Please weigh in on photon noise

Know it or not, it is there. For shot noise, it is a Poisson process with some expected value.

In what way is that a problem?

Some do. From the wikipedia page: "Despite the newly abstract situation, this definition [of expected value] is extremely similar in nature to the very simplest definition of expected values, given above, as certain weighted averages." Mean = average in this case.

Also: "In probability theory, the expected value (also called expectation, expectancy, mathematical expectation, mean, average, or first moment) is a generalization of the weighted average. Informally, the expected value is the arithmetic mean of a large number of independently selected outcomes of a random variable."

It is the mean (with respect to the probability measure) of the process. Not of the outcome of each experiment. Of course, with a low sample count, we would have a poor estimate of it.

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?

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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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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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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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