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Re: derivation: counts in an interval



John Denker wrote:

The probability of recording m counts in time t
(given an average counting rate of r) is given by
g(m,t) = (rt)^m exp(-rt) / m!

That is the Poisson distribution. Before attempting to use
it let me restate the original problem. The average waiting
time (between counting two consecutive Geiger pulses)
was found to be 0.325 seconds, as described yesterday
morning. This makes r=1/0.325=3.08 cnts/s. Predict
the distribution of m when t=0.2 seconds.

I will make predictions in terms of an experiment in
which counting outcomes were recorded 2676 times.
Why 2676 and 0.2 s? Because that was the actual
experiment (performed to verify predictions). It
showed, for example, that 123 times (out of 2676)
nothing was recorded in 0.2 seconds, that 467 times
only one pulse was recorded in 0.2 seconds, etc.
The task was to predict the distribution of counts
from the distribution of waiting times.

The above formula (with r*t=3.08*0.2=0.615) makes
the following predictions:

m=0 --> 1445 outcomes (instead of experimental 123)
m=1 --> 890 outcomes (instead of experimental 467)
m=2 --> 274 outcomes (instead of experimental 674)
m=3 --> 56 outcomes (instead of experimental 604)
m=4 --> 7 outcomes (instead of experimental 503)
m=5 --> 1 outcome (instead of experimental 172)
m=6 --> 0 outcome (instead of experimental 85)

The formula would roughly match experimental outcomes
if r*t=15 were used, instead of 0.616. I am puzzled;
something is not right somewhere.
Ludwik Kowalski

John Denker wrote:

At 07:11 AM 7/9/00 -0400, I wrote:

The probability of recording m counts in time t given an average counting
rate of r is given by
g(m,t) = (rt)^m exp(-rt) / m!

For those of you who might be wondering where that came from.... It's
easy. It's even easier if we do it step by step:

Say that the interval t is made up of "slots" of size dt.
number of slots: t/dt
average counting rate: r
probability of a count in a given slot: r dt
probability of no count in a given slot (1 - r dt)
probability of no count in the whole interval: (1 - r dt)^[t/dt]
... which in the limit dt -> 0 is just exp(-rt)
... so g(0,T) = exp(-rt)

====

probability of no count in the first (t/dt-1) slots followed
by a count in the final slot: (1 - r dt)^[t/dt - 1] (r dt)
then apply a shuffle factor to get the
probability of one count in SOME slot: (1 - r dt)^[t/dt - 1] (r dt) t/dt
... which is just rt exp(-rt)

===

prob. of no count in the first (t/dt-m) slots followed by a count
in each of the last m slots: (1 - r dt)^[t/dt - m] (r dt)^m
then apply a shuffle factor to get the
prob. of m counts somewhere: (1 - r dt)^[t/dt - m] (r dt)^m (t/dt)^m / m!
... note in the limit dt->0 we can ignore the possibility of
collisions, i.e. multiple counts in the same slot.
bottom line: g(m,t) = (rt)^m exp(-rt) / m!