There is a Poisson process I'm observing for a window $4.5$ units in length. I'm conditioning on the fact that the first event happened within $1.5$ units from the start of the time window. Meaning that if the first event doesn't happen before $1.5$ units, the observation would be discarded. In my valid observation (where the condition above was first satisfied), the first event happened at time, $t=1$. And the second event didn't happen until the end of the interval. I want the rate, $\mu$ that will maximize the likelihood of making these observations.
My attempt
We know that the first event has to happen before $t=1.5$. The only question is, where inside this $1.5$ unit window the first event happens. This is a truncated exponential distribution. The PDF of this is given by (divide regular PDF by CDF at $1.5$):
$$f_{T_1}(t) = \frac{\mu e^{-\mu t}}{1-e^{-1.5 \mu}}$$
And the second event from there (time taken to it: $T_2$) is just an exponential distribution with rate $\mu$. And we know the second event fell outside the interval, so $T_2>3.5$.
The likelihood function for these two observations becomes:
$$L(\mu) = f_{T_1}(t=1) P(T_2>3.5)$$ $$ = \frac{\mu e^{-\mu}}{1-e^{-\mu}} e^{-3.5 \mu}$$ $$ = \frac{\mu}{e^{4.5 \mu} - e^{3.5 \mu}}$$
Now, this likelihood function doesn't seem to have any maxima. It starts out really large at $\mu=0$ and then just decreases from there. Does this suggest that $\mu=0$ is the best rate that satisfies these requirements? Or did I make a mistake somewhere?
Though the result is unintuitive, it is correct! (With the caveat that you should write "$\mu$ has no MLE" rather than "the MLE of $\mu$ is $0$).
To see why, consider the simpler experiment, where you're just observing a $\mathrm{Poisson}(\mu)$ process until some point $C$ and don't see anything. In this case, it's easier to see that that as you decrease $\mu$, the probability of observing an event in $[0, C]$ decreases, and so the maximum likelihood estimate of $\mu$ is $0$.
In this more involved case, we first observe an event $T_1$ before observing that $T_1 + T_2 > C$. Now, if $T_1$ were not truncated, then the likelihood of observing $T_1 = 1$ would be $L(\mu; T_1 = 1) \rightarrow 0$ as $\mu \rightarrow 0$. This would counteract the fact that the likelihood of $T_2 + 1 > C$ grows to $1$ as $\mu \rightarrow 0$, and so produce a strictly positive MLE for $\mu$.
However, since $T_1$ is truncated by the conditioning, $(T_1|T_1 \leq 3/2) \xrightarrow{\mathrm{d}}\mathrm{Uniform}(0, 1.5)$, and so $L(\mu; T_1 = 1) \rightarrow 2/3$ instead of $0$. This allows $L(\mu; T_1 = 1, T_2 > 3.5) \rightarrow 2/3$ as $\mu \rightarrow 0$, and indeed, the likelihood function is decreasing in $\mu$.
Now, it would be wrong to say that $\hat{\mu}_{\mathrm{MLE}} = 0$ since in the $\mu = 0$ case, the event $\{T_1 \leq 3/2\}$ cannot happen. This is not a zero-probability event that can be handled smoothly for the purposes of conditioning, and so $0$ must be excluded from the parameter space for this experiment.
Thus, $L(\mu; T_1 = 1, T_2 > 3.5)$ has no maximum in the parameter space $(0, \infty)$, and so the MLE doesn't exist.