I have a point process defined in two parts as follows. Consider first the main process which we call $A$ which is homogeneous Poisson process with conditional intensity

$$\lambda(t) = \mu$$

For every arrival time $x_i$ that $A$ produces we run another process which we call $B$. The conditional intensity of this process assuming $t\geq x_i$ is

$$\nu(t) = \gamma e^{-(t-x_i)}$$

The combined process simply combines the list of arrival times from $A$ and $B$.

This is easy enough to simulate if you are given the parameters $\mu$ and $\gamma$.

Given some real data listing arrival times, how can one compute the MLE for the two parameters of the combined point process in an efficient way?

There seems to be a combinatorial explosion that I can't see how to avoid. I understand there is unlikely to be a closed form solution.

To give some more details of the difficulty in doing this. If you know which points are caused by process $A$ then you can compute the likelihood and use an optimization procedure to find the parameters $\mu$ and $\gamma$ which give the MLE. The problem is that you don't know that and trying all $2^n$ possibilities (where $n$ is the number of points in the data) is too much. It is possible that something from the HMM training literature might help but I am not expert enough in that to be able to tell.

For a homogeneous Poisson process alone on a time interval $(0,T)$ which has $n$ points, the MLE is of course $n/T$.


Cross-posted to https://cstheory.stackexchange.com/questions/25295/an-efficient-method-to-find-the-mle-of-the-combination-of-two-point-processes

  • 1
    Have you tried the EM algorithm (HMM training comment)? Here the missing data would be Z_i =1 if point i came from a copy of B, and 0 otherwise. Given an initial (μ,γ) guess, You could compute E(Z_i) iteratively for all i. Using these, re-estimate (μ,γ). This should converge to a local maximum without combinatorial explosion. Just a comment, no proof. – guest Jul 21 '14 at 10:35
up vote 2 down vote accepted

If I'm not mistaken, there are always infinitely many solutions to this problem.

Informally: Suppose you plant an average of $\mu$ (fertile) parent plants per year, and each parent yields an average of $\gamma$ (infertile) children per year. An outside observer will then see the number of plants grow as $O(T^2)$. However, any specific growth rate could either be attributed to high a $\mu$ and a low $\gamma$, or a low $\mu$ and a high $\gamma$.

To bring this a bit closer to your situation, suppose you plant $\mu T$ parent plants at epochs

$$ \frac{1}{\mu}, \frac{2}{\mu}, \frac{3}{\mu}, \ldots, \frac{(\mu T - 1)}{\mu}, T. $$

Then the expected number of children that parent number $i$ will have before you stop observing is

$$ \gamma \left( T - \frac{i}{\mu} \right). $$

The total number of children in the interval $[0,T]$ is therefore given by the "triangular" sum

$$ \sum_{i=1}^{\mu T} \gamma \left( T - \frac{i}{\mu} \right) \ =\ \gamma \frac{ T (\mu T - 1) }{2}, $$

and the total expected number of data points in $[0,T]$ is

$$ N \ =\ \textrm{parents} \,+\, \textrm{children} \ =\ \mu T \, + \, \gamma \frac{ T (\mu T - 1) }{2}. $$

For large $N$, the strong law of large numbers ensures that any pair of values that satisfy this relationship should be close to one of the infinitely many MLE parameter settings that explains the data.

I realize this doesn't actually answer your question as you posed it, but I hope it helps a bit.

  • 1
    This is an interesting observation. However, are you considering the inter-arrival times? It seems intuitively that a low $\mu$ and high $\gamma$ will give a very clustered process where a high $\mu$ and low $\gamma$ will give an un-clustered one (basically a homogeneous Poisson process). – Lembik Jul 22 '14 at 15:17
  • @Lembik No, I don't think so. Have a look at this image: drive.google.com/file/d/0B7-4xydn3MXJT29Bcl9XcmItTXc/…. It shows ten sample paths from a process with $(\mu,\gamma)=(10,20)$ in red, and 10 sample paths from one with $(\mu, \gamma) = (17.5, 10)$. They should have the same expected behavior by the argument above – and indeed, they seem to do. – Mathias Jul 22 '14 at 22:16

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.