/18\). Thus

E[X] =\sum_{j=1}^{18}E[I_j] =18\left(\frac{11}{18}\right) =11.

Because the entire population is drawn, every sample contains all \(11\) successes. Hence \(X=11\) on every outcome, and its expectation is indeed \(11\). This check confirms that the formula behaves correctly when the sample is a census.

Worked Example: No Draws and No Successes

Consider a population of \(10\) objects with \(K=0\) successes, and suppose \(n=4\) objects are drawn. For each position, the success probability is \(0/10=0\), so each indicator has expectation \(0\). Therefore,

E[X] =\sum_{j=1}^{4}0 =0 =\frac{4\cdot 0}{10}.

There are no successes in the population, so the count is always zero, as the calculation confirms. If instead \(n=0\), there are no indicators to add and \(X=0\) as well; the formula gives \(0\cdot K/N=0\) for any allowed \(K\).

Why Independence Is Not Needed

A common mistake is to reason that the indicator variables must be independent before their expectations can be added. That condition is unnecessary: linearity of expectation applies to any finite collection of random variables with finite expectations. Here each indicator is bounded between 0 and 1, so its expectation is finite, and the sum has finitely many terms.

The indicators do not always have the same dependence behavior across all parameter choices. If \(0<K<N\) and \(n\geq2\), the indicators at two distinct draw positions are dependent. To see this, there are \(K(K-1)\) ordered ways to put successes in the first two positions out of \(N(N-1)\) ordered pairs of distinct objects. Hence

P(I_1=1,I_2=1)=\frac{K(K-1)}{N(N-1)} \quad\text{whereas}\quad P(I_1=1)P(I_2=1)=\left(\frac{K}{N}\right)^2.

For \(0<K<N\), these quantities are unequal: their difference is \(-K(N-K)/(N^2(N-1))\), which is nonzero. In contrast, when \(K=0\) or \(K=N\), every indicator is constant and the indicators are independent; when \(n\leq1\), there is no pair of distinct draw positions to compare. These qualifications matter at the boundary cases. In every case, the mean derivation remains valid because it uses linearity rather than independence.

Key takeaway: Write the success count as a sum of draw indicators. Each position has success probability \(K/N\), so each indicator has mean \(K/N\), and adding the \(n\) expectations gives \(E[X]=nK/N\). No independence assumption is needed.

Check Your Understanding

Use the indicator representation and the stated sampling conditions to answer each question.

A population has \(25\) objects, \(10\) of them successes. A sample of \(6\) is drawn without replacement. What is the expected number of successes? Show the indicator calculation.
Explain why \(E[I_j]=K/N\) for a specified draw position, even though the objects are sampled without replacement.
Can a hypergeometric count have a noninteger mean? Explain what such a mean represents.
For which conditions in this model are all draw indicators constant? What does that imply about their independence?
Why is independence not needed to derive the expectation of the sum of the indicators?