Deriving the Hypergeometric Mean With Indicators

Why the Hypergeometric Mean Has a Simple Form

A sample drawn without replacement has dependent outcomes: after a success is drawn, the composition of the remaining population changes. Yet the expected number of successes in the sample has a particularly simple form. The key is to count success indicators, one for each draw position, and add their expectations. This approach works without treating the draws as independent.

The result is useful whenever a fixed-size sample is selected from a finite population containing a known number of successes. In an actuarial setting, for example, it can describe the expected number of flagged files in a hypothetical audit sample. The formula gives an average over repeated samples; it does not predict the exact count in any one sample.

Definition: Suppose a population contains $N$ distinct items, of which $K$ are designated successes. A simple random sample of size $n$ is selected without replacement, so every subset of $n$ population items is equally likely. The number $X$ of sampled successes has a hypergeometric distribution with population size $N$, success count $K$, and sample size $n$, where $N>0$, $0\leq K\leq N$, and $0\leq n\leq N$. The parameters $N$, $K$, and $n$ are integers.

Here “success” is just the label for the category being counted. It need not mean a favorable outcome. The essential feature of the model is sampling without replacement from a finite population, with a fixed number $K$ of successes. We will use $E[X]$ for the expected value of the sample success count.

Turn the Count Into Indicators

Imagine ordering the $n$ sampled items by draw position. For each position $j$, define an indicator $I_j$: it equals 1 if the item in position $j$ is a success, and 0 otherwise. Every sampled success contributes exactly one 1, and every nonsuccess contributes 0. Therefore, the total count is the sum of the indicators.

I_j = \begin{cases} 1, & \text{if draw } j \text{ is a success},\\ 0, & \text{if draw } j \text{ is a nonsuccess}, \end{cases} \qquad X=\sum_{j=1}^{n} I_j.

An indicator has expectation equal to the probability that it equals 1. Each draw position is equally likely to contain any one of the $N$ population items. Since $K$ of those items are successes, the probability that position $j$ is a success is $K/N$. Thus $E[I_j]=K/N$ for every $j$.

Theorem: For a hypergeometric count $X$ under the definition above, $$ E[X]=\frac{Kn}{N}. $$ The formula also applies at the boundary cases $K=0$, $K=N$, $n=0$, and $n=N$, provided $N>0$.

Derivation: Add the Expected Indicators

Linearity of expectation says that the expected value of a finite sum is the sum of the expected values. It does not require the summands to be independent. Applying it to the indicator representation gives the mean directly.

Represent the count.
For each draw position $j$, let $I_j$ indicate whether that position contains a success. Then $X=\sum_{j=1}^{n}I_j$.

Find each indicator expectation.
Every position has probability $K/N$ of containing a success, so $E[I_j]=1(K/N)+0(1-K/N)=K/N$.

Use linearity of expectation.
Sum the expectations of the $n$ indicators. Independence is not needed for this step.

\begin{aligned} 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{Kn}{N}. \end{aligned}

The probability of success on a later draw, conditional on the earlier draws, may change. That fact does not alter the calculation: each position has the same unconditional probability $K/N$ of being a success. More importantly, linearity of expectation only needs those individual expectations. It does not require the indicators to be independent.

Worked Examples

Worked Example: Expected Flags in a File Audit

A hypothetical insurer has 18 claim files selected for a quality review. Five files are flagged for a particular documentation issue. An auditor selects 6 files uniformly at random without replacement. Let $X$ be the number of flagged files selected. Find $E[X]$, and find the expected number of unflagged files in the sample.

Model and parameters. This is a hypergeometric model with $N=18$, $K=5$, and $n=6$. The sample is a simple random sample without replacement. No interest-rate or mortality assumption is involved.

Method. For each of the 6 draw positions, define an indicator for whether the selected file is flagged. Each indicator has expectation $5/18$. Adding the six expectations gives $$ E[X]=6\left(\frac{5}{18}\right)=\frac{30}{18}=\frac{5}{3}=1.6667. $$ The expected number of flagged files is therefore $1.6667$, to four decimal places.

There are $18-5=13$ unflagged files. Let $U$ be the number of unflagged files selected. By the same indicator argument, $$ E[U]=6\left(\frac{13}{18}\right)=\frac{78}{18}=\frac{13}{3}=4.3333. $$ As a check, $X+U=6$ for every sample, so $E[X]+E[U]=5/3+13/3=6$, as required.

Common mistake avoided. The mean $1.6667$ is an expected count, not a claim that a particular sample contains a fractional file. The actual value of $X$ is an integer; the mean averages its values across repeated samples.

Worked Example: Choosing a Sample Size for a Target Mean

A warehouse has 40 distinct sensor units, of which 10 have a specified calibration issue. A technician chooses $n$ units uniformly at random without replacement. What sample size gives an expected issue count of 3.5? Also find the expected number without the issue in that sample.

Model and parameters. The issue count has a hypergeometric distribution with $N=40$, $K=10$, and sample size $n$, where $0\leq n\leq40$. The sample is selected without replacement. No interest-rate or mortality assumption is involved.

Method. Set the hypergeometric mean equal to the requested target and solve for $n$: $$ \frac{Kn}{N}=3.5 \quad\Longrightarrow\quad \frac{10n}{40}=3.5 \quad\Longrightarrow\quad \frac{n}{4}=3.5 \quad\Longrightarrow\quad n=14. $$ The sample size is an allowed integer and satisfies $0\leq14\leq40$.

