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Poisson random number generator

7 vues (au cours des 30 derniers jours)
Ahmed raheem
Ahmed raheem le 6 Fév 2012
Commenté : Torsten le 20 Jan 2022
Hi all please i need to know how to generate a Poisson distributed random variable without using the built-in function (poissrnd).

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Andreas Goser
Andreas Goser le 6 Fév 2012
If this is an acadamic exercise - you can look at the literature refererence
% References:
% [1] Devroye, L. (1986) Non-Uniform Random Variate Generation,
% Springer-Verlag.
  1 commentaire
Zhuofan Zheng
Zhuofan Zheng le 26 Mar 2018
great thanks

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Plus de réponses (8)

Derek O'Connor
Derek O'Connor le 6 Fév 2012
Dirk Kroese has excellent notes here: http://www.maths.uq.edu.au/~kroese/mccourse.pdf, which are based on his book:
D.P. Kroese, T. Taimre, Z.I. Botev: Handbook of Monte Carlo Methods. John Wiley & Sons, 2011.
His notes and book have lots of Matlab examples.
  1 commentaire
Ahmed raheem
Ahmed raheem le 7 Fév 2012
thank you for your help....
this one is quite helpful...

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Derek O'Connor
Derek O'Connor le 7 Fév 2012
I prefer this:
% -------------------------------------------------------------
function S = PoissonSamp(lambda,ns);
% -------------------------------------------------------------
% Generate a random sample S of size ns from the (discrete)
% Poisson distribution with parameter lambda.
% Derek O'Connor, 6 Feb 2012. derekroconnor@eircom.net
%
S = zeros(ns,1);
for i = 1:ns
k=1; produ = 1;
produ = produ*rand;
while produ >= exp(-lambda)
produ = produ*rand;
k = k+1;
end
S(i) = k;
end

Derek O'Connor
Derek O'Connor le 3 Mai 2012
I would like to thank Kang Wook Lee of Berkeley for pointing out an error in the code above. The last line should be S(i) = k-1;
% -------------------------------------------------------------
function S = PoissonSamp(lambda,ns);
% -------------------------------------------------------------
% Generate a random sample S of size ns from the (discrete)
% Poisson distribution with parameter lambda.
% Fixed error: changed S(i) = k; to S(i) = k-1;
% Derek O'Connor, 3 May 2012. derekroconnor@eircom.net
%
S = zeros(ns,1);
for i = 1:ns
k=1; produ = 1;
produ = produ*rand;
while produ >= exp(-lambda)
produ = produ*rand;
k = k+1;
end
S(i) = k-1;
end
  2 commentaires
PhD Student
PhD Student le 10 Juil 2019
an End is missing
Ian Van Giesen
Ian Van Giesen le 24 Juin 2020
Why was the last line changed? Was it to make the events where probability for 'success' a null event? Sorry in advance for a confusing question, still wrapping my head around this, but many thanks for the code!

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Derek O'Connor
Derek O'Connor le 24 Juil 2012
This is a cleaner fix of PoissonSamp
% -------------------------------------------------------------
function S = PoissonSamp3(lambda,ns);
% -------------------------------------------------------------
% Generate a random sample S of size ns from the (discrete)
% Poisson distribution with parameter lambda.
% Fixed error:
% CHANGED k = 1; produ = 1; produ = produ*rand
% TO k = 0; produ = rand;
% Derek O'Connor, 24 July 2012. derekroconnor@eircom.net
%
S = zeros(ns,1);
for i = 1:ns
k = 0;
produ = rand;
while produ >= exp(-lambda)
produ = produ*rand;
k = k+1;
end
S(i) = k;
end
  3 commentaires
John D'Errico
John D'Errico le 10 Juil 2019
Because the function name is spelled poissrnd, not poisrnd.
Binlin Wu
Binlin Wu le 1 Avr 2021
Modifié(e) : Binlin Wu le 2 Avr 2021
Would just like to make some minor changes to accept any array size:
function S = PoissonSamp3(lambda,varargin);
% -------------------------------------------------------------
% Generate a random sample S of size ns from the (discrete)
% Poisson distribution with parameter lambda.
% Fixed error:
% CHANGED k = 1; produ = 1; produ = produ*rand
% TO k = 0; produ = rand;
% Derek O'Connor, 24 July 2012. derekroconnor@eircom.net
%
nn = [varargin{:}];
S = zeros([nn,1]);
for i = 1:numel(S(:))
k = 0;
produ = rand;
while produ >= exp(-lambda)
produ = produ*rand;
k = k+1;
end
S(i) = k;
end

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Richard Willey
Richard Willey le 6 Fév 2012
Mark Steyvers has written a nice book titled "Computational Statistics with MATLAB" which can be downloaded from
The first chapter has a very good section describing inverse transform sampling which provides everything you need to know.

Ahmed raheem
Ahmed raheem le 7 Fév 2012
First of all, I'd like to thank you all for your cooperation with me. Based on Dirk Kroese, I wrote the following code to generate the Poisson random variable:
The m-file code:
n=1;
lambda=500;
for i=1:10000
x=rand(1);
a=1;
a=a*x;
if a>=exp(-lambda)
n=n+1;
continue
else
X(i)=n-1;
end
end
  • would you please check it if it's correct or not? i feel it is not correct.

Derek O'Connor
Derek O'Connor le 7 Fév 2012
@Ahmed, you're correct, it is not correct.
The function below is a Matlab translation of Kroese's algorithm. It seems to work ok but needs to be thoroughly tested. You should do this and let us know the results. Note that Poisson(L) ~ Norm(L,L), for large L.
The one thing I don't like about Kroese is his awful algorithm and programming style. Why does he use GOTOs instead of proper WHILEs etc?
% -------------------------------------------------------------
function X = Poisson(lambda);
% -------------------------------------------------------------
% Generate a random value from the (discrete) Poisson
% distribution with parameter lambda.
% Derek O'Connor, 6 Feb 2012. derekroconnor@eircom.net
%
k=1; produ = 1;
produ = produ*rand;
while produ >= exp(-lambda)
produ = produ*rand;
k = k+1;
end
X = k;
  1 commentaire
Ahmed raheem
Ahmed raheem le 7 Fév 2012
thank you Derek.
it is performing well now...
i putted the code inside a (for loop) and the result was ok.
thanks again.

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Ahmed raheem
Ahmed raheem le 7 Fév 2012
This is the final code:
% -------------------------------------------------------------
function X = Poisson(lambda,n); % n represents the number of iterations
% -------------------------------------------------------------
% Generate a random value from the (discrete) Poisson
% distribution with parameter lambda.
% Derek O'Connor, 6 Feb 2012. derekroconnor@eircom.net
%
for i=1:n;
k=1; usave=1;
usave = usave*rand;
while usave >= exp(-lambda)
usave = usave*rand;
k = k+1;
end
X(i) = k;
end
hist(X)
  2 commentaires
Abdulramon Adeyiola
Abdulramon Adeyiola le 20 Jan 2022
@Ahmed raheem I tried using your code to generate 100 samples from Poisson distribution with parameter 1, but the mean of the resulting samples was way above 1. This is strange!
Torsten
Torsten le 20 Jan 2022
L = exp(-lambda);
for i=1:n
k=0; usave=1;
while usave > L
k = k+1;
usave = usave*rand;
end
X(i) = k-1;
end

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