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On 10/19/21 10:35 AM, Paul Nord wrote:
the sum of Poisson distributions is still a Poisson distribution.
That's true and important.
Taking the next step down that road: The intensity (λ) of the
combined process is the sum of the contributing intensities.
The following may be obvious to everybody on this list, but if
I were explaining this to students I would assign the following
calculation. Start with the standard formula for the Poisson
probability.
P(λ, k) = λ^k exp(—λ) / k! [1]
Now suppose we divide the time axis into bins of infinitesimal
size. What is the probability of finding an event in some small
interval of time?
There is no chance of k>1, so the only thing worth calculating is
the case where k=1. Then differentiate with respect to t. In the
case where λ=rt for some constant r, the answer is simply:
dP/dt = r [2]
That should be super-obvious in retrospect.
In the case we have been discussing, the rate is itself a function
of time, given by the radioactive decay law (and background). In
this case I assert that λ=rt needs to be replaced by
λ = ∫r dt [3]
This is the correct way to implement the aforementioned idea that
the intensity of the overall process is the sum of the contributing
intensities. So the general result is:
dP/dt = r(t) [4]
which should still be obvious, at least in retrospect.
Let's be clear: [3] is not an integral over the Poisson probability.
In our example it is the integral of the radioactive decay rate,
integrated over the interval of interest. The result is the intensity
λ for that interval, which we can then plug into the Poisson formula,
along with the observed k for that interval.
More generally, r(t) could be anything, radioactive or otherwise.
The math is the same no matter what.
I've done this enough times that I can write down equation [3]
without even thinking about it ... but for students it's not exactly
the first thing that comes to mind.
There's a really fundamental principle here: The intensity (λ) of
the combined process is the sum of the intensities of the contributing
processes.
================
Of course if the data comes to us in timestamped form (as opposed to
binned form) we don't need the Poisson formula at all. The only thing
we care about is the instantaneous rate r(t). This is the case that
MacKay considered in his lecture #10.
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