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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 $\hat{\beta }=\log (133/67)\simeq 0.6857$.

For nonnegative $y_i$ summing to $m$, divide the joint Poisson mass by $P(M=m)=e^{-{\lambda }_{+}}{\lambda }_{+}^m/m!$:This is independent Poisson conditioning.

For independent aggregate Poisson distributions with means $n\lambda$ and $n\lambda e^{\beta }$, conditioning on the total eliminates $\lambda$. Independent Poisson conditioning gives the displayed binomial distribution. Its conditional likelihood is proportional to $e^{\beta Y_1}(1+e^{\beta }{)}^{-m}$, also the profile likelihood after maximizing over the unrestricted positive baseline rate. Unequal exposures replace the binomial success probability by $n_1{e}^{\beta }/(n_0+n_1{e}^{\beta })$.