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A pooled event count can hide very different exposure patterns. This guide shows how conditioning within each stratum turns that problem into a transparent exact calculation—and where the common-ratio assumption can fail.
Open the chapter contents · Probability root · Gold Standard of Education · How X Works
Chapter contents
Choose the route closest to the question, or read from Chapter 1 for the complete argument. The teaching chapters remain open; these grouped routes form the index.
Route 1 · Build the model
Route 2 · Work the exact test
Route 3 · Stress the assumptions
Route 4 · Check limits and evidence
Route 5 · Transfer and continue
CHAPTER 1 OF 12 · BUILD THE MODEL
1. One pooled rate ratio can hide several exposure clocks
Suppose events are counted in three clinics, factories or neighbourhoods. Each stratum has an exposed group, a comparison group and its own amount of person-time. Simply adding all events and all exposure can be tempting, but pooling changes the weights and can let a large stratum dominate. The scientific question is narrower: is there one rate ratio, rho, that is reasonably shared across strata?
The data in stratum i are two Poisson counts, X_i and Y_i, with exposure times T_1i and T_0i. Their rates are lambda_1i and lambda_0i. The common-effect model says lambda_1i / lambda_0i = rho while allowing the baseline rate lambda_0i to differ by stratum. That distinction matters: the strata need not have the same background event rate.
This article extends the Probability root and the earlier guide to exact Poisson rate-ratio inversion. The new problem is not two rates once, but two rates repeated across sparse strata.
CHAPTER 2 OF 12 · BUILD THE MODEL
2. Conditioning removes one nuisance rate per stratum
Condition on the total N_i = X_i + Y_i in each stratum. Under a proposed common ratio rho, X_i has a binomial distribution with N_i trials and probability p_i(rho) = rho T_1i / (rho T_1i + T_0i). The unknown baseline rate disappears from that conditional distribution.
The probabilities are not generally equal across strata because the exposure ratios differ. That is why the exact calculation is not an ordinary binomial test on the grand total. The statistic S = sum X_i is a sum of independent binomials with stratum-specific probabilities. Its null distribution can be built by multiplying probability polynomials or by dynamic-programming convolution.
Conditioning gives a clean finite-sample reference distribution, but it also discards information in the random totals N_i. This is the same central trade-off discussed in conditional tests and the cost of conditioning.
CHAPTER 3 OF 12 · WORK THE EXACT TEST
3. Worked example: three sparse strata
Consider the following original teaching data. Rates are events per unit of recorded person-time; the units could be hundreds of person-hours or person-years, provided they are used consistently.
| Stratum | Exposed events X | Comparison events Y | Exposed time T1 | Comparison time T0 |
|---|---|---|---|---|
| A | 2 | 1 | 80 | 100 |
| B | 1 | 2 | 50 | 120 |
| C | 3 | 0 | 100 | 90 |
Each stratum has three total events, so N_A = N_B = N_C = 3. The observed exposed-side total is S = 2 + 1 + 3 = 6. Under rho = 1, the exposed-side probabilities are 80/180 = 0.444, 50/170 = 0.294 and 100/190 = 0.526. The null expected value of S is therefore 3(0.444 + 0.294 + 0.526) = 3.795, not 4.5.
The exposure clocks have already changed the null. That is the first transfer check: equal rates do not imply a one-half event probability when person-time is unequal.
CHAPTER 4 OF 12 · WORK THE EXACT TEST
4. Convolution turns the strata into one exact distribution
For stratum i, write the probability polynomial [(1-p_i) + p_i z] raised to N_i. Multiplying the three polynomials makes the coefficient of z^s equal to P(S=s). For the worked example under rho = 1, the probabilities for S = 0 through 9 are approximately 0.0064, 0.0448, 0.1366, 0.2388, 0.2634, 0.1897, 0.0892, 0.0264, 0.00445 and 0.00033.
The observed upper-tail probability P(S at least 6) is about 0.1203. That is not small enough, by itself, to reject equal rates at a conventional five-percent one-sided level. A two-sided exact test requires a declared ordering rule; merely doubling a tail is not automatically the same as probability ordering or likelihood-ratio ordering.
| Quantity | Worked value | Interpretation |
|---|---|---|
| Observed S | 6 | Exposed-side events across all strata |
| Null E(S) | 3.795 | Exposure-adjusted expectation when rho = 1 |
| P(S at least 6) | 0.1203 | Exact upper-tail probability by convolution |
| Conditional MLE of rho | 2.859 | Ratio solving sum N_i p_i(rho) = 6 |
CHAPTER 5 OF 12 · WORK THE EXACT TEST
5. Exact confidence limits invert the chosen family of tests
A confidence interval is obtained by changing rho, rebuilding the convolution distribution and finding the values where the selected lower and upper tail rules reach their error limits. The interval and test must use compatible tail definitions. Otherwise a value may be rejected by the test yet remain inside its advertised interval.
Sparse data make the coverage function step-shaped. Endpoints can jump when an attainable value of S crosses the tail boundary. Conservative coverage is not a software bug; it follows from discreteness. Randomisation or mid-p adjustments can smooth the size-power trade-off, but they change the inferential contract. See exact confidence-interval inversion and mid-p tests before mixing methods.
