Title doesn't make a lot of sense due to the complicated explanation and the fact I tried to be brief.

Basically, I have a Poisson process, and I have been given the information that an event occurred between 9:10:00pm and 9:10:30pm. How do I go about finding the probability that it occurred within the last 10 seconds (i.e. between 9:10:20pm and 9:10:30pm)? (I have the expected number of events at 1 per 5 minutes)

I've tried searching for the probabilities of events in overlapping Poisson intervals, but they don't seem to apply and I can't find anything regarding sub-intervals like the one in this question.

This question is for an assignment due tomorrow and I originally thought I had a correct answer (the probability for an attack within 10 seconds) but the probability was far too low at 0.0322 and I only just realised it was wrong.

It's not just 1/3 is it?

  • $\begingroup$ we don:t need to worry about the 9:10 - all that matters is 0-30 and 20-30 - B = (1 event 0-30) and A = P(no event 0-20 and 1 event 20-30) - then Bayes Theorem P(A|B) = P(A and B) / P(B) = P(A) / P(B) (since B includes A) $\endgroup$ – Cato May 8 '17 at 10:52
  • $\begingroup$ The answer to this question should help you. $\endgroup$ – Ritz May 8 '17 at 10:58

let's say that number of events in 10 seconds is poisson with parameter $\lambda$

then you should know that for 30 seconds it is poisson with $3\lambda$

then P(1 event in 30 seconds) = $3\lambda exp(-3\lambda)$

p(1 event 20-30 seconds only) = P(0 events 0-20)P(1 event 20-30) = $exp(-2\lambda) \times \lambda exp(-\lambda) = \lambda exp(-3\lambda)$

A = 1 event 20-30 only B = 1 event 0-30

P(A|B) = P(A and B) / P(B)

P(A and B) = P(A) (since A is a subset of B)


P(A|B) = $\lambda exp(-3\lambda) / (3\lambda exp(-3\lambda)) = \frac{1}{3}$

as you suspected, the answer turns out to be 1/3 , which I think is justified by symmetery, even though the first event might seem to be weighted towards the beginning of the period, if we know there IS only one event, it is equally likely to be anywhere.

  • $\begingroup$ So it's a property of a Poisson process that it is equally likely to be anywhere in that interval? I thought it might be but I couldn't figure out how. Thank you $\endgroup$ – user3744605 May 8 '17 at 11:25
  • $\begingroup$ a single event is equally likely to be anywhere, but the first event is skewed towards the beginning of the interval. In this case though, we knew there was only a single event. $\endgroup$ – Cato May 8 '17 at 11:35
  • $\begingroup$ Ok that makes sense. What about the expected value of the time until the attack occurred during this interval (once again knowing there is only one event)? $\endgroup$ – user3744605 May 8 '17 at 11:39
  • $\begingroup$ Just because if it's equally likely to be anywhere, where is the expected value? Do you assume it's exponentially distributed over the 30 seconds, or is the expected value halfway, at 15 seconds, or is it something else entirely? $\endgroup$ – user3744605 May 8 '17 at 11:43

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