MLE for a homogeneous Poisson process?

Question

If we have a data set consisting of event times $\{t_1, t_2, \ldots, t_N\}$ and would like to model this as a Poisson process with intensity $\lambda$, how do we do it? Intuitively, I would expect that we can calculate the average waiting time $w = \frac{1}{N-1}\sum_{i=2}^N (t_i-t_{i-1})$ and then set $\hat{\lambda} = 1/w$?

Is this correct? If so, how is it justified? Do we have to consider fitting a Poisson distribution to the number of events after doing some sort of binning?

Answer 1

You can use Maximum Likelihood Estimation, either with synchronous data (time-binned data) or asynchronous data (time-stamped data). The likelihood function changes accordingly.

For time-binned (or synchronous) data, you can simply use the joint Poisson probability mass function for your observed counts as the likelihood function:

$ L = \prod^{K}_{i=1} \frac{\lambda^{x_i}}{x_i!} \exp(-\lambda) $,

where $K$ is the number of bins, $x_i$ the count of events in bin $i$, and $\lambda$ the constant intensity that you want to estimate.

For asynchronous data, the likelihood is specified as follows:

$L = \left[ \prod^{N(T)}_{i=1} \lambda^*(t_i) \right] \exp\left[-\int^{T}_{0}\lambda^*(s) ds \right] $,

where $N(T)$ is the number of points at end-of-sample time $T$, and $\lambda^*(t)$ is the conditional intensity function, which is simply the constant $\lambda^*(t)=\lambda$ for the homogeneous Poisson process.

In some cases including the homogeneous Poisson process, there are closed-form solutions for both cases (take logs, set derivative with respect to $\lambda$ equal to zero, and solve for $\lambda$). Otherwise the log-likelihood can be optimised numerically.

For more background on theory and estimation, these are good references:

• Lecture notes on temporal point processes by Rassmussen
• Daley, D. J.; Vere-Jones, D., An introduction to the theory of point processes. Vol. I: Elementary theory and methods., Probability and Its Applications. New York, NY: Springer. xxi, 469 p. (2003). ZBL1026.60061.

Answer 2

For the homogeneous Poisson process with rate $\lambda$ the likelihood function can be written as

$L(\lambda)=\prod\limits_{n=1}^{N}\dfrac{\left(\lambda.(t_n-t_{n-1})\right)^1.e^{-\lambda(t_n-t_{n-1})}}{1!}$, with $t_0=0$, s.t., the log-likelihood

$l(\lambda)=\sum\limits_{n=1}^{N}ln(\lambda) + ln(t_n-t_{n-1})-\lambda(t_n-t_{n-1})$

For MLE, setting $\dfrac{dl}{d\lambda}=0$, we get,

$\sum\limits_{n=1}^{N}\dfrac{1}{\lambda} - (t_n-t_{n-1})=0$

$\implies \dfrac{N}{\lambda_{MLE}} = \sum\limits_{n=1}^{N}(t_n-t_{n-1})$

$\quad \quad = t_N-t_{N-1}+t_{N-1}-t_{N-2}+\ldots+t_2-t_1+t_1-t_0=t_N-t_0$

$\quad \quad =t_N$

$\lambda_{MLE}=\dfrac{N}{t_N}$

Answer 3

Expanding from @sandipan-dey's answer for homogenous process: