I kinda think we're saying the same thing, sort of. You don't like the Gaussian assumption, and neither do I. If you make the reasonable Poisson assumptions, then you don't get the Ispot-Iback=Iobs for the best estimate of Itrue. Except as an approximation for large values, but we are talking about the case when Iback>Ispot, where the Gaussian approximation to the Poisson no longer holds. The sum of two Poisson variates is also Poisson, which also can never be negative, unlike the Gaussian.
So I reiterate: the Ispot-Iback=Iobs equation assumes Gaussians and hence negativity. The Ispot-Iback=Iobs does not follow from a Poisson assumption. On Jun 21, 2013, at 1:13 PM, Ian Tickle <[email protected]> wrote: > On 21 June 2013 17:10, Douglas Theobald <[email protected]> wrote: >> Yes there is. The only way you can get a negative estimate is to make >> unphysical assumptions. Namely, the estimate Ispot-Iback=Iobs assumes that >> both the true value of I and the background noise come from a Gaussian >> distribution that is allowed to have negative values. Both of those >> assumptions are unphysical. > > Actually that's not correct: Ispot and Iback are both assumed to come from a > _Poisson_ distribution which by definition is zero for negative values of its > argument (you can't have a negative number of photons), so are _not_ allowed > to have negative values. For large values of the argument (in fact the > approximation is pretty good even for x ~ 10) a Poisson approximates to a > Gaussian, and then of course the difference Ispot-Iback is also approximately > Gaussian. > > But I think that doesn't affect your argument. > > Cheers > > -- Ian
