Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 46 / 1 / b

Let $Y_0$ and $Y_1$ be the aggregate counts in the control and treated groups. Assuming independent physicians, their Poisson distributions have means $12\lambda$ and $12\lambda e^{\beta }$. Condition on $E=\{Y_0+Y_1=m\}$. By independent Poisson conditioning,The PDF's full table gives $Y_0=67$, $Y_1=133$ and $m=200$. Thus the Two-group Poisson ratio conditional likelihood isConditioning the individual observations instead gives a multinomial distribution with cell probabilities $(1-p)/12$ in the first group and $p/12$ in the second; its parameter-dependent likelihood is the same. No factor involves $\lambda$. The conditional estimates are $\hat{p}=133/200$ and $\hat{\beta }=\log (133/67)\simeq 0.6857$.