3.5 - Relative Risk

Consider for example the yearly risk of cancer death to be .005 in one population and .01 in another population; most health officials would consider this difference to be important, even though \(\delta = .005\) is very small. This suggests the use of the relative risk of a "success" which is the ratio of the relevant conditional probabilities,

Like \(\delta\), this is a function only of the parameters of \(P(Z | Y )\), so likelihood-based inferences about \(\rho\) will be the same under Poisson, multinomial, or product-multinomial (\(n_{i+}\)s fixed) sampling.

The natural (maximum-likelihood) estimate of \(\rho\) uses observed data:

\(\hat{\rho}=r=\dfrac{n_{11}/n_{1+}}{n_{21}/n_{2+}}\)

Because \(\rho\), and thus \(r\), are non-negative, a normal approximation for \(\log r\) tends to work better than for \(r\) itself. So in this case, it makes sense to transform the data using the natural log (ln) which will allow us to use a less skewed distribution. The approximate variance of \(\log r\) is

\(V(\log r)\approx \dfrac{1-\pi_{11}/\pi_{1+}}{n_{1+}\pi_{11}/\pi_{1+}}+\dfrac{1-\pi_{21}/\pi_{2+}}{n_{2+}\pi_{21}/\pi_{2+}}\)

which is estimated by

\begin{align}
\hat{V}(\log r)&= \dfrac{1-n_{11}/n_{1+}}{n_{1+}n_{11}/n_{1+}}+\dfrac{1-n_{21}/n_{2+}}{n_{2+}n_{21}/n_{2+}}\\
&= \dfrac{1}{n_{11}}-\dfrac{1}{n_{1+}}+\dfrac{1}{n_{21}}-\dfrac{1}{n_{2+}}\\
\end{align}

An approximate 95% confidence interval for \(\log \rho\) is

\(\log r\pm 1.96\sqrt{\hat{V}(\log r)}\)

and the corresponding interval for \(\rho\) is found by exponentiating the endpoints.

Equivalence to statistical independence testing

Testing \(H_0 \colon \rho = 1\) is equivalent to testing for independence in a \(2 \times 2\) table. For our current example, \(r = \dfrac{0.12}{0.22}= 0.5523\) and has corresponding 95% CI \((0.3209, 0.9506)\). Thus, at 0.05 level, we can reject the independence model and furthermore can specifically say that the skiers who took vitamin C are less likely (nearly two times less likely) to get a cold in comparison with the skiers who did not take vitamin C. Notice that we could have computed \(\rho = \dfrac{0.22}{0.12} = 1.8\), which would lead to the same conclusion.

Keep in mind that relative risk is a better measure of association than the difference in proportions when cell probabilities are close to 0 and 1, i.e., when they are in the tails of the probability distribution. Relative risk has far more application in medical literature and epidemiology than difference of proportions.