Re: [Phys-L] Coincidence Statistics
Sadly I do not presently have access to the most-developed bots: Clude-3 and
ChatGPT-4.
I find Claude's response attractive. I appreciate Web-Search's addition of
citations - none of which have I checked.
This is to show the hit & miss shape in which I find the accessible bots.
Brian
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Claude opines in this way:
Here is how I would think about this problem:
- You have two independent Poisson processes, each with an average rate of
0.1 events per 10 seconds.
- So in 100 seconds, you expect each process to generate 10 events on
average.
- Since the processes are independent, the number of joint events
(coincidences) follows a binomial distribution. Specifically, if X and Y are
the number of events from process 1 and 2, the number of coincidences Z = X ∩ Y
~ Binomial(X + Y, p)
- Here p, the probability of a coincidence, is just 1/100 since you have 100
time bins.
- So the expected number of coincidences is E[Z] = E[X + Y] * p = (10 + 10)
* (1/100) = 0.2
So the coincidence rate Rc would be 0.2 coincidences per 100 seconds, or 1
coincidence every 500 seconds on average.
The formula Rc = R1R2t is not quite right here, since it double counts the
overlapping events. The binomial model properly accounts for the overlaps. Let
me know if this makes sense!*************************************Claude-2 says
this:
Here are a few thoughts on calculating the expected rate of coincident events:
-
You are correct that the occurrence of each event follows a Poisson
distribution with a mean rate of 10 events per 100 seconds.
-
The timing resolution is 1 second bins. So there are 100 possible bins where an
event can occur.
-
For any given bin, the probability that an event from process 1 occurs in that
bin is R1*dt = 0.1 * (1/100) = 0.001
-
Similarly, the probability that an event from process 2 occurs in that same bin
is R2*dt = 0.001
-
The probability that both occur in the same bin is the product of these
individual probabilities: (R1dt) * (R2dt) = 0.001 * 0.001 = 0.000001
-
There are 100 bins, so the expected number of coincident events is:
Number of bins * Probability of coincidence per bin= 100 * 0.000001 = 0.01
So the expected rate of coincident events is 0.01 per 100 seconds, or 1 per
10,000 seconds.
The equation Rc = R1R2t you showed seems conceptually correct, but the values
are a bit off. I think t should be the timing resolution (1/100 seconds) rather
than the full observation time. With that adjustment, the equation gives the
same result.
Let me know if this makes sense! The key is thinking through the probability of
a coincidence occurring in a small time bi
****************************************************
Meanwhile ChatGPT i offers this:
The formula you mentioned, Rc = R1 * R2 * t, gives you the expected rate of
observing coincident events when you have two independent processes with rates
R1 and R2, and a timing resolution of t.
In your case, you have two processes occurring once every 10 seconds, which
translates to rates of R1 = 0.1 events per second and R2 = 0.1 events per
second.
Since your timing resolution is 1 second (t = 1), you can substitute these
values into the formula:
Rc = R1 * R2 * t
= 0.1 * 0.1 * 1
= 0.01
Therefore, the expected rate of observing coincident events in your case is
0.01 events per second, which corresponds to 1 event every 100 seconds.
***********************************************************************
Here is a similar response from Web-Search. This one offers some cites:
The rate at which you can expect to observe coincident events can be determined
using the formula Rc = R1 * R2 * t, where Rc is the coincidence rate, R1 and R2
are the rates of the two stochastic processes, and t is the timing resolution
[2].
In your case, the rates of the two processes are both 0.1 events per second,
and the timing resolution is 1 second. Plugging these values into the formula,
we get:
Rc = 0.1 * 0.1 * 1 = 0.01 events per second
So, you can expect to observe approximately 0.01 coincident events per second,
or 1 event every 100 seconds [2].
Learn more:
- ShieldSquare Captcha
- quantum mechanics - Coincidence detectors in Bell tests: How close is
close enough? - Physics Stack Exchange
- Lab Experiment 9: Gamma-Ray Coincidence Counting Techniques | Mirion
On Thursday, November 16, 2023 at 05:40:12 PM CST, Paul Nord via Phys-l
<phys-l@mail.phys-l.org> wrote:
Subtle statistics question is bothering me. Assume that I have two
stochastic processes that take place once every 10 seconds. If I watch for
100 seconds, I expect to see the "thing" happen 10 times (a Poisson
distribution around 10, actually).
My "timing resolution" for watching these is 1 second. So, I'll put them
into 100 bins depending on the moment in the 100 second observation window
that the event happened.
What is the rate at which I will expect to observe coincident events?
I'm looking at a lab writeup that suggests the coincidence rate Rc will be
a function of the two rates (R1 and R2) and the timing resolution (t).
Rc = R1*R2*t
In my case:
Rc = 0.1 * 0.1 * 1
or 1 event every 100 seconds
My sense is that the birthday paradox should apply here and the time
between pulses will have an exponential probability density.
Or that the binning described above is actually 1/2 second timing
resolution. Two pulses of 0.5 second length would overlap if they were
just under 0.5 seconds apart.
Thoughts?
Paul
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