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A perceptron for finding a hyper-exponential distribution.

Recently I have been looking at some data, jitter data for spike trains, which may have a hyper-exponential distribution:

p(t)=\sum p_i s_ie^{-s_i t}

The idea is that there is a probability p_i of event type i which is in turn exponentially distributed with rate s_i. The p_is sum to one. It is hard, even with a ton of data, to fit the parameters, I thought I might try using a perceptron as a way to do this. I started by changing the sum to an integral so

p(t)=\int_0^\infty f(s) s e^{-st} ds

which looks a lot like a Laplace transform, though it is hard to know what to do with that. Now, this means

P(t_1\le t \le t_2)=\int_0^\infty f(s) \left[e^{-st_2}-e^{-st_1}\right] ds

I imagine a situation where f(x) is compactly supported and can be sensibly discretized f_i=f(s_i). Thinking of the stuff in square brackets as input at the input nodes and the corresponding f_i as weights, the predicted P(t_1\le t\le t_2) is the output. The corresponding data values were found by interpolation with the (a-1)th and (b+1)th points and

p=P(t_1\le t\le t_2) = \sum f_i \left[e^{-s_it_2}-e^{-s_it_1}\right] \delta s

was calculated. The error is now


with n the number of points. The learning rule is applied

f_i\leftarrow f_i-\eta E \left[e^{-s_it_2}-e^{-s_it_1}\right] \delta s

Evolve until happy.

It didn't work really, starting with some known $f(s)$ it evolves until the error is small and the predicted $p(t)$ looks a lot like the real one, but $f(s)$ doesn't look much like the input. The lesson seems to be that there are lots of ways to produce more-or-less the same distribution.

The code is at

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