2.4 Misclassification and its effects
When a binary exposure or disease is measured with error, the \(2 \times 2\) table is misclassified, and the measures of association computed from it are biased. Whether the error depends on the other variable is the key distinction.
Definition 2.4.1. Misclassification of exposure is nondifferential if its sensitivity and its specificity are the same for cases and non-cases; misclassification of disease is nondifferential if they are the same for exposed and unexposed. Otherwise it is differential.
Exposure taken from records made before the disease appeared, such as employment files or prescriptions, is usually nondifferentially misclassified. Exposure recalled at interview after diagnosis may not be: a mother whose child has a birth defect searches her memory of the pregnancy more thoroughly than the mother of a healthy child. This is recall bias, the commonest form of differential misclassification.
Theorem 2.4.2 (Nondifferential misclassification of disease). In a cohort study, let disease be classified with sensitivity Se and specificity Sp in both exposure groups, with \(J = \text {Se} + \text {Sp} - 1 > 0\). The observed risks are \(p_i^* = (1 - \text {Sp}) + J p_i\) for \(i = 0, 1\), and
- 1.
- the observed risk difference is \(J\) times the true one;
- 2.
- the observed risk ratio lies between \(1\) and the true risk ratio, and equals it if \(\text {Sp} = 1\).
Proof. The expression for \(p_i^*\) is (2.1) applied within exposure group \(i\). Then \(p_1^* - p_0^* = J(p_1 - p_0)\), which is (1). For (2), write \(s = 1 - \text {Sp} \geq 0\) and suppose \(p_1 > p_0\). Then \(p_1^*/p_0^* = (s + Jp_1)/(s + Jp_0) > 1\), and \[ \frac {s + Jp_1}{s + Jp_0} \leq \frac {p_1}{p_0} \iff p_0 (s + Jp_1) \leq p_1 (s + Jp_0) \iff s\,p_0 \leq s\,p_1 , \] which holds, with equality if and only if \(s = 0\). The case \(p_1 < p_0\) is the same with the inequalities reversed. □
The second part has a practical consequence. When the effect is measured as a ratio, specificity matters far more than sensitivity: a case definition that misses some cases but admits no false ones leaves the risk ratio unbiased. Studies therefore often use strict case definitions, accepting that some true cases are lost.
Theorem 2.4.3 (Nondifferential misclassification of exposure). In a case-control study, let exposure be classified with sensitivity Se and specificity Sp among both cases and controls, with \(J = \text {Se} + \text {Sp} - 1 > 0\), and let \(q_1, q_0 \in (0, 1)\) be the true probabilities of exposure among cases and controls. Then the observed odds ratio lies between \(1\) and the true odds ratio.
Proof. The observed exposure probabilities are \(q_i^* = f(q_i)\), where \(f(q) = a + Jq\) with \(a = 1 - \text {Sp}\); note \(1 - f(q) = b + J(1 - q)\) with \(b = 1 - \text {Se}\), and \(a + b = 1 - J\). Let \(g(\theta ) = \operatorname {logit} f(q)\) where \(\theta = \operatorname {logit} q\). The log odds ratios are \(\theta _1 - \theta _0\) (true) and \(g(\theta _1) - g(\theta _0)\) (observed), so it is enough to show \(0 < g'(\theta ) \leq 1\); the mean value theorem then gives \(g(\theta _1) - g(\theta _0) = g'(\xi )(\theta _1 - \theta _0)\), of the same sign and no larger in size.
Since \(d\theta /dq = 1/\{q(1 - q)\}\) and \(d \operatorname {logit} f/dq = J/\{f(1 - f)\}\), \[ g'(\theta ) = \frac {J\,q(1 - q)}{f(q)\{1 - f(q)\}} . \] This is positive. Expanding the denominator, \[ f(1 - f) = (a + Jq)\{b + J(1 - q)\} = ab + aJ(1 - q) + bJq + J^2 q(1 - q) . \] Now \(aJ(1 - q) \geq aJ\,q(1 - q)\) and \(bJq \geq bJ\,q(1 - q)\), so \begin {align*} f(1 - f) &\geq (a + b)J\,q(1 - q) + J^2 q(1 - q) \\ &= (1 - J + J)\,J\,q(1 - q) = J\,q(1 - q) , \end {align*}
and \(g'(\theta ) \leq 1\). □
So nondifferential misclassification of a binary exposure biases the odds ratio towards the null, and a study that finds an association in spite of it has, if anything, understated it. Three cautions keep this result in its place. It is a statement about the expected value; in a particular small study random error can carry the estimate past the true value. It concerns a binary exposure; with three or more categories misclassification between adjacent levels can bias the middle category away from the null. And it assumes the errors in exposure are independent of any errors in disease.
Example 2.4.4. In a cohort study the true risks are \(p_1 = 0.10\) and \(p_0 = 0.05\) (\(\RR = 2\), \(\RD = 0.05\)). Disease is ascertained with \(\text {Se} = 0.80\) and \(\text {Sp} = 0.95\), so \(J = 0.75\). The observed risks are \begin {gather*} p_1^* = 0.05 + 0.75 \times 0.10 = 0.125, \\ p_0^* = 0.05 + 0.75 \times 0.05 = 0.0875 , \end {gather*} giving an observed risk ratio of \(1.43\) and a risk difference of \(0.0375 = 0.75 \times 0.05\).
In a case-control study with true \(q_1 = 0.4\) and \(q_0 = 0.2\) (\(\OR = 2.67\)), exposure is classified with \(\text {Se} = 0.8\) and \(\text {Sp} = 0.9\) in both groups. Then \(q_1^* = 0.1 + 0.7 \times 0.4 = 0.38\), \(q_0^* = 0.1 + 0.7 \times 0.2 = 0.24\), and the observed odds ratio is \((0.38/0.62)/(0.24/0.76) = 1.94\).
If instead the misclassification is differential (cases recall exposure with sensitivity \(0.95\), controls with \(0.75\), specificity \(0.9\) in both) and the exposure has no effect, \(q_1 = q_0 = 0.3\), then \(q_1^* = 0.355\) and \(q_0^* = 0.295\), and the observed odds ratio is \(1.32\): an association created entirely by the way exposure was measured.
Correcting for misclassification
When Se and Sp are known, from a validation study in which a subsample is measured both ways, the expected observed counts can be inverted. Among \(m\) subjects of whom \(x^*\) are classified exposed, the expected number truly exposed is \[ x = \frac {x^* - (1 - \text {Sp})\,m}{J} , \] which is Theorem 2.2.4 written in counts. Applying it separately to cases and controls (with their own Se and Sp if the misclassification is differential) gives a corrected table.
Example 2.4.5. A case-control study with \(200\) cases and \(200\) controls observes \(76\) exposed cases and \(48\) exposed controls, an odds ratio of \((76 \times 152)/(124 \times 48) = 1.94\). If exposure is known to be classified with \(\text {Se} = 0.8\), \(\text {Sp} = 0.9\), the corrected numbers exposed are \[ \frac {76 - 0.1 \times 200}{0.7} = 80 \quad \text {(cases)}, \qquad \frac {48 - 0.1 \times 200}{0.7} = 40 \quad \text {(controls)}, \] and the corrected odds ratio is \((80 \times 160)/(120 \times 40) = 2.67\), recovering the second part of Example 2.4.4. The correction is only as good as the values of Se and Sp; small errors in them, especially in Sp when exposure is rare, can produce impossible (negative) corrected counts.
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