6.3 Matching
Matching chooses the controls so that their distribution of some variable, usually a strong confounder such as age or sex, is the same as the cases’. In frequency matching the controls are sampled within strata in proportion to the cases; in individual matching each case is given one or more controls with the same value (same sex, age within two years).
Matching in a case-control study does not by itself remove confounding by the matching variable. It makes the cases and controls alike on that variable, but the controls are then no longer a random sample of the base, and their exposure distribution is distorted towards that of the cases. The matched variable must be allowed for in the analysis, by stratification on the matched sets. What matching achieves is efficiency: each stratum of the analysis contains both cases and controls, so none is wasted.
Overmatching means matching on a variable that is associated with exposure but is not a confounder, typically a consequence of exposure or a close correlate of it. It brings the exposure distributions of cases and controls together for no gain, losing efficiency and, if the variable lies on the causal path, biasing the odds ratio towards \(1\).
Matched pairs
With one control per case, each pair falls into one of four types according to the exposure of the case and of the control.
| Control exposed | Control unexposed | |
| Case exposed | \(n_{11}\) | \(n_{10}\) |
| Case unexposed | \(n_{01}\) | \(n_{00}\) |
Concordant pairs (\(n_{11}\), \(n_{00}\)) carry no information about the association; the analysis rests on the discordant pairs.
Theorem 6.3.1. Suppose that within each matched pair the odds of exposure among cases are \(\psi \) times those among controls, with the baseline odds free to vary between pairs. Given that a pair is discordant, the probability that it is the case who is exposed is \(\psi /(1 + \psi )\), whatever the pair’s baseline odds. Hence, conditionally on \(n_{10} + n_{01}\), \(n_{10}\) is binomial with this probability, the conditional maximum likelihood estimate of the odds ratio is \[ \hat \psi = \frac {n_{10}}{n_{01}} , \] and the test of \(\psi = 1\) is McNemar’s test, \(X^2 = (n_{10} - n_{01})^2/(n_{10} + n_{01})\) on one degree of freedom.
Proof. In pair \(i\) let the control’s probability of exposure be \(\pi _i\), with odds \(\omega _i = \pi _i/(1 - \pi _i)\), and the case’s odds \(\psi \omega _i\), so the case’s probability is \(\psi \omega _i/(1 + \psi \omega _i)\). The exposures are independent, so \begin {align*} P(\text {case exposed, control not}) &= \frac {\psi \omega _i}{1 + \psi \omega _i} \cdot \frac {1}{1 + \omega _i} , \\ P(\text {control exposed, case not}) &= \frac {1}{1 + \psi \omega _i} \cdot \frac {\omega _i}{1 + \omega _i} . \end {align*}
Given discordance, the first has probability \(\psi \omega _i/(\psi \omega _i + \omega _i) = \psi /(1 + \psi )\); \(\omega _i\) cancels. The discordant pairs are therefore independent Bernoulli trials with success probability \(\psi /(1 + \psi )\), and the binomial maximum likelihood estimate \(n_{10}/(n_{10} + n_{01})\) of this probability gives \(\hat \psi = n_{10}/n_{01}\). Under \(\psi = 1\) the probability is \(\tfrac 12\), and the score test of a binomial proportion equal to \(\tfrac 12\) is McNemar’s statistic. □
Conditioning eliminates the pair-specific odds \(\omega _i\), which could not be estimated, since each pair has only two people. Ignoring the matching and computing the crude odds ratio from the four margins does not estimate \(\psi \) in general, and is biased towards \(1\) when the matching variable is related to exposure.
Example 6.3.2. In \(100\) matched pairs, the case alone is exposed in \(30\) pairs and the control alone in \(12\). Then \(\hat \psi = 30/12 = 2.5\), and McNemar’s statistic is \((30 - 12)^2/42 = 7.71\), \(P = 0.005\). The approximate interval uses \(\text {SE}(\log \hat \psi ) = \sqrt {1/30 + 1/12} = 0.342\), giving \((1.28,\ 4.88)\). An exact interval comes from the Clopper–Pearson interval for \(\pi = \psi /(1 + \psi )\) with \(30\) successes in \(42\), which is \((0.554,\ 0.843)\), transformed by \(\psi = \pi /(1 - \pi )\) to \((1.24,\ 5.36)\).
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