Hypergeometric Sampling and Its Support

Sampling Without Replacement From a Finite Population

A count from sampling without replacement is constrained by what is actually in the population. If a population contains only a few marked units, a sample cannot contain more marked units than that. Conversely, if the sample is so large that it cannot be filled using only unmarked units, then some marked units must be included. The support records all the count values allowed by these constraints.

The basic model has a finite population with two categories: marked and unmarked. We fix the sample size before drawing, and each draw removes a unit from the available population. The number of marked units in the completed sample is the random variable of interest.

A simple random sample of size \(n\) means that every subset of \(n\) distinct population units is an equally likely sample. In particular, every such subset has positive probability. This condition makes the support a question of which counts are feasible: if at least one sample subset has count \(x\), then \(x\) belongs to the support. The sample itself is a subset of units; \(X\) is only its count of marked units.

Deriving the Support

There are two direct upper limits on \(X\). The sample contains only \(n\) units, and the population contains only \(K\) marked units. Thus \(X\) cannot exceed either \(n\) or \(K\). There are also two lower limits. A count cannot be negative, and if \(n\) is greater than the number \(N-K\) of unmarked units, the extra sampled units must be marked.

The lower limit accounts for the shortage, if any, of unmarked units needed to fill the sample. The upper limit accounts for both the sample size and the total number of marked units. Both limits matter: using only \(0\leq X\leq n\), for example, can include counts that the population cannot supply.

For completeness, the feasibility argument can be checked through its inequalities. If \(x\geq\max(0,n-(N-K))\), then \(x\geq 0\) and \(n-x\leq N-K\). If \(x\leq\min(n,K)\), then \(x\leq K\) and \(n-x\geq 0\). Thus \(x\) marked units and \(n-x\) unmarked units can be selected. The conditions \(x\leq K\) and \(n-x\leq N-K\) ensure those selections are available in the population. Their union has \(x+(n-x)=n\) distinct units, so it is a valid sample. Every valid sample subset has positive probability under simple random sampling, so \(P(X=x)>0\). The upper and lower bounds already proved rule out all other values. \(\square\)

Worked Examples

Support Size and Boundary Cases

The endpoints also tell us how many distinct integer values the count can take. Write \[ L=\max(0,n-(N-K)),\qquad U=\min(n,K). \] Because the theorem establishes that every integer from \(L\) through \(U\) is attainable, the support contains \(U-L+1\) values. There is no gap between the endpoints.

To justify the count, list the integers in the support in order: \(L,L+1,\ldots,U\). There is one initial value \(L\), and each increase by 1 adds one value. The number of increases is \(U-L\), so the total number is \(1+(U-L)=U-L+1\). The interval is nonempty: a sample of size \(n\) can be formed from the full population, so at least one count is feasible; alternatively, this follows from the theorem's feasibility argument. A singleton occurs precisely when \(U-L=0\), or \(L=U\).

Several boundary cases make the formula especially transparent:

• If \(n=0\), the sample is empty and the support is \(\{0\}\).
• If \(K=0\), there are no marked units, so the support is \(\{0\}\) for any allowed \(n\).
• If \(K=N\), every unit is marked, so the support is \(\{n\}\).
• If \(n=N\), the entire population is sampled, so the support is \(\{K\}\).

These cases are all included in the same endpoint formula; they do not require a separate distribution model. For instance, when \(K=0\), the upper endpoint \(\min(n,0)\) is 0, while the lower endpoint is also 0 because \(n\leq N=N-K\).

Why the Support Matters

The support identifies which values a count can actually take before any probability calculation is attempted. It is useful for checking whether a proposed value is possible, for setting limits in later calculations, and for recognizing when a sampling plan forces the count to a particular value. The support depends on all three quantities \(N\), \(K\), and \(n\), not on the sample size alone.

A frequent mistake is to say that \(X\) can be any integer from 0 to \(n\), because the sample has \(n\) places. That reasoning ignores the composition of the population. If the population has only 2 marked units, a sample cannot contain 3 marked units, even if \(n\geq 3\). If instead the sample is larger than the number of unmarked units, a count of zero marked units may be impossible.

For example, with \(N=9\), \(K=7\), and \(n=6\), the shortcut \(0\leq X\leq6\) would include 0, 1, 2, and 3. But only 2 unmarked units exist, so a sample of 6 must include at least 4 marked units. The correct support is \(\{4,5,6\}\), not \(\{0,1,2,3,4,5,6\}\). Always check both how many marked units can be selected and how many unmarked units are available to fill the rest of the sample.