Why the Sampling Design Matters
A predictive dataset may contain only a fraction of the records in the population it is intended to describe. How that fraction was selected affects what can be inferred from the sample. If every eligible record had the same known chance of selection, the sample can support one kind of population estimate; if records were selected at different rates, the analysis needs to account for those differences. If the selection chances are unknown, a design-based population estimate may not be available at all.
This tutorial focuses on the route from population records to the observed sample: the sampling design and each record’s inclusion probability. It does not assume that the sampled records automatically represent the target population. Instead, it makes explicit which records could be selected and how likely they were to be included. This distinction is aligned with the SOA Exam PA (Predictive Analytics) syllabus.
The target population and sampling frame need not be identical. For example, a hypothetical analysis might target all active policies during a year but draw its sample from a system extract that omits policies administered on a separate platform. Sampling probabilities describe selection from the frame; they cannot, by themselves, recover records that never entered the frame. That is why defining the population and frame precisely is part of specifying a sampling design.
A probability sample uses a specified random procedure so that the relevant inclusion probabilities can be calculated. In a nonprobability sample, such as one assembled from volunteers or convenient records without a known random selection procedure, those probabilities generally cannot be calculated from the design. A large sample size does not, on its own, supply the missing probabilities.
Common Probability Sampling Designs
Different designs give records different routes into the sample. The central calculation is always the chance that a particular record is selected, but the calculation depends on the actual stages and rules of selection.
| Design | How selection works | Inclusion probability |
|---|---|---|
| Simple random sampling without replacement | Select \(n\) records uniformly from \(N\) frame records. | \(\pi_i=n/N\) for every record. |
| Stratified sampling | Divide the frame into strata, then sample within each stratum. | With \(N_h\) records and \(n_h\) selected in stratum \(h\), \(\pi_i=n_h/N_h\) for a record in that stratum. |
| Cluster sampling | Select groups, such as clinics or depots, and include some or all records within selected groups. | For a record, combine the probabilities of the relevant selection stages. |
| Systematic sampling | Choose a random start and then select records at a fixed interval from an ordered frame. | Depends on the design; with a random start among \(k\) positions and interval \(k\), each record’s first-order probability is \(1/k\). |
Strata are groups defined before sampling, with sampling performed within each group. The number sampled in each stratum can be proportional to its population size or deliberately disproportionate. In the latter case, records in some strata have higher inclusion probabilities than others. Cluster sampling instead selects groups as the first sampling stage. If the design then selects records within each chosen cluster, a record must pass both selection stages to be included.
For systematic sampling, the statement about probability \(1/k\) assumes a frame size divisible by \(k\), a uniformly selected random start among the \(k\) possible starting positions, and selection every \(k\)th record. The ordering of the frame still matters: a pattern in the ordering that aligns with the interval can make the realized sample unusual. The inclusion probability alone does not describe every feature of a design.
Worked Example: Simple Random Sampling of Policy Records
A hypothetical insurer has a frame of 1,200 eligible policy records and selects 120 by simple random sampling without replacement. The payment amount recorded for the sampled policies totals $84,000. Find the inclusion probability for one policy, its inverse-probability weight, and an estimate of total payments for all 1,200 policies.
Method. Under simple random sampling without replacement, every frame record has the same inclusion probability, \(n/N\).
There are \(N=1{,}200\) records and \(n=120\) selected records, so \(\pi_i=120/1{,}200=0.10\). Each policy therefore has a 10% chance of selection.
The inverse-probability weight is \(1/\pi_i=1/0.10=10\). In this design, each sampled policy represents 10 frame policies for estimation.
The estimated population total is the sampled payment sum multiplied by the weight: \(84{,}000\times10=\$840{,}000\).
Check. The sample is one-tenth of the frame, so multiplying its total by 10 scales the observed sum to the full frame size. The estimate is $840,000. The common mistake this avoids is treating the sample sum of $84,000 as if it were already the total for all 1,200 policies.
Using Inclusion Probabilities to Estimate a Total
Suppose a finite population contains \(N\) records with fixed values \(y_1,\ldots,y_N\), and the sampling design gives every record a known positive inclusion probability \(\pi_i\). Let \(I_i=1\) if record \(i\) is included and \(I_i=0\) otherwise. The inverse-probability estimator of the population total is the Horvitz–Thompson estimator:
The weight \(1/\pi_i\) is larger for a record that was less likely to be selected. The estimator adds each observed value after scaling it to account for its selection chance. When the population size \(N\) is known, dividing \(\widehat{T}_{HT}\) by \(N\) estimates the population mean. These design-based results use the specified random selection mechanism; they do not require the fixed record values themselves to follow a particular probability distribution.
The expectation is taken over repeated samples drawn using the same design, while the population values \(y_i\) remain fixed. Therefore,
This establishes that the estimator is unbiased over repeated samples from the design. It does not mean that a particular sample’s estimate must equal the true total. Nor does it establish that a sample drawn from an incomplete frame estimates a target population outside that frame without additional assumptions. The proof depends on each target record being represented in the frame and having a known positive probability of selection.
