Past exam of the mathematical and Computational Biology course of the University of Cambridge / 2024 / Paper 1 / A1

A probability density function must have integral one. For $\theta >-1$,The restriction $\theta >-1$ is essential: for $\theta \le -1$ the integral diverges and no positive normalizing constant exists.

For a moment of order $q$ with $\theta +q>-1$, integration of the probability density function givesThus the expected value and variance areThe cumulative distribution function on $(0,1)$ is $F(x)=x^{\theta +1}$. Its unique median $m$ satisfies $F(m)=1/2$, giving

For the independent and identically distributed observations, multiply the probability density functions. On the admissible parameter range $\theta >-1$, the likelihood function and log-likelihood areThese expressions presume that every observation lies strictly between zero and one; an observation outside that support would give zero likelihood under this model.

Write $s={\sum }_i\log x_i<0$. The score function isSetting it to zero gives the maximum-likelihood estimatorThis lies above $-1$. The observed Fisher information isConsequently the log-likelihood is strictly concave throughout its domain. Its stationary point is the unique global maximum, not merely a candidate local maximum. Indeed the log-likelihood tends to minus infinity as either $\theta \downarrow -1$ or $\theta \to \infty$.

The sample mean is $\bar{x}=0.7$. The method of moments equates this to the expected value from part (b):Hence the estimate is $\boxed{\tilde{\theta }=4/3\approx 1.33}$.

Here $s=\log (0.4\times 0.7\times 0.8\times 0.9)=-1.601469743$. The maximum-likelihood estimate and estimated standard error from the observed Fisher information areThe 95% Wald confidence interval is thereforeThe whole reported confidence interval lies in $\theta >-1$, though its lower endpoint is close to the boundary. With only four observations, the normal approximation underlying this Wald confidence interval need not provide accurate 95% coverage.

Keep the true parameter fixed and imagine repeatedly collecting four independent and identically distributed observations under the same model. Compute the maximum-likelihood estimate, its estimated standard error, and the Wald confidence interval afresh each time. The intended coverage statement is that approximately 95% of those random confidence intervals contain the fixed true parameter, provided the normal approximation is adequate.

After these particular observations have been collected, both endpoints are fixed. The frequentist statement does not assign 95% probability to the fixed parameter lying between these particular endpoints. Nor does the confidence interval describe the range containing 95% of future observations. The small sample makes the nominal 95% coverage an approximation whose accuracy is not guaranteed.

Both candidate models fit one parameter, but their parameter symbols have different meanings. For the exponential distribution in model B,Its second derivative is $-N/{\theta }_B^2<0$, so this is its maximum-likelihood estimate. The maximized log-likelihoods, including their normalization terms, areUse the corrected Akaike information criterion because $N=4$ is small relative to the number $k=1$ of fitted parameters:The correction is $2$ for each model, givingPrefer model A, whose corrected Akaike information criterion is smaller by $7.67252$. The ordinary Akaike information criterion also gives that ranking because the two models receive equal penalties. This compares the two proposed models; it does not establish that either is a correct description of the process. The usual corrected Akaike information criterion formula is a small-sample model-selection convention, rather than an exact finite-sample guarantee for these density families.