= Solution
For $n$ observed complete independent exponential gaps, let $T_n=\sum_{i=1}^n\tau_i$. The <likelihood function> is $L(\lambda)=\lambda^ne^{-\lambda T_n}$, and the <log-likelihood> has derivative $n/\lambda-T_n$ and second derivative $-n/\lambda^2<0$. Hence the unique <maximum-likelihood estimator> is
$$
\boxed{\widehat\lambda=n/T_n=1/\overline\tau.}
$$
Alternatively, 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 $\lambda^{\sum Y_j}e^{-m\lambda}$, giving
$$
\boxed{\widehat\lambda=\frac1m\sum_{j=1}^mY_j=N_m/m.}
$$
It is unbiased, with <variance> $\lambda/m$. A zero observed count gives the boundary estimate zero, or the supremum as $\lambda\downarrow0$ 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.
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