Relative risk from group compositions ($\mathrm{RR}=\frac{a(1-p)}{p(1-a)}$)

Let $G$ be the event of membership in the first of two groups and $A$ the event of selection. Write $p=P(G)$, $a=P(G\mid A)$ and $q=P(A)>0$, with $0<p,a<1$. Bayes' theorem gives the conditional probabilities $r_1=P(A\mid G)=aq/p$ and $r_0=P(A\mid G^c)=(1-a)q/(1-p)$. Their risk ratio is the displayed expression: the unknown overall selection probability cancels. Equivalently, it is the ratio of the group-membership odds after selection to those before selection.

When both selection probabilities lie strictly between zero and one, the selection odds ratio is $[r_1/(1-r_1)]/[r_0/(1-r_0)]$ and generally depends on $q$. Thus compositions alone do not identify that odds ratio or provide sample sizes for a standard error. When $a<p$, the first group's observed selection rate is lower; this comparison alone does not establish a causal effect of group membership.

Suppose $0<p_L\le p\le p_U<1$ and $0<a_L\le a\le a_U<1$. The formula for relative risk from group compositions increases with $a$ and decreases with $p$, since $a/(1-a)$ increases and $(1-p)/p$ decreases. This monotonicity proves the displayed bounds. If percentages are rounded to the nearest whole percentage point, each corresponding proportion lies within $0.005$ of its displayed value. These bounds describe rounding uncertainty; they are not a confidence interval for sampling uncertainty.