Sparse Matched-Set Exact Inference Workshop

From Matched Sets to Exact Inference

In a matched case-control study, each set contains one case and one or more controls selected to match that case on specified characteristics. When exposure is sparse, a large-sample approximation can be unreliable, and a small number of matched sets may carry most of the information about the exposure effect. Conditional inference offers a direct route: condition on there being one case in each set, write the probability of each possible case assignment, and enumerate the resulting sample space.

We use a binary exposure \(x_{ij}\), where \(x_{ij}=1\) means member \(j\) of set \(i\) is exposed. Let \(\beta\) be the logistic exposure effect and write \(\theta=\exp(\beta)\), the matched odds ratio. The conditional probability below follows from the matched-set logistic model after conditioning on one case per set; the set-specific intercept cancels.

Worked Example: Constructing the Conditional Likelihood

For these data, the exposure score is \(T=\sum_i x_{i,\text{case}}\), the number of exposed cases. Its observed value is \(T=2\). Given the exposure pattern in each set, every possible response allocation has probability proportional to \(\theta^T\), with a normalizing denominator that depends on the covariates and \(\theta\). This makes it possible to derive the distribution of \(T\) by counting allocations.

To verify the distributional form, consider a complete allocation \(a\) that selects one case member in every set. Its probability is the product of the set probabilities, hence $$ P_\theta(a)= \frac{\theta^{T(a)}}{\prod_i\sum_j\theta^{x_{ij}}}. $$ The denominator is the same for every allocation with the same observed covariate pattern. Summing over the allocations with \(T(a)=t\) gives \(c_t\theta^t\) divided by that common denominator. Summing over all allocations shows that the denominator equals \(\sum_t c_t\theta^t\), proving the stated distribution. \(\square\)

Worked Example: Enumerating the Exact Sample Space

An Exact Test and Its Conditional Justification

For a prespecified alternative that exposure increases the odds of being the case, an exact upper-tail test of \(H_0:\theta=1\) uses \(P_{\theta=1}(T\geq T_{\text{obs}})\). The test is exact in the sense that its null distribution is calculated from the conditional sample space, not approximated by a continuous distribution.

The calibration of the upper-tail p-value follows directly from discreteness. Under any fixed null value, let \(q(t)=P(T\geq t)\). Since \(q(t)\) decreases as \(t\) increases, the event \(q(T)\leq a\) consists of an upper tail of possible values. Its probability is the probability of that tail, which is at most \(a\); the inequality may be strict because the distribution places probability at discrete points. Thus a test that rejects when \(q(T)\leq a\) has null rejection probability no greater than \(a\).

For the observed maximum score in this example, a probability-ordering two-sided p-value and a doubled one-sided p-value need not agree in general. Here the two point masses at \(T=0\) and \(T=2\) are equal under the null, so probability-ordering includes both and gives \(4/9\); doubling the smaller one-sided tail also gives \(4/9\). The convention should be stated when a two-sided exact p-value is reported.

Worked Example: Inverting the Test for a Confidence Interval

Here is the coverage argument behind this inversion. For a finite conditional exponential family \(P_\theta(T=t)=c_t\theta^t/Z(\theta)\), the distribution shifts toward larger \(T\) as \(\theta\) increases. In fact, for any nondecreasing function \(g\), $$ \frac{d}{d\log\theta}E_\theta[g(T)] =\operatorname{Cov}_\theta(g(T),T)\geq0. $$ The equality follows by differentiating the finite sum defining the expectation. The covariance is nonnegative because if \(T'\) is an independent copy of \(T\), then \(2\operatorname{Cov}(g(T),T)=E[(g(T)-g(T'))(T-T')]\geq0\). Taking \(g(T)=I(T\geq t)\) proves that upper tails are nondecreasing in \(\theta\); applying the same argument to lower-tail indicators shows lower tails are nonincreasing. \(\square\)

To see why inversion gives at least 95% coverage, set \(\alpha=0.05\) and form the lower endpoint from the upper-tail equation \(P_\theta(T\geq t)=\alpha/2\), and the upper endpoint from the lower-tail equation \(P_\theta(T\leq t)=\alpha/2\), using parameter-space boundaries when an equation has no solution. At the true parameter, the probability that the lower endpoint exceeds the truth is at most \(\alpha/2\): that event requires an upper-tail p-value no greater than \(\alpha/2\), whose probability is at most \(\alpha/2\) by the discrete-tail argument. The probability that the upper endpoint falls below the truth is likewise at most \(\alpha/2\). By adding these two error probabilities, the interval covers the true parameter with probability at least \(1-\alpha=0.95\). Discreteness can make coverage larger than 95%.

Why Sparsity and Allocation Checks Matter

When matched sets are sparse, the conditional sample space may be small enough to enumerate, but it is easy to count it incorrectly. In the example, B has two exposed members. Those are two distinct possible case assignments, even though they have the same exposure value. Treating them as one allocation would change the coefficient of \(\theta\), and hence the exact distribution and p-value.

A second useful check is to verify the normalizing polynomial in two ways: multiply the set-specific sums, \((\theta+2)(2\theta+1)\), and add the score-specific allocation weights, \(2+5\theta+2\theta^2\). They agree. Also check that probabilities are nonnegative and sum to one at the null value. These elementary checks are especially valuable when one or two allocations can materially change an exact result.