Past exam of the options II course of the University of Oxford / 2013 / O5 / d / i

The final cell is a grouped tail. Thus use twelve probabilitiesMaximize the grouped Grouped Poisson likelihoodthen take $\boxed{{\hat{\pi }}_i=p_i(\hat{\theta })}$ for all twelve cells. For example, the interior likelihood equation isbecause $p_{11}^{\prime }(\theta )=p_{10}(\theta )$. It can be solved numerically, with no numerical work required here.

It would be incorrect to treat the last six observations as exactly eleven. If the underlying ungrouped particle counts were available, their sample mean would be the usual Poisson maximum likelihood estimate. For the data as supplied, the grouped likelihood handles the unknown tail counts directly.