 Solution

For $n$ observed complete independent exponential gaps, let $T_n={\sum }_{i=1}^n{\tau }_i$. The likelihood function is $L(\lambda )={\lambda }^n{e}^{-\lambda T_n}$, and the log-likelihood has derivative $n/\lambda -T_n$ and second derivative $-n/{\lambda }^2<0$. Hence the unique maximum-likelihood estimator isAlternatively, observe $m$ unit-year increments $Y_j=N_j-N_{j-1}$. These are independent Poisson counts of mean $\lambda$. Their joint likelihood is proportional to ${\lambda }^{\sum Y_j}{e}^{-m\lambda }$, givingIt is unbiased, with variance $\lambda /m$. A zero observed count gives the boundary estimate zero, or the supremum as $\lambda \downarrow 0$ if the parameter space is restricted to positive rates.

The Poisson observation scheme and rate likelihood distinction matters when comparing the techniques. For a fixed observation horizon, conditional on the total number of events the event locations have a distribution independent of $\lambda$. Thus exact timestamps add no rate information to the total count under the homogeneous Poisson model; unit-interval counts already contain that total. The correct fixed-horizon likelihood includes the final unfinished waiting interval, and gives count divided by the full exposure, not by the sum of only completed gaps.

The first calculation instead stops after a fixed number $n$ of events, so its exposure $T_n$ is random. Its estimate has mean $n\lambda /(n-1)$ for $n>1$ and an infinite mean for $n=1$, unlike the unbiased fixed-horizon count estimate. For $n>2$ its variance is $n^2{\lambda }^2/[(n-1{)}^2(n-2)]$, asymptotic to ${\lambda }^2/n$. There is no universally more informative sampling technique without specifying the stopping rule and exposure. Counts are sufficient and convenient for estimating a truly constant rate; timestamps additionally help diagnose departures from constant intensity or independent gaps.