I have a discreet random variable X known to be Poisson distributed. This represents the number of observations in a certain time window, day one day.

Assuming there are no other factors dependent on say the time of day, how do I derive the distribution for multiples and fractions of my one day? What is the distribution of observations seen for every 6 hours or every 5 days?


1 Answer 1


If the events are independent - following a Poisson process, for example, then it doesn't matter how you divide the time up -- they'll also be Poisson.

But for them to be Poisson with common mean (i.e. so you can treat the set of values as coming from one Poisson distribution) you'd need the event process to have a constant rate (homogeneous Poisson process).

As described here, probability function for the number of events in a time period of length $\tau$ is:

$P [N(t+ \tau) - N(t) = k] = \frac{e^{-\lambda \tau} (\lambda \tau)^k}{k!} \qquad k= 0,1,\ldots.$

  • $\begingroup$ So, they would be Poisson with λ simply scaled proportionally to the window, i.e. 5λ for 5 days and λ/4 for 6 hours? $\endgroup$
    – SoftMemes
    Nov 8, 2014 at 14:06
  • 1
    $\begingroup$ Yes, if the event rate is constant. I'll add a link. $\endgroup$
    – Glen_b
    Nov 8, 2014 at 14:27

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