I forgot to add probably the most important. James is correct, the expected
value of u, the true mean, given a single observation k is indeed k+1 and
k+1 is also the mean square error of using k+1 as the estimator of the true
mean.

Cheers,
Filipe

On Wed, 13 Oct 2021 at 23:17, Filipe Maia <[email protected]> wrote:

> Hi,
>
> The maximum likelihood estimator for a Poisson distributed variable is
> equal to the mean of the observations. In the case of a single observation,
> it will be equal to that observation. As Graeme suggested, you can
> calculate the probability mass function for a given observation with
> different Poisson parameters (i.e. true means) and see that function peaks
> when the parameter matches the observation.
>
> The root mean squared error of the estimation of the true mean from a
> single observation k seems to be sqrt(k+2). Or to put it in another way,
> mean squared error, that is the expected value of (k-u)**2, for an
> observation k and a true mean u, is equal to k+2.
>
> You can see some example calculations at
> https://colab.research.google.com/drive/1eoaNrDqaPnP-4FTGiNZxMllP7SFHkQuS?usp=sharing
>
> Cheers,
> Filipe
>
> On Wed, 13 Oct 2021 at 17:14, Winter, Graeme (DLSLtd,RAL,LSCI) <
> [email protected]> wrote:
>
>> This rang a bell to me last night, and I think you can derive this from
>> first principles
>>
>> If you assume an observation of N counts, you can calculate the
>> probability of such an observation for a given Poisson rate constant X. If
>> you then integrate over all possible value of X to work out the central
>> value of the rate constant which is most likely to result in an observation
>> of N I think you get X = N+1
>>
>> I think it is the kind of calculation you can perform on a napkin, if
>> memory serves
>>
>> All the best Graeme
>>
>> On 13 Oct 2021, at 16:10, Andrew Leslie - MRC LMB <
>> [email protected]> wrote:
>>
>> Hi Ian, James,
>>
>>                       I have a strong feeling that I have seen this
>> result before, and it was due to Andy Hammersley at ESRF. I’ve done a
>> literature search and there is a paper relating to errors in analysis of
>> counting statistics (se below), but I had a quick look at this and could
>> not find the (N+1) correction, so it must have been somewhere else. I Have
>> cc’d Andy on this Email (hoping that this Email address from 2016 still
>> works) and maybe he can throw more light on this. What I remember at the
>> time I saw this was the simplicity of the correction.
>>
>> Cheers,
>>
>> Andrew
>>
>> Reducing bias in the analysis of counting statistics data
>> Hammersley, AP <https://www.webofscience.com/wos/author/record/2665675> 
>> (Hammersley,
>> AP) Antoniadis, A
>> <https://www.webofscience.com/wos/author/record/13070551> (Antoniadis, A)
>> NUCLEAR INSTRUMENTS & METHODS IN PHYSICS RESEARCH SECTION A-ACCELERATORS
>> SPECTROMETERS DETECTORS AND ASSOCIATED EQUIPMENT Volume 394
>> Issue 1-2
>> Page 219-224
>> DOI 10.1016/S0168-9002(97)00668-2
>> Published JUL 11 1997
>>
>> On 12 Oct 2021, at 18:55, Ian Tickle <[email protected]> wrote:
>>
>>
>> Hi James
>>
>> What the Poisson distribution tells you is that if the true count is N
>> then the expectation and variance are also N.  That's not the same thing as
>> saying that for an observed count N the expectation and variance are N.
>> Consider all those cases where the observed count is exactly zero.  That
>> can arise from any number of true counts, though as you noted larger values
>> become increasingly unlikely.  However those true counts are all >= 0 which
>> means that the mean and variance of those true counts must be positive and
>> non-zero.  From your results they are both 1 though I haven't been through
>> the algebra to prove it.
>>
>> So what you are saying seems correct: for N observed counts we should be
>> taking the best estimate of the true value and variance as N+1.  For
>> reasonably large N the difference is small but if you are concerned with
>> weak images it might start to become significant.
>>
>> Cheers
>>
>> -- Ian
>>
>>
>> On Tue, 12 Oct 2021 at 17:56, James Holton <[email protected]> wrote:
>>
>>> All my life I have believed that if you're counting photons then the
>>> error of observing N counts is sqrt(N).  However, a calculation I just
>>> performed suggests its actually sqrt(N+1).
>>>
>>> My purpose here is to understand the weak-image limit of data
>>> processing. Question is: for a given pixel, if one photon is all you
>>> got, what do you "know"?
>>>
>>> I simulated millions of 1-second experiments. For each I used a "true"
>>> beam intensity (Itrue) between 0.001 and 20 photons/s. That is, for
>>> Itrue= 0.001 the average over a very long exposure would be 1 photon
>>> every 1000 seconds or so. For a 1-second exposure the observed count (N)
>>> is almost always zero. About 1 in 1000 of them will see one photon, and
>>> roughly 1 in a million will get N=2. I do 10,000 such experiments and
>>> put the results into a pile.  I then repeat with Itrue=0.002,
>>> Itrue=0.003, etc. All the way up to Itrue = 20. At Itrue > 20 I never
>>> see N=1, not even in 1e7 experiments. With Itrue=0, I also see no N=1
>>> events.
>>> Now I go through my pile of results and extract those with N=1, and
>>> count up the number of times a given Itrue produced such an event. The
>>> histogram of Itrue values in this subset is itself Poisson, but with
>>> mean = 2 ! If I similarly count up events where 2 and only 2 photons
>>> were seen, the mean Itrue is 3. And if I look at only zero-count events
>>> the mean and standard deviation is unity.
>>>
>>> Does that mean the error of observing N counts is really sqrt(N+1) ?
>>>
>>> I admit that this little exercise assumes that the distribution of Itrue
>>> is uniform between 0.001 and 20, but given that one photon has been
>>> observed Itrue values outside this range are highly unlikely. The
>>> Itrue=0.001 and N=1 events are only a tiny fraction of the whole.  So, I
>>> wold say that even if the prior distribution is not uniform, it is
>>> certainly bracketed. Now, Itrue=0 is possible if the shutter didn't
>>> open, but if the rest of the detector pixels have N=~1, doesn't this
>>> affect the prior distribution of Itrue on our pixel of interest?
>>>
>>> Of course, two or more photons are better than one, but these days with
>>> small crystals and big detectors N=1 is no longer a trivial situation.
>>> I look forward to hearing your take on this.  And no, this is not a
>>> trick.
>>>
>>> -James Holton
>>> MAD Scientist
>>>
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