For uncertainty calculations, the probabilities that two records are included together can also matter. Define \(\pi_{ij}=P(i\text{ and }j\text{ are both included})\). In general, records’ inclusion events need not be independent, so knowing only the individual \(\pi_i\) values is not enough to describe all the sampling dependence or calculate every design-based variance.
Worked Example: Unequal Rates in a Stratified Sample
A hypothetical portfolio contains 800 East-region records and 200 West-region records. A stratified design selects 80 East records and 40 West records by simple random sampling within each region. The sample mean payment is $500 in the East and $700 in the West. Estimate the portfolio total and compare it with the unweighted sample-mean estimate.
Method. Calculate a separate inclusion probability for each stratum, then use the Horvitz–Thompson estimator, or equivalently multiply each stratum’s sample mean by its population size.
For the East, \(\pi_E=80/800=0.10\), so each sampled East record has weight \(1/0.10=10\). Its estimated stratum total is \(800\times\$500=\$400{,}000\).
For the West, \(\pi_W=40/200=0.20\), so each sampled West record has weight \(1/0.20=5\). Its estimated stratum total is \(200\times\$700=\$140{,}000\).
Adding the stratum estimates gives \(\widehat{T}_{HT}=\$400{,}000+\$140{,}000=\$540{,}000\). The known population size is \(800+200=1{,}000\), so the corresponding estimated population mean is \(\$540{,}000/1{,}000=\$540\).
For comparison, the unweighted sample mean is \((80\times\$500+40\times\$700)/120=(\$40{,}000+\$28{,}000)/120=\$566.67\), rounded to the nearest cent. Multiplying that by 1,000 gives $566,666.67. This gives the West, which was sampled at twice the rate, more influence than its share of the population warrants.
Check. The weighted total is $540,000 and the estimated mean is $540. The common mistake this avoids is pooling the sample records without accounting for the unequal inclusion probabilities. The unweighted calculation is not the design-weighted estimate.
Multistage Selection and Unknown Probabilities
In a multistage design, a record’s inclusion probability is the probability of passing every selection stage required for inclusion. For example, if a clinic is selected first and a patient is then sampled within that clinic, the patient’s probability is the clinic-selection probability multiplied by the conditional probability of patient selection, provided that the second-stage probability is defined conditional on the clinic being selected. Omitting a stage gives the wrong record-level probability and therefore the wrong weight.
The next example illustrates this calculation. Its payment amounts and population are hypothetical.
Worked Example: Two-Stage Sampling of Clinic Records
A hypothetical study has 40 clinics, each with 25 eligible records. It selects 8 clinics by simple random sampling without replacement, then selects 5 records by simple random sampling within each selected clinic. The payment amounts for all sampled records sum to $16,000. Find a record’s inclusion probability, its weight, and the estimated total payment across all 1,000 records.
Method. Multiply the first-stage clinic-selection probability by the conditional second-stage record-selection probability.
A particular clinic is selected with probability \(8/40=0.20\). Given that clinic is selected, a particular record in it is selected with probability \(5/25=0.20\). Thus a particular record is included with probability \(0.20\times0.20=0.04\).
The record’s weight is \(1/0.04=25\). There are \(8\times5=40\) sampled records, and the Horvitz–Thompson estimate is the observed sum multiplied by the common weight: \(\$16{,}000\times25=\$400{,}000\).
Check. The frame contains \(40\times25=1{,}000\) records, and the design samples 40 records, so the sample fraction is \(40/1{,}000=0.04\), matching the calculated inclusion probability. The estimated total is $400,000. The common mistake this avoids is using only the clinic probability of 0.20 and a weight of 5, despite sampling only some records within each selected clinic.
A convenience sample or a sample assembled through volunteers is different: unless a known selection mechanism supplies the inclusion probabilities, weights cannot be calculated by simply taking their reciprocals. A post hoc guess at each record’s selection chance is not the same as a probability specified by the sampling design. Claims about population totals based on such a sample require other methods and assumptions; the design-based calculation above cannot be justified by the sample size alone.
Check Your Understanding
For each design, calculate the requested probability or estimate. Answers are given so you can check the arithmetic.
- A frame has 600 records, and 30 are selected by simple random sampling without replacement. What is one record’s inclusion probability and inverse-probability weight? (Answer: \(\pi_i=0.05\); weight \(=20\).)
- A stratified sample has 900 North records and 100 South records. It samples 90 North records and 30 South records. The sample means are $40 in the North and $70 in the South. Estimate the total using the stratified design. (Answer: $43,000.)
- A hypothetical design has 40 depots with 25 contracts each. It selects 8 depots, then 5 contracts per selected depot. The sampled contract payments sum to $16,000. Find the record inclusion probability and estimated total payment. (Answer: \(\pi_i=0.04\); estimated total \(=\$400{,}000\).)
- A systematic design has 240 records and selects every 10th record, using a uniformly random start from positions 1 through 10. What is each record’s first-order inclusion probability and inverse-probability weight? (Answer: \(\pi_i=0.10\); weight \(=10\).)
- A team selects records from volunteers but has no documented random procedure or selection probabilities. Can it calculate design-based inverse-probability weights from the information given? (Answer: No; the inclusion probabilities are unknown, so their reciprocals cannot be calculated from the stated design.)