Expected value is population mean while mean of measured values is sample mean? At least for a short exposure in dim light, the sample is random? For a long enough exposure, or for multiple exposures combined, the measured mean approaches the population mean?
April 29, 2023
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Yes, I had to think through that also.
If we know that the input is a Poisson process (which with Photon 'Noise' we do), and as a consequence we know that the probability of exactly one e- being generated in a tiniest interval of time (dt) is approximately (𝝺p)dt, the result is a Poisson process too, right? Of course assuming Independent and stationary increments of time and all that.
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That depends on what you're expecting.
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That's a circular argument. It says that the expected value of a Poisson process is its expected value.
It's a problem in that our expectation might not align with a simple mean - at whatever is the integration period that you choose for the mean.
'Some' do a lot of things. Doesn't mean they're right.
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Mean is sometimes called expected value. Sample mean is used to estimate population mean.
I suggest that a single exposure is a random sample of the photons striking a sensor. For a short exposure in dim light, there is a larger probability that the sample value will vary from the population value. A longer exposure would yield a measurement that has a larger probability of being close to the expected value because it is a larger sample. The mean of several random samples will approach the population mean as the number of samples or size of samples increases. If a human eye also takes a random sample of the photons and that is compared to the sensor data, then the longer exposure is more likely to match the expected value.
online.stat.psu.edu/stat414/lesson/24/24.4 "the mean (or expected value, if you prefer) of the sample mean"
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I don't think so.
I think that you're misidentifying what is the process.
Yes, but you need to apply what's in the textbook to the right problem. Try looking at it from a communication theory POV.
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This is the Photographic Science and Technology forum, not the Pure Mathematics and Statistics forum, so here's a somewhat handwavy explanation of why a constant arrival rate (probability density of arrival with respect to time) over a finite time implies a Poisson distribution. 😁
We have a constant probability density of arrivals, 𝝺. 𝝺 has dimensions T^-1.
How many arrivals do we get in a time interval of length t ?
Let's divide t into some huge number, N, of equal subintervals, so that the probability of more than one arrival in time t/N is vanishingly small.
The probability of an arrival in any subinterval is 𝝺t/N.
The distribution of the number of arrivals in time t is Binomial(N,𝝺t/N), with mean 𝝺t, and variance N.(𝝺t/N).(1-lambda t/N) ~𝝺t.
As N->infinity, the probability of an an arrival in any subinterval -> 0, but the expected number of arrivals remains constant. The distribution approaches Poisson(𝝺t).
Some folks like to use the letter 𝝺 for the mean of a Poisson distribution. Others use it for the "rate parameter" of the Poisson process.
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Apologies, I don't know your maths background.
Poisson is a limiting case of Binomial(N,p), as N->∞, p->0, but Np remains finite. This is a standard result.
See, for example, en.wikipedia.org/wiki/Poisson_limit_theorem
Or put:poisson limit of binomial
into a search engine.A maybe less handwavy way is to turn: "the probability of an arrival in an infinitesimal interval of length dt is 𝝺dt" into a differential equation involving the cumulative distribution function of the waiting time for the next arrival, showing that the waiting time has an exponential distribution, and then use the Erlang distribution (distribution of sums of exponentialy distributed variables) to show that the number of arrivals in a given time has a Poisson distribution.
This might be interesting: researchers.ms.unimelb.edu.au/~hpkeeler@unimelb/PoissonPointProcess.pdf
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"the probability of an arrival in an infinitesimal interval of length dt is 𝝺dt" - where 𝝺 is a constant - is the defining characteristic of a (homogeneous) Poisson Process.
You could dress that up in measure theory, but that would not change the basic meaning, and would be an obfuscation in this context.
It follows directly from "the probability of an arrival in an infinitesimal interval of length dt is 𝝺dt" that the number of arrivals in any given finite interval has a Poisson distribution.
That's dimensionally incorrect.
The distribution of the number of arrivals in the compound process in a time interval of (finite) duration t is Poisson(𝝺pt).
No.
A Poisson process does not, of itself, have any distribution.
A Poisson distribution can be derived from the (Poisson) arrival process, when considering the arrivals over some (finite) interval.
Maybe there's some misunderstanding about the word "process".
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If it's a Poisson process then it's not 'noise' - it's just the output of the process.
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Strictly a mis-definition. The mean is only the expected value if you're expecting a mean.
