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I tried finding this on the internet, but with little success. How do I get a generalized expression involving rate parameter and time parameter for the Poisson process? The expression,

$P(n) = \frac{\lambda^ne^{-\lambda}}{n!}$

consider a fixed interval. I want a generalized model where time is also incorporated in the expression. Any leads? (Thanks in advance)

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  • $\begingroup$ incorporated how? Is this so you can include an exposure time, so that $\lambda$ is the rate per unit time and you want to write the Poisson for the count over some non-unit time $t$? $\endgroup$
    – Glen_b
    Commented Apr 19, 2016 at 16:00
  • $\begingroup$ @Glen_b, yes, I want to write the count over some non-unit time $\endgroup$
    – akashrajkn
    Commented Apr 19, 2016 at 16:03
  • $\begingroup$ Are you assuming the rate is constant over time? $\endgroup$
    – jbowman
    Commented Apr 19, 2016 at 16:48
  • $\begingroup$ @jbowman, well, I am doing hypothesis testing, and I need to consider both the cases - constant rate and rate with updation after certain intervals $\endgroup$
    – akashrajkn
    Commented Apr 19, 2016 at 16:50

1 Answer 1

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For an exposure time $t$, where that $\lambda$ is the rate per unit time, the Poisson pmf for the count is

$p_{\lambda t}(x) = \frac{\exp(-\lambda t) (\lambda t)^x}{x!}$

This follows directly from a change of units.

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  • $\begingroup$ Since 'x' is the number of observations in the entire time period, shouldn't we have a term like (x/t) ? $\endgroup$
    – akashrajkn
    Commented Apr 20, 2016 at 11:41
  • $\begingroup$ @akashrajkn No. The explanation why is given in my final sentence above -- it follows directly form a change of units. Take your new time unit to be the length of exposure (the observation time) $t$. Then the Poisson rate per (new) time unit is $\mu=\lambda t$ and we're back to a plain old Poisson with the parameter being $\mu$ (equivalently, $\lambda t$. $\endgroup$
    – Glen_b
    Commented Apr 20, 2016 at 11:57

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