I suppose it comes from one's perspective -- from a modeller's point of 
view, a zero rate is certainly not a special case! The error was 
surprising, which made me baulk at handling it.

>From a purity argument, I agree with you -- Poisson(0) is not defined, 
however from a usefulness argument, simply returning 0 is what people 
(especially those coming from most other languages) would expect (after 
all, the limit of P(0) as lambda -> 0 is 0), and returning infinity for 
skewness etc. rather than an error also makes a geometric sense.

However, I'll probably handle it as David Gonzales' nice one liner for the 
moment.

Thanks for your advice!

Regards,
Jamie

On Monday, 15 September 2014 16:43:13 UTC+1, John Myles White wrote:
>
> Why not just write out an explicit check in your code for the special 
> case? Here's an example of how one might do that:
>
> n_samples = 10
> n_lambdas = 3
> lambdas = [0.0, 0.1, 0.2]
>
> samples = Array(Int, n_samples, n_lambdas)
>
> for (i, lambda) in enumerate(lambdas)
> for sample in 1:n_samples
> if rate == 0.0
> samples[sample, i] = 0
> else
> samples[sample, i] = rand(Poisson(lambda))
> end
> end
> end
>
> In my mind, this is the best possible approach to this kind of problem: 
> you want slighly unusual behavior, so you should make the place where your 
> goals are slightly unusual as explicit as possible so that anyone who reads 
> your code will understand exactly what you're doing and exactly where 
> you're breaking from textbook definitions.
>
> As for precedent, I believe the list of systems you've provided breaks 
> down as follows:
>
> RNG's take in numbers, not objects:
>
> * R
> * Octave
> * GSL
>
> RNG's take in objects, not numbers:
>
> * SciPy
> * Julia
>
> So it seems like the split in behavior is perfectly explicable in terms of 
> the types of arguments you provide to RNG's. This seems like a very good 
> argument for not changing the behavior of Distributions.
>
>  -- John
>
> On Sep 15, 2014, at 8:25 AM, spaceLem <[email protected] <javascript:>> 
> wrote:
>
> 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 <[email protected]> 
>> 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 <[email protected]>:
>>
>>>
>>>
>>> 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
>>>
>>
>

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