CHAPTER 6 OF 12 · STRESS THE ASSUMPTIONS
6. Sparse-stratum weighting is produced by the model
The conditional likelihood does not assign a homemade weight such as total exposure or inverse variance and then pretend it is exact. Each stratum contributes through N_i and p_i(rho). A stratum with zero total events contributes no conditional information about rho. A stratum with events on only one side can still influence the likelihood, but its contribution remains bounded by its total event count.
This is why “more person-time” is not synonymous with “more information.” If a very large stratum records no events, the conditional test has nothing to allocate after conditioning on N_i = 0. An unconditional analysis may use that information differently because it models the totals rather than fixing them.
For sparse follow-up data, Greenland and Robins showed why Mantel–Haenszel-type estimators can remain consistent while some weighted least-squares alternatives are biased. The lesson is not that one named estimator always wins; it is that weights must follow the sampling model.
CHAPTER 7 OF 12 · STRESS THE ASSUMPTIONS
7. Changed context: one stratum has a different effect
Imagine that stratum C reflects a different mechanism and its true rate ratio is much larger than the ratios in A and B. A common-ratio test can still produce one summary, but the number answers an average-effect question that may no longer be scientifically useful.
Check homogeneity before interpreting the common estimate as universal. With very sparse strata, a homogeneity test may have little power. Use subject knowledge, study design and visible stratum-specific estimates as well as a p-value. Martin and Austin developed exact common-rate-ratio limits and exact homogeneity tests for series of person-time tables; their method makes this distinction explicit.
A practical audit asks: if the most unusual stratum is removed, does the estimate move a little, or does the scientific conclusion reverse? Sensitivity is information, not an instruction to delete inconvenient data.
CHAPTER 8 OF 12 · STRESS THE ASSUMPTIONS
8. Changed context: many zero-event strata
When many strata have N_i = 0, the conditional analysis may collapse to a much smaller effective dataset. If the zero-event pattern is scientifically meaningful—perhaps because surveillance duration differs or risk is genuinely low—an unconditional Poisson model, hierarchical model or prespecified aggregation may be more informative.
Do not add arbitrary half-events merely to make ratios finite without explaining the estimand that the correction changes. A zero in one cell is not the same as no information in the entire stratum. Keep exposure times, surveillance rules and ascertainment visible.
The transfer question is simple: what became fixed by conditioning, and what evidence disappeared with it?
CHAPTER 9 OF 12 · CHECK LIMITS AND EVIDENCE
9. Limits and misconceptions
Exact does not mean assumption-free. The Poisson model assumes that the count model and exposure measurement are appropriate; conditioning assumes independent strata and a common rho under the tested model. Overdispersion, recurrent-event dependence, migration between strata or shared shocks can invalidate that reference distribution.
Common does not mean identical baseline rates. The method allows baseline rates to differ. Conversely, a significant common-effect test does not prove homogeneity. It only addresses the proposed null using the specified ordering rule.
Finally, exact tail control can be conservative. Unconditional procedures may improve power while retaining finite-sample validity, but nuisance-parameter handling must be explicit. Berger–Boos p-values provide one principled route when a nuisance parameter cannot be eliminated cleanly.
CHAPTER 10 OF 12 · CHECK LIMITS AND EVIDENCE
10. Evidence and verification route
The central exact stratified construction is documented by Martin and Austin, Exact estimates for a rate ratio. Greenland and Robins’ Estimation of a common effect parameter from sparse follow-up data explains sparse-data consistency and weighting. Gu, Ng, Tang and Schucany compare conditional exact, likelihood-ratio and approximate tests in Testing the ratio of two Poisson rates.
Reproducibility checks are short: preserve the event and exposure ledger; recompute every p_i(rho); confirm the polynomial coefficients sum to one; verify the observed tail from those coefficients; and invert the same tail rule used by the test. Record software version and whether the reported interval is conditional, unconditional, mid-p or randomised.
CHAPTER 11 OF 12 · TRANSFER AND CONTINUE
11. Frequently asked questions
Can I just pool all events and exposure?
Only if that estimand and weighting are justified. Pooling can obscure stratum-specific baselines and effect heterogeneity.
Why do strata have different binomial probabilities under one rho?
Because p_i(rho) includes the stratum’s exposed and comparison person-times. One rate ratio does not erase unequal exposure.
What does a zero-total stratum contribute conditionally?
Nothing to the allocation statistic after conditioning on N_i = 0. It may still matter to an unconditional model or to study-design interpretation.
Is the conditional maximum-likelihood estimate enough?
No. Report uncertainty, the tail or ordering rule, a homogeneity check and the exposure ledger that makes the estimate interpretable.
CHAPTER 12 OF 12 · TRANSFER AND CONTINUE
12. Final transfer check and connected reading
Before accepting a common exact rate ratio, answer five questions: Were events and exposure measured comparably? Is the common-ratio model scientifically defensible? Which quantities were conditioned on? Which two-sided ordering rule was used? Does the conclusion survive a changed-context check on the most influential stratum?
Continue through the Probability root, the two-rate guide on conditional counts and unconditional coverage, and the broader Gold Standard of Education. The learning goal is not to memorise a named test. It is to make the exposure clock, conditioning step and exact reference distribution visible enough to transfer into a new study.
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