The KL divergence (and other comparisons) between this edge case and any standard instance of the Poisson distribution are problematic because the inclusion of this point in parameter space changes the support of the distribution from the entire set of non-negative integers to the set containing only zero.
-- John On Sep 15, 2014, at 8:42 AM, Andreas Noack <andreasnoackjen...@gmail.com> wrote: > This is not a big deal to me, but what exactly would break by allowing λ=0? > Returning inf for skewness and kurtosis and an error for entropy doesn't seem > problematic to me. I don't do estimation of KL divergences, but it appears a > bit strange to that extending the parameter space with a point could make a > difference as the degenerate case can be approximated arbitrarily well with > the non degenerate distribution (except for floating point rounding). > > Med venlig hilsen > > Andreas Noack > > 2014-09-15 11:25 GMT-04:00 spaceLem <space...@minioak.com>: > Hi John, thanks for your response. > > If you don't think silently accepting lambda = 0 is good practice, how best > should I approach this? The only other thing I can think of is writing my own > function wrapper specifically to catch the case where lambda = 0. > > I don't know how other people work with random numbers, but my suspicions are > that being able to get a random number when lambda = 0 might be more common > than needing skew and kurtosis for that case (although I do accept these are > all problematic cases, ballooning to infinity). The Poisson distribution is > very important to modellers, and zero rates do need to be handled somehow > (currently R, Octave, and the GSL in C++ all accept lambda = 0, so the > textbook definition isn't necessarily the popular one! On the other hand, > Scipy returns an error). > > Regards, > Jamie > > On Monday, 15 September 2014 16:03:32 UTC+1, John Myles White wrote: > I’m not so sure that we should follow the lead of Octave and R here. Neither > of those languages reify distributions as types, so changes to their RNG’s > don’t affect other operations on those same distributions. > > In contrast, the proposed change here would break a lot of other code in > Distributions that assumes that every Poisson object defines a non-delta > distribution. At a minimum, all of the following functions would need to have > branches inserted into them to work around the lambda = 0 case: > > * entropy > * kurtosis > * skewness > > In addition, every person who ever wrote code in the future that worked with > Poisson objects would need to know that our definition of the Poisson > distribution contradicted the definition found in textbooks. This would > affect people writing code to estimate KL divergences, for example. > > — John > > On Sep 15, 2014, at 7:33 AM, Andreas Noack <andreasno...@gmail.com> wrote: > >> I can see that R accepts zero as rate parameter so maybe we should do the >> same. Could you open a pull request to Distributions.jl with that change? >> >> Regarding the vectorized version, the answer is that you can do almost what >> you want with a comprehension, i.e. something like X += >> dX*[rand(Poisson(r*dt)) for r in rates]. >> >> Med venlig hilsen >> >> Andreas Noack >> >> 2014-09-15 10:05 GMT-04:00 spaceLem <spac...@minioak.com>: >> >> >> Hi all, >> >> I work in disease modelling, using use a mix of C++ and Octave. I'm fairly >> new to Julia, although I've managed to get a basic model up and running, and >> at 1/5.5 times the speed of C++ I'm pretty impressed (although hoping to >> close in on C++). I'd love to be able to work in one language all the time, >> and I'm feeling that I'm not far off. >> >> I have two questions regarding random numbers and the Poisson distribution. >> One algorithm I use has a number of possible events, each with an associated >> event rate. From these events, you choose a time step dt, then the number of >> times each event happens is Poisson distributed with lambda = rate of the >> event * dt. In Octave I could write code along these lines (simplified to >> get the gist of things): >> >> rates = 50*rand(1,6); >> rates(3) = 0; >> dt = 0.1; >> K = poissrnd(rates*dt); % = [1 6 0 4 3 4] >> >> where K is an array giving the number of times each event occurs. In Julia, >> I would write >> >> using Distributions >> rates = 50rand(6) >> rates[3] = 0 >> dt = 0.1 >> K = zeros(Float64,6) >> for i = 1:6; K[i] = rand(Poisson(rates[i] * dt)); end >> >> This gives: ERROR: lambda must be positive >> in Poisson at >> /Users/spacelem/.julia/Distributions/src/univariate/poisson.jl:4 >> in anonymous at no file:1 >> >> Which brings me to my first question: how best to handle when the events >> have zero rates (which is not uncommon, and needs to be handled)? The >> correct behaviour is that an event with zero rate occurs zero times. >> >> I found that editing poisson.jl (mentioned in the error), and changing line >> 4 from if l > 0 to if l >= 0, then the error goes away and when events have >> zero rates, they correctly occur zero times. I know the Poisson distribution >> is technically only defined for lambda > 0, but it really does make sense to >> handle the case lambda = 0 as returning 0. Somehow I feel that editing that >> file was probably not the correct thing to do (although it was great that I >> was able to), but I'd like to follow good practice, and I'm going to run >> into problems if I ever need to share my code. >> >> And my second question: in Octave I can specify an array of lambdas and get >> back an array of random numbers distributed according to those event rates. >> Is it possible to do the same in Julia? (I can write rand(6) and get a >> vector of uniformly distributed random numbers, but I need the for loop to >> do the same for other distributions). In Octave you can pretty much write >> something like X += dX * poissrnd(rates * dt); all in one line (where X is a >> vector of populations, and dX is the event rate / state change matrix), and >> it would be nice to be as elegant as that in Julia. >> >> Thank you very much in advance. >> >> Regards, >> Jamie >> > >
