Inclusion Indicators in Sampling Without Replacement

Why Represent a Sample With Indicators?

A sample drawn without replacement has a fixed size: selecting one unit affects which other units can be selected. Inclusion indicators make that constraint explicit. Rather than describing the sample only as a set, we attach a random variable to every unit in the population and record whether it was selected.

This representation is useful because many sampling quantities can be assembled from the indicators. The essential first step is to understand their individual and pairwise behavior. In particular, the indicators are generally dependent: their covariances record the competition induced by a fixed sample size.

Definition and Sampling Setup

Consider a finite population of $N$ distinct units, labelled $1,\ldots,N$. A simple random sample without replacement of size $n$ means that each of the $\binom{N}{n}$ subsets of size $n$ has the same probability of being selected. The sample size satisfies $0\leq n\leq N$. The order in which units might be drawn does not affect inclusion.

Definition: For population unit $i$, its inclusion indicator is the random variable $$ I_i = \begin{cases} 1, & \text{if unit } i \text{ is in the sample},\\ 0, & \text{if unit } i \text{ is not in the sample}. \end{cases} $$ Thus $I_i$ is Bernoulli distributed, with its success probability determined by the sampling design. For every realized sample, the indicators satisfy $\sum_{i=1}^{N} I_i=n$.

The last identity is deterministic: every possible sample contains exactly $n$ units. It is a useful check on calculations and will also explain why the indicators cannot be independent in a nontrivial fixed-size design.

Inclusion Probabilities and Pairwise Covariances

For any specified unit, the chance of inclusion is the fraction of size-$n$ samples that contain it. For two distinct specified units, the chance that both are included is the fraction containing both. These counting arguments give the marginal and joint probabilities.

Theorem: In a simple random sample without replacement from $N\geq 2$ units, for $0\leq n\leq N$ and distinct $i,j$, $$ E[I_i]=P(I_i=1)=\frac{n}{N}, \qquad \operatorname{Var}(I_i)=\frac{n}{N}\left(1-\frac{n}{N}\right), $$ and $$ P(I_i=1,I_j=1)=\frac{n(n-1)}{N(N-1)}, \qquad \operatorname{Cov}(I_i,I_j) =-\frac{n(N-n)}{N^2(N-1)}. $$

To prove the marginal probability, count the subsets that contain unit $i$. Once $i$ is included, the other $n-1$ units can be chosen from the $N-1$ remaining units. Therefore

P(I_i=1)=\frac{\binom{N-1}{n-1}}{\binom{N}{n}}=\frac{n}{N}.

The equality also holds at $n=0$ and $n=N$, by direct interpretation of the sampling design. Since $I_i$ takes only the values zero and one, $I_i^2=I_i$. Consequently,

\operatorname{Var}(I_i) =E[I_i^2]-E[I_i]^2 =E[I_i]-E[I_i]^2 =\frac{n}{N}\left(1-\frac{n}{N}\right).

For the joint probability, both specified units must be in the sample, so the remaining $n-2$ units are chosen from the other $N-2$. If $n<2$, the joint probability is zero; the formula below gives zero as well. Thus

P(I_i=1,I_j=1) =\frac{\binom{N-2}{n-2}}{\binom{N}{n}} =\frac{n(n-1)}{N(N-1)}.

For $n=0$ or $n=1$, the probability is zero directly, with the same result from the expression. Now use the definition of covariance and subtract the product of the marginal probabilities:

\begin{aligned} \operatorname{Cov}(I_i,I_j) &=E[I_i I_j]-E[I_i]E[I_j]\\ &=\frac{n(n-1)}{N(N-1)}-\frac{n^2}{N^2}\\ &=\frac{nN(n-1)-n^2(N-1)}{N^2(N-1)}\\ &=-\frac{n(N-n)}{N^2(N-1)}. \end{aligned}

This proves the stated formulas. The covariance is strictly negative when $0<n<N$: including one unit leaves fewer sample places for the others. At $n=0$ or $n=N$, every indicator is constant and the covariance is zero. $\square$

Worked Calculations

Worked Example: Two Units in a Sample of Size Two

A simple random sample of size $n=2$ is taken from $N=6$ units. Find the expectation and variance of one inclusion indicator, and the joint inclusion probability and covariance for two distinct units.

For either specified unit $i$,

E[I_i]=\frac{2}{6}=\frac{1}{3}, \qquad \operatorname{Var}(I_i)=\frac{1}{3}\left(1-\frac{1}{3}\right)=\frac{2}{9}.

For two distinct units $i$ and $j$, the probability that both are selected is

P(I_i=1,I_j=1)=\frac{2(2-1)}{6(6-1)}=\frac{2}{30}=\frac{1}{15}.

