1.5 Association and causation
An association between exposure and disease, however strong, is not by itself evidence that the exposure causes the disease. To say precisely what causation would mean, imagine each person’s outcome under each exposure.
Definition 1.5.1 (Potential outcomes). For each person let \(Y^1\) be the disease indicator that would be observed if the person were exposed and \(Y^0\) the indicator if unexposed. The exposure has a causal effect on a person if \(Y^1 \neq Y^0\). In a population the causal risk ratio and causal risk difference are \[ \frac {P(Y^1 = 1)}{P(Y^0 = 1)} \qquad \text {and}\qquad P(Y^1 = 1) - P(Y^0 = 1) , \] comparing the risk if everyone were exposed with the risk if no one were.
Only one of \(Y^1, Y^0\) is ever seen for a given person: under consistency, the observed outcome is \(Y = Y^1\) for the exposed and \(Y = Y^0\) for the unexposed. The observed (associational) risk ratio \(P(Y = 1 \mid E)/P(Y = 1 \mid \bar E)\) compares different people, and equals the causal risk ratio only under a condition on how exposure was assigned.
Definition 1.5.2. Exposure is exchangeable with respect to the outcome if \((Y^1, Y^0)\) is independent of \(E\): the exposed and unexposed would have had the same distribution of outcomes under either exposure.
Theorem 1.5.3. Under consistency and exchangeability the associational risk ratio and risk difference equal the causal ones.
Proof. By consistency \(P(Y = 1 \mid E) = P(Y^1 = 1 \mid E)\), and by exchangeability \(P(Y^1 = 1 \mid E) = P(Y^1 = 1)\). Likewise \(P(Y = 1 \mid \bar E) = P(Y^0 = 1 \mid \bar E) = P(Y^0 = 1)\). The associational measures are therefore built from the same two numbers as the causal ones. □
Randomisation makes the exposure independent of everything that exists before it, including \((Y^1, Y^0)\), so a randomised trial delivers exchangeability by design. An observational study does not, and when exchangeability fails the associational measure is said to be confounded. Confounding is treated formally in a later section. The practical response, used throughout the course, is to find variables \(C\) (age, sex, social class) such that exchangeability holds within the strata of \(C\), and to compare like with like there.
Judging causation from observational evidence
Because exchangeability can never be verified from observational data, causal conclusions there rest on judgement. Hill (1965) listed aspects of an association that make a causal reading more or less plausible: its strength (a large risk ratio is harder to explain away by confounding), consistency across studies and populations, specificity, temporality (exposure must precede disease, the only one that is necessary), a biological gradient (dose–response), plausibility, coherence with other knowledge, experimental evidence and analogy. They are not criteria to be scored. Each has exceptions, and a modern reading treats them as questions about bias: could confounding, selection or measurement error produce an association this strong, this consistent, with this gradient?
Example 1.5.4 (The ecological fallacy). Three regions have the following proportions \(x\) of the population exposed and annual disease risks \(r\):
| Region | A | B | C |
| Proportion exposed \(x\) | \(0.2\) | \(0.5\) | \(0.8\) |
| Disease risk \(r\) | \(0.01\) | \(0.02\) | \(0.03\) |
The least squares line through the three points is \(r = 0.00333 + 0.0333\,x\). Read naively, a wholly unexposed region (\(x = 0\)) has risk \(0.00333\) and a wholly exposed one (\(x = 1\)) has risk \(0.0367\), an ecological risk ratio of \(0.0367/0.00333 = 11\).
Now suppose that within each region the risk is the same in exposed and unexposed people: \(0.01\) in A, \(0.02\) in B, \(0.03\) in C. The data above are exactly what this produces, yet the exposure has no effect on anyone. The regional risks differ for some other reason that happens to rise with the proportion exposed, such as the age structure of the regions. Group-level data cannot separate the two explanations, because they do not record who among the exposed became ill.
Proposition 1.5.5. Let region \(j\) have exposure proportion \(x_j\) and risks \(p_{1j}\), \(p_{0j}\) among its exposed and unexposed. If \(p_{1j} = p_1\) and \(p_{0j} = p_0\) in every region, then the regional risks lie exactly on the line \(r = p_0 + (p_1 - p_0)x\), and the ecological risk ratio equals \(p_1/p_0\).
Proof. The risk in region \(j\) is the average of the two group risks weighted by group size, \(r_j = p_{0j}(1 - x_j) + p_{1j}\,x_j\). With common \(p_1, p_0\) this is \(p_0 + (p_1 - p_0)x_j\), linear in \(x_j\) with intercept \(p_0\) and value \(p_1\) at \(x = 1\). □
The ecological analysis is sound when the baseline risk \(p_{0j}\) does not vary with the exposure proportion, and the effect does not either. Both are assumptions about individuals that group data cannot check. In Example 1.5.4 \(p_{0j}\) rises with \(x_j\), so the condition fails.
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