= Solution
Among the 12 pairs with control time first, the eight in which that first time is an event contribute $(1+\beta)^{-1}$ each. Among the eight pairs with treated time first, the four in which that first time is an event contribute $\beta/(1+\beta)$ each. Later events in one-eye risk sets contribute one. Hence
$$
L(\beta)\propto
\left(\frac1{1+\beta}\right)^8
\left(\frac\beta{1+\beta}\right)^4
=\frac{\beta^4}{(1+\beta)^{12}}.
$$
Differentiating the log likelihood gives $4/\beta-12/(1+\beta)=0$, and therefore
$$
\boxed{\widehat\beta=\frac12.}
$$
Without randomization, treatment side can be associated with prognosis. Always treating the left eye confounds treatment with systematic left-right differences; choosing the worse eye creates severe <confounding by indication>, baseline imbalance, and possible <regression toward the mean>. The within-patient comparison then no longer identifies a treatment effect without stronger adjustment assumptions.
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