A clerk in a shop has noticed two customers arrived at the shop between 12:00 and 12:45. Another clerk noticed only one customer between 12:15 and 13:00. Assuming a Poisson distribution on the number of customers visiting the shop (with parameter $\lambda$ being the avg. amount of patrons/hour), use the two observations to estimate $\lambda$ using MLE.
I don't know how to set up the likelihood function - we have two observations so it should be their joint density, but what is that? I read somewhere that the densities should be should be $Poiss(T_1 \lambda)$ and $Poiss(T_2 \lambda)$ where $T_1, T_2$ are the respective time intervals, but I don't quite understand where that comes from, particularly how to incorporate these times.
An idea I have here (not sure if it makes sense as an MLE) would be to define an event $A$ to be the event where the first guy notices two customers at 12-12:45 and the second guy notices one customer between 12:15-13 - essentially $A$ is what really happened, and since we want our estimate to be close to the truth, we should choose it so that the probability of $A$ happening is maximal. But I'm having trouble actually computing this probability. Is this even the right way to proceed? Any tips would be really appreciated.
We know that for a poisson process with rate $\lambda$ the probability that $k$ customers arrive within time $\tau$ is
$$P[N(t+\tau) - N(t) = k]=\frac{e^{-\lambda\tau}(\lambda\tau)^k}{k!}$$
so the probability that 2 customers arrive in the first fifteen minutes, no customers arrive in the next half hour and 1 customer arrives in the last half hour equals
$$\frac{e^{-\lambda/4}(\lambda/4)^2}{2!}\frac{e^{-\lambda/2}(\lambda/2)^0}{0!}\frac{e^{-\lambda/4}(\lambda/4)^1}{1!}=\frac{e^{-\lambda}(\lambda/4)^3}{2}$$
The probability of the other event (1,1,0) will be
$$\frac{e^{-\lambda/4}(\lambda/4)^1}{1!}\frac{e^{-\lambda/2}(\lambda/2)^1}{1!}\frac{e^{-\lambda/4}(\lambda/4)^0}{0!}=\frac{e^{-\lambda}(\lambda^2/8)}{1}$$
Since the two probabilities are disjoint the probability of the total is the sum of the probabilities, the total probability will be
$$\frac{e^{-\lambda}\lambda^3}{128}+\frac{e^{-\lambda}\lambda^2}{8}$$
I'll let you verify the algebra and grind through the maximization, but I get $\lambda \approx 2.1$ which is within the realm of plausibility.