Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-46/1/b/solution

Let and be the aggregate counts in the control and treated groups. Assuming independent physicians, their Poisson distributions have means and . Condition on . By independent Poisson conditioning,
The PDF's full table gives , and . Thus the Two-group Poisson ratio conditional likelihood is
Conditioning the individual observations instead gives a multinomial distribution with cell probabilities in the first group and in the second; its parameter-dependent likelihood is the same. No factor involves . The conditional estimates are and .

New to topics? Read the docs here!