To my knowledge the signal to noise ratio when talking about shot noise, is defined as the mean signal divided by the standard deviation. In the Poisson distribution, which can be used to model shot noise, the mean and the variance are equal and thus giving us the equation N/N^(1/2).

I have a couple questions regarding this:

  1. The first one being that, since N/N^(1/2) is equal to N^(1/2) we can see that the rate at which the SNR improves with increasing exposure/signal diminishes. I understand that graphically this is quite obvious. However, I cannot seem to wrap my head around it, since the shot noise has a y = N^(1/2) function, meaning that the increase in noise is diminishing as the signal increases and the mean signal has a function of y=N , so the signal is linearly increasing with increasing signal. So why does the rate of increase in SNR decrease?

  2. How do other noise elements factor into this, such as dark current noise (read noise) and is it fair to assume that in most cases (other than very low exposure) shot noise is the most dominant (visibly dominant) form of noise in cinematography/photography.

  3. As the mean signal value increases the Poisson distribution begins to look more and more like a Gaussian distribution and as we know the SNR continuously improves. However, when I look at the graphs, the more and more the signal increases the more the spread of the curve seems to increase, which to me visually would suggest more noise? (graph below from wikipedia)

enter image description here

  • \$\begingroup\$ Point 3) actually it makes sense as variance increases with increasing mean value. \$\endgroup\$
    – vannira
    Commented Dec 18, 2023 at 17:35

1 Answer 1


Your formula makes no sense to me... poisson noise is equal to the square root of the signal. So SNR = S/√S.

  1. It doesn't. The rate of SNR improvement increases at higher exposures.
  2. Yes, shot noise is the most relevant form of noise in most photography... at least with modern sensors, especially since other forms of noise (read noise, PRNU) are commonly measured and zero corrected for.
  3. Yes, there is more noise; it is just less in relation to the qtty of signal... not sure what the graph is in relation to.
  • \$\begingroup\$ Thank you for the quick response. The 2nd point makes a lot of sense and is very interesting, are there perhaps any links you may have to this? \$\endgroup\$
    – vannira
    Commented Oct 20, 2023 at 6:28
  • \$\begingroup\$ Regarding the first point, I first wanted to apologize for my notation it looks a bit confusing (I didn’t know you glycolic write square roots) and I forgot to explain what N is, being the mean signal. However, I don’t think that the rate of SNR improvement cause when you plot the graph SNR=√N the function is a square root, so as exposure increases the rate of improvement in SNR decreases. ARRI also briefly talks about it in their paper on dynamic range. Hence why my original question in part one. \$\endgroup\$
    – vannira
    Commented Oct 20, 2023 at 6:31
  • \$\begingroup\$ To the third part, those are just general graphs I found of how the the Poisson distribution begins to model the Gaussian distribution more and more as the mean increases. It’s not specific with respect to SNR or shot noise but since these distributions are used to model these scenarios I thought it was a valid general representation. I understand that noise does increase with more exposure, yet SNR increases, however it seems to me that those graphs (e.g. from purple to blue) would show decreasing SNR too. \$\endgroup\$
    – vannira
    Commented Oct 20, 2023 at 6:35
  • \$\begingroup\$ @vannira, I think you must have something mixed up; a 100:10 SNR ratio (10 signal: 1 noise) is better than a 25:5 (5 signal: 1 noise); as the signal increases the ratio improves, the rate of noise increase is reduced relative the the signal. But that is not the same as a decrease in the rate of SNR improvement. It does get to the point where additional increases of signal make negligible qualitative differences because the signal already overwhelms the noise level. And with digital image sensors there is the photosite full well capacity limit(s). \$\endgroup\$ Commented Oct 20, 2023 at 12:46
  • 1
    \$\begingroup\$ @vannira I think Steven is referring to the linear relationship on a log-log graph. That is, noise (in dB; i.e., logarithmic scale) is linear with respect stops of ISO (again, logarithmic scale). \$\endgroup\$
    – scottbb
    Commented Nov 19, 2023 at 2:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.