There are $40-10=30$ units without the issue. Their expected number in the sample is $$ 14\left(\frac{30}{40}\right)=14\left(\frac{3}{4}\right)=10.5. $$ The two expected counts add to $3.5+10.5=14$, the sample size. This checks both calculations.

Common mistake avoided. Do not use the number of units without the issue as $K$ when the target is the number with the issue. Here $K=10$ for the requested count; $30$ is used only for the complementary count.

Worked Example: Combining Samples From Two Lots

A hypothetical review draws 8 records without replacement from a lot of 24 records containing 6 flagged records. It also draws 10 records without replacement from a separate lot of 30 records containing 3 flagged records. Let $T$ be the total number of flagged records in both samples. Find $E[T]$.

Model and parameters. Let $X_1$ and $X_2$ be the flagged counts from the first and second lots. Their hypergeometric parameters are $(N,K,n)=(24,6,8)$ and $(30,3,10)$, respectively. Each lot is sampled uniformly without replacement. No interest-rate or mortality assumption is involved.

Method. Write each lot’s count as a sum of draw indicators. For the first lot, each of its 8 indicators has expectation $6/24=1/4$, so $$ E[X_1]=8\left(\frac{6}{24}\right)=8\left(\frac14\right)=2. $$ For the second lot, each of its 10 indicators has expectation $3/30=1/10$, so $$ E[X_2]=10\left(\frac{3}{30}\right)=10\left(\frac1{10}\right)=1. $$ Since $T=X_1+X_2$, linearity of expectation gives $$ E[T]=E[X_1]+E[X_2]=2+1=3. $$ The expected total is 3 flagged records.

Check. The first sample contributes 8 records and the second contributes 10, so the combined total is 18 records; an expected flagged count of 3 is within the possible range from 0 to 18. Independence between the two samples is not needed to add the expectations, although the scenario specifies separate lot samples.

Common mistake avoided. Do not combine the lots into one population and use a single common success fraction. Their success fractions differ: $6/24=1/4$ and $3/30=1/10$. Calculate each sample’s expected count using its own lot parameters, then add.

What the Mean Does—and Does Not—Say

The mean $Kn/N$ is the sample size multiplied by the population success fraction. It is therefore a useful benchmark: a sample of size $n$ is expected to contain the same fraction of successes as the population, on average. The result is not a statement that every sample mirrors the population exactly. Random samples can contain more or fewer successes.

The formula also handles the boundary cases naturally. If $K=0$, then $E[X]=0$, since no success exists in the population. If $K=N$, then $E[X]=n$, since every sampled item is a success. If $n=0$, the sample has no successes. If $n=N$, the whole population is selected and $E[X]=K$. These checks are useful for catching an incorrect choice of parameters.

A related distinction is between the expected count and the probability of at least one success. The expected count measures the average number of successes, while the latter is a probability for a particular event. They are not generally equal.

Common Mistakes and an Exam Trap

Assuming independence is needed. The draw indicators are dependent under sampling without replacement. The mean derivation remains valid because linearity of expectation does not require independence.
Confusing a mean with a probability. $Kn/N$ is an expected count. It can be greater than 1, and it can be noninteger.
Using the wrong population count. $K$ must count the items in the category whose sample total is being studied. For the complementary category, use $N-K$.
Forgetting the fixed-size sample assumption. The formula applies to a sample of $n$ selected items without replacement from the stated finite population. It is not a general mean formula for every sampling scheme.

For a concrete trap, suppose $N=10$, $K=4$, and $n=3$. The correct mean is $$ E[X]=\frac{4(3)}{10}=1.2. $$ A mistaken calculation might use the probability of at least one success as the expected number of successes. The probability of no successes is $\binom{6}{3}/\binom{10}{3}=20/120=1/6$, so $$ P(X\geq1)=1-\frac16=\frac56=0.8333. $$ That is a probability, not the mean count. The corrected mean is $1.2$, obtained by adding the expectations of the three draw indicators. The two numbers answer different questions.

Key takeaway: For a fixed-size sample without replacement from a population of $N>0$ items containing $K$ successes, represent the count as a sum of $n$ draw indicators. Each indicator has expectation $K/N$, so linearity of expectation gives $E[X]=Kn/N$, whether or not the draws are independent.

Check Your Understanding

For each question, assume sampling is uniform without replacement wherever a sample is specified. Answers are given for checking.

A shipment contains 36 components, 9 of which are marked for inspection. A sample of 8 is selected. What is the expected number of marked components? (Answer: $2$)
A review samples 10 files from a population of 50, of which 15 are flagged. Find the expected number flagged and the expected number not flagged. (Answer: $3$ flagged and $7$ not flagged)
A finite population has $N=28$ items, including $K=7$ successes. What sample size gives an expected success count of 2? (Answer: $n=8$)
From one lot with $N=16$, $K=4$, a sample of 6 is taken; from a separate lot with $N=25$, $K=10$, a sample of 5 is taken. What is the expected combined number of successes? (Answer: $3.5$)
A population has $N=20$ items. Find the expected success count in a sample of 12 if $K=0$, and then if $K=20$. (Answer: $0$ when $K=0$, and $12$ when $K=20$)

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