In imaging, one often does flat-field and dark-current correction. The dark current is assumed to be some fixed value and has a Poisson-like shot-noise. Therefore, the value of any pixel follows to be $P(n)$, where $n$ is the amount of dark-current.

My question now is, when I correct an image for this dark-current, I measure the dark-current for some time (exposure time). I usually measure my Signal for the same exposure time and finally subtract the dark-current measurement from the signal. In the resulting, corrected image, what does the noise look like?

For simplicity, assume, that I have no signal and measure two distinct dark-frames and subtract them from each other.

On a further note: I understand that the relative error decreases, when measuring the signal over a longer time (as $n$ increases and the error goes like $\sqrt{n}$. It therefore makes sense to take the dark-current image for a longer time and normalize it afterwards to the exposure time of the signal. I would also be interested in the resulting distribution in that case.


1 Answer 1


The answer to your simple question is the Skellam distribution, that is the difference between two identical and independent Poisson distributions.

For the further note, let's say that you are measuring your signal over time $t_s$, you calibrate dark current estimate over time $t_c$, and the true dark current distribution over unit time is distributed according to $P(\lambda)$. Then, abusing the notation a little, the noise will be distributed according to

$$ P(\lambda t_s) - \frac{t_s}{t_c} P(\lambda t_c). $$

I believe this distribution does not have a simplified representation (you can check this question). On the positive side, given the independence, it is easy to compute its mean ($=0$) and variance ($=\lambda (t_s + t_s / t_c)$), which might be enough for you.


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