Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 7

Let $M={\max }_i{X}_i$. For an observed sample in the nonnegative orthant, the likelihood function under the uniform distribution isIt is zero for $\theta <M$, and decreases with $\theta$ on $[M,\infty )$. Thus the maximum likelihood estimation problem is solved at the smallest admissible value:Under the model $M>0$ with probability one. The all-zero sample is a null event on which the likelihood function has no finite maximizer in the parameter space $\theta >0$.

For $0\le t\le 1$, independence and the uniform distribution giveTherefore the cumulative distribution function isIt contains no $\theta$, so $M/\theta$ is a pivotal quantity.

Differentiating the cumulative distribution function in part (ii) gives the probability density function $n{t}^{n-1}$ on $(0,1)$. HenceThus an unbiased estimator based on the sample maximum is

Each observation from the uniform distribution has mean $\theta /2$. By linearity of expectation, $\mathbb{E}{\sum }_i{X}_i=n\theta /2$, so another unbiased estimator is

The pivotal quantity $U=M/\theta$ lies in $[0,1]$ and satisfies $\mathbb{P}(U\le t)=t^n$. ThusRearranging these inequalities yields the exact confidence intervalFor every $\theta >0$, its coverage probability is exactly $1-\alpha$; the result uses the finite-sample probability distribution, with no normal approximation.

The log-likelihood function for the Laplace distribution isThus maximum likelihood estimation is equivalent to minimizing the sum of absolute values. Write $n=2m+1$ and sort the data as $Y_{(1)}\le \cdots \le Y_{(n)}$. Between observations, if $r$ values lie to the left of $\kappa$, the derivative of ${\sum }_i|Y_i-\kappa |$ is $r-(n-r)=2r-n$. It is negative before the middle observation and positive after it. The minimum occurs at the sample median:This minimizer remains unique for odd $n$ even when some observations coincide, since fewer than half the observations can lie strictly on either side of the middle value.