Re: scalling of poisson distribution

"Nick" <tulse04-news1@xxxxxxxxxxx> writes:

I think I found the answer myself, and I post it because it might help others.

Actually, the problem was that when you double the time interval (for
example), then the probablitiy for 2 events is:

p(2,2) = p(1,1) * p(1,1) + 2 * p(2,1)

so the combination of the two possible cases, of both events happening
in one of the halve intervals or both happening in one of them. It's
these addional combinations that skew it all.

Also I found when you try and scale the whole thing what happens is the
distribution becomes sharper as you scale up. Which makes a lot of
sense: the longer you measure, the more shure you can be of the mean
value.

My misconception arose because I understood that neuronal _firing rates_
are distributed poisson like, but what is the case is that actual
sequence of spikes, given a mean firing rate, approximates a poisson
process. Then the rates again will have a different distribution, which
probably depends a lot on the circumstances (a model assumes gamma, but
that is just because gamma is conjugate to the poisson).

Cheers,

Daniel

"Daniel Oberhoff" <daniel.oberhoff@xxxxxxxxxxxxxxxxx> wrote in message
news:m1mz0wik8y.fsf@xxxxxxxxxxxxxxxxxxxxxxxxxxxx
Hi all,

I am having some trouble wrapping my head around the poisson
distribution. It is supposed to be a good model for firing rate
distributions of neurons which makes sense to me. But rates are always
expressed in some unit (per second, minute, whatever), but the choice of
unit should not alter the distribution. In the case of the poisson
distribution however it seems to. Or in other words:

p = poiss(n,l) (where l = <x>)

does not scale according to:

p -> c * l, n -> c * n, l -> c * l

which means the distribution is different depending on how long I
count??? Even more confusng since I read on Wikipedia that the poisson
process is memoryless, so doubling the time interval I should just
expect double the occurences???

can anyone explain this to me?

It is a long time since I did this, but if the mean is lambda for period t,
then the mean is lambda for period 2t

The probability (x,2*lambda) = (exp(-lambda)*lambda^x)/x factorial

and Sum of the probability (x,2*lambda) (x=0,1,...) is 1

I think that the fact that the distribution is discrete rather than
continuous comes into it. After all, if we take it the other way round, if
we have x=0,1,2,... in unit time t and we then halve the unit to t/2, we
can't divide the x values by 2. We are still looking at discrete values of
x. If it was a continuous variable then we could truly scale the
distribution.

Nick
.

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