Variances of photon numbers and electrons In optical sensors (e.g., photodiodes), the number of electrons $N$ generated by photons follows Poisson distribution. e.g., $Var[N] = N$.The number of photons $P$ also follows Poisson distribution. e.g., $Var[P] = P$. $P$ and $N$ are related by quantum efficiency $k$, $N = kP$.
Then, $Var[N] = k^2 Var[P] = k^2P = kN \neq N$.
What mistakes did I make here?
 A: In order to definitively answer you question we need to know what is your model. Pick one of three:
Given $P\sim Pois(\lambda_P)$

*

*$N=kP$

*$N\sim Pois(k\lambda_P)$

*$N\sim Pois(kP)$
1
If $N=kP$ then electron count is not from Poisson distribution. Multiplying Poisson numbers by a constant you don't get a set of larger Poisson numbers. It's easy to see with a counter example. Suppose, $k=2$, then you'll never get 1 in kP sequence which will start with 0,2,4 etc. Hence, kP is not Poisson, and the equalities that you're trying to derive are irrelevant.
2
If you defined Poisson distribution of electrons as with intensity $\lambda_N=k\lambda_P$, i.e. $N\sim Pois(\lambda_N)$ where $\lambda_N=k\lambda_P$, then you have all it defined nicely: and you get $E[N]=Var[N]=k\lambda_P=kE[P]=kVar[P]$
3
On the other hand if you define electron's distribution as $N\sim Pois(\lambda_N)$ where $\lambda_N=kP$, then it's quite more difficult because the intensity of Poisson is random variable itself now. Hence, N is from Poisson only conditionally on observing P photons, unconditionally N is not from Poisson anymore. In other words if you don't know how many photons were sent your way exactly then observing only N will shows that it's not from Poisson distribution.
