Past exam of the mathematical and Computational Biology course of the University of Cambridge / 2025 / Paper 1 / 2

Using the probability density function of the Erlang distribution, the expected value isHere $u=\theta x$, and two applications of integration by parts give ${\int }_0^{\infty }{u}^2{e}^{-u}du=2{\int }_0^{\infty }u{e}^{-u}du=2$. The boundary terms vanish because exponential decay dominates these powers.

Put $S={\sum }_{i=1}^N{x}_i$. Since the observations are independent random variables, their joint probability density function is a product. Therefore the likelihood function and log-likelihood areThese expressions assume positive observed values, as occurs with probability one under this continuous Erlang distribution.

The score function and the observed Fisher information areSolving the score equation gives the maximum-likelihood estimateThe positive observed Fisher information means that the log-likelihood is strictly concave on $\theta >0$. Its unique stationary point is therefore the global maximum. Equivalently, the score function changes from positive to negative at this point. At the estimate, $j(\hat{\theta })=S^2/(2N)$.

Test the null hypothesis $H_0:{\theta }_m={\theta }_f=\theta$ against the alternative hypothesis that the two rate parameters differ. Under $H_0$, the pooled maximum-likelihood estimate is ${\hat{\theta }}_0=16/36.8\simeq 0.434783$. Under the alternative, the estimates are ${\hat{\theta }}_m=8/22\simeq 0.363636$ and ${\hat{\theta }}_f=8/14.8\simeq 0.540541$.

In the likelihood-ratio test statistic, the data-only terms cancel, and both fitted models have total $\hat{\theta }S=16$. Consequently,The alternative hypothesis adds one parameter, so Wilks theorem gives an approximate chi-squared distribution with one degree of freedom. Do not reject the common-rate model at the 5% level, since $T<3.841$. The corresponding approximate p-value is $0.43$. The fitted expected values are $5.5$ and $3.7$ days, but four observations per group supply little evidence that the underlying expected values differ; failure to reject does not establish equality. The chi-squared distribution calibration is asymptotic and should be treated cautiously for these small samples.