Their covariance is therefore

\operatorname{Cov}(I_i,I_j) =\frac{1}{15}-\left(\frac{1}{3}\right)^2 =\frac{3}{45}-\frac{5}{45} =-\frac{2}{45}.

The general covariance formula gives the same value: $-2(6-2)/(6^2\cdot5)=-8/180=-2/45$. The negative sign matches the sampling constraint: if $i$ is selected, only one of the five remaining units can accompany it.

Worked Example: Checking the Fixed-Size Constraint

Take a simple random sample of size $n=4$ from $N=10$ units. Calculate the variance of each indicator and the covariance between distinct indicators, then verify that these values agree with the fact that the total number selected is fixed.

For any $i$,

\operatorname{Var}(I_i)=\frac{4}{10}\left(1-\frac{4}{10}\right) =\frac{2}{5}\cdot\frac{3}{5} =\frac{6}{25}.

For any distinct $i,j$,

\operatorname{Cov}(I_i,I_j) =-\frac{4(10-4)}{10^2(10-1)} =-\frac{24}{900} =-\frac{2}{75}.

The sum $S=\sum_{i=1}^{10}I_i$ equals 4 for every possible sample, so $\operatorname{Var}(S)=0$. Expanding its variance using the ten individual variances and the $10\cdot9$ ordered pairs of distinct indicators gives

\begin{aligned} \operatorname{Var}(S) &=\sum_{i=1}^{10}\operatorname{Var}(I_i) +\sum_{i\ne j}\operatorname{Cov}(I_i,I_j)\\ &=10\left(\frac{6}{25}\right) +90\left(-\frac{2}{75}\right)\\ &=\frac{12}{5}-\frac{12}{5}=0. \end{aligned}

The negative pairwise terms exactly offset the individual variances. The result is a check both on the covariance calculation and on the fixed sample size.

Worked Example: Conditioning on One Unit’s Inclusion

A sample of size $n=2$ is selected from $N=5$ units. Find the probability that unit 2 is included given that unit 1 is included. Then compare it with the unconditional inclusion probability for unit 2.

Given that unit 1 is included, the second sample place is filled by one of the other four units, each with equal probability. Exactly one of those four is unit 2, so

P(I_2=1\mid I_1=1)=\frac{1}{4}.

Unconditionally, unit 2 has inclusion probability

P(I_2=1)=\frac{2}{5}.

The conditional probability is smaller. This is dependence, not a contradiction: conditional on selecting unit 1, just one place remains for the four other units. The joint probability confirms the calculation:

P(I_1=1,I_2=1) =P(I_1=1)P(I_2=1\mid I_1=1) =\frac{2}{5}\cdot\frac{1}{4} =\frac{1}{10} =\frac{2(2-1)}{5(5-1)}.

For comparison, multiplying the unconditional probabilities as if the indicators were independent would give $(2/5)^2=4/25$, not $1/10$.

What the Indicator Structure Tells Us

The marginal distribution of each $I_i$ is simple, but that does not make the full collection independent. The joint inclusion probability is smaller than the product of marginal inclusion probabilities whenever $0<n<N$:

\frac{n(n-1)}{N(N-1)} < \left(\frac{n}{N}\right)^2.

Indeed, subtracting the right-hand side from the left gives the strictly negative covariance already derived. A common mistake is to assign each unit an independent Bernoulli indicator with success probability $n/N$. That construction gives the correct marginal inclusion probability, but its sum is random rather than identically equal to $n$; it does not represent simple random sampling of a fixed size.

Key takeaway: Under simple random sampling without replacement, each unit has inclusion probability $n/N$, while every pair of distinct indicators has covariance $-n(N-n)/[N^2(N-1)]$. The negative dependence is precisely consistent with the fixed total $\sum_i I_i=n$.

The indicator representation also provides a compact way to record which units enter any sample. Once the marginal and pairwise behavior is known, expectations and variances of quantities built from these indicators can be organized systematically. The fixed-size identity remains an important diagnostic: any proposed joint model for the indicators must respect it.

Check Your Understanding

Use the indicator definition and the simple-random-sampling formulas to answer these questions.

For $N=12$ and $n=3$, what are $E[I_i]$ and $\operatorname{Var}(I_i)$?
For two distinct units in that same design, calculate their joint inclusion probability and covariance.
Why does the identity $\sum_{i=1}^{N} I_i=n$ rule out independent indicators when $0<n<N$?
In a sample of size $n=1$ from $N=7$ units, what is $P(I_i=1,I_j=1)$ for distinct $i,j$, and why?
What changes in the indicator variances and pairwise covariances when $n=N$?