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

Ignoring a factor independent of the probabilities, the likelihood function is $L(\pi )={\prod }_i{\pi }_i^{n_i}$, so its log is $\ell (\pi )={\sum }_i{n}_i\log {\pi }_i$ subject to ${\sum }_i{\pi }_i=1$. A Lagrange multiplier gives $n_i/{\pi }_i=\lambda$. Summing ${\pi }_i=n_i/\lambda$ makes $\lambda =n$.

ThusSince all counts are positive, a zero fitted cell probability would have zero likelihood. On the interior the log-likelihood is strictly concave, so this stationary point is its unique global maximum.

The unrestricted likelihood is maximized at the empirical proportions, while the null likelihood is maximized at ${\pi }_i(\hat{\theta })$. Twice the log of their ratio isLarge values mean the fitted null likelihood is much smaller than the unrestricted maximum, so rejection is in the upper tail.

For a regular identifiable scalar parameter lying in the interior, positive true cell probabilities, and sufficiently large expected counts, Wilks theorem gives an approximate chi-squared distribution with $(k-1)-1=k-2$ degrees of freedom under the null. The unrestricted probability vector has dimension $k-1$ and the fitted null has dimension one. At significance level $\alpha$, reject if $\Lambda (\hat{\theta })>F_{k-2}^{-1}(1-\alpha )$, or report the approximate p-value $1-F_{k-2}(\Lambda )$.

Cells with very small expected counts may require a prespecified pooling scheme or simulation calibration; the asymptotic reference is not an exact finite-sample law. For a regular null with $d$ fitted parameters, the corresponding degrees of freedom are $k-1-d$.

For a fully specified parameter value, do not maximize within the null. Substitute ${\theta }_0$ directly:Under the same regularity and large-count conditions, compare with ${\chi }_{k-1}^2$, since no null parameter has been estimated. Reject in its upper tail at the desired significance level. Fixing the parameter increases the reference degrees of freedom by one relative to part (b).

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.

There are twelve cells and one fitted parameter, so the likelihood-ratio reference is approximately ${\chi }_{10}^2$. The supplied value givesThus do not reject the Poisson goodness-of-fit model at the conventional 5% level; it would also not be rejected at 10%. The data do not provide strong evidence of a departure from this model. This is a compatibility conclusion, not proof that the Poisson model is true. A fully specified Poisson mean would instead use eleven degrees of freedom, which is not the fitted situation here.