S/N = sqrt(k)
That's the easy case. How do you do the same thing if the data is drawn from a Gaussian? It's basically the same kind of an issue, except you need to choose where you want to optimize the S/N. For the case from work, we largely ignore everything outside of 2\sigma, so I've used that as the critical point. Now, for N total data points, how large of a box do you need centered on the critical point to achieve that S/N value? Conveniently, this is just an exercise in error functions:
k/N = normcdf(critical + binsize/2.0) - normcdf(critical - binsize/2.0)
k/N = 0.5 * (erf(sqrt(2.0) * (critical + binsize/2.0)) - erf(sqrt(2.0) * (critical - binsize/2.0)))
So, you choose the S/N value for the critical point, determine the k/N value, and then find the binsize that achieves that (such as via the following figure).
This is for critical = 2\sigma. |