I was wondering how would Poisson statistics be used in a photodetectors to account for the number of events that I'm missing in my experiment. Say I have a material (scintillator, ...) that emits $n$ photons per particle and a Photomultiplier tube (PMT) or any other photodetector with quantum efficiency Q. If I have $N$ particles arriving to the detector, how would I use Poisson statistics to calculate the number of particles that I'm missing?

  • $\begingroup$ Do you mean missing due to the quantum efficiency? Or for some other (perhaps unknown) reason? $\endgroup$ – Bob Jacobsen May 18 '18 at 6:26

The convolution of a Poisson with a binomial is a Poisson - which makes life easy. $n$ will be Poisson, and $N$ will be Poisson with mean $nQ$.

  • $\begingroup$ Thanks for the answer. $N$ is not a Poisson distribution, it is actually the real number of particles arriving to the detector, what I want to find out is $N_d$, the number of detected particles if I have a Poissonian distributed quantum efficiency $\endgroup$ – Juanjo May 18 '18 at 6:46
  • $\begingroup$ For each particle the number of photons is Poisson with mean $nQ$ So there is a probability $e^{-nQ}$ that no photons will be observed and you will miss it, meaning you have $Ne^{-nQ}/1-e^{-nQ}$ missing particles. If your detector has a higher threshold (how many photons are needed to qualify as a signal above background) adjust accordingly. $\endgroup$ – RogerJBarlow May 18 '18 at 6:56
  • $\begingroup$ Sorry @RogerJBarlow, I came back to this question and there is something I'm not getting. If the probability of detecting 0 photons is $e^{-nQ}$, the missing particles I have should be $Ne^{-nQ}$, right? where is the $1/1-e^{-nQ}$ factor coming from? $\endgroup$ – Juanjo Jun 23 '18 at 6:57

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