# Calculating the probability of a sum of Poisson-distributed random variables

In Java, I have a set of users, each of which has a Poisson-distributed demand with a known mean:

import org.apache.commons.math3.distribution.PoissonDistribution;

public class User {
private int mean;
private PoissonDistribution dist;
public void setMean(int mean) {
this.mean = mean;
this.poissonDistribution = new PoissonDistribution(mean);
}
}

...
User u1 = new User(); u1.setMean(20);
User u2 = new User(); u2.setMean(30);
User u3 = new User(); u3.setMean(40);


Now I'd like to calculate the probability that all Users have a cumulated demand below a certain value:

double probabilityBelowX = calculateCumulatedProbability(50, u1, u2, u3); // <- what must this method look like?


I am stuck at the question, how to solve this problem in Java. Am I missing something in the math package? I know that the demands of the customers are independent, so according to my knowledge about the Poisson distribution I can simply add up the values. But I only have cumulativeProbability(int x) for every single one of the user demands, but not for several at once.

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If the demands are $X_1$,$X_2$, & $X_3$, & your certain value is $k$, do you want to calculate (1) the probability that the total demand is less than a certain value $\mathrm{P}(X_1+X_2+X_3<k)$, or (2) the joint probability that demand for each user is less than a certain value $\mathrm{P}(X_1<k,X_2<k,X_3<k)$? – Scortchi Jan 11 '13 at 9:31
@Scortchi the first one: I want to know the probability that the aggregated demand is below a given value: $P(X_1 + X_2 + X_3 < k)$ – DaDaDom Jan 11 '13 at 9:36

## 1 Answer

If you assume that customers demands are independant and Poisson, the total demand will be a Poisson distribution as well with parameter equal to the sum of the individual Poisson parameters (see wikipedia for example).

A C++ code example (I am not so familiar with Java):

double cumulative(double x, User * users, int nusers)
{
double lambda = 0.;
for (unsigned i = 0; i < nusers; ++i) lambda += users[i].mean();
return PoissonDistribution(lambda).cumulativeProbability(x);
}

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Thank you, I will try this in my Java code and give a feedback once I got it implemented. I just didn't know or wasn't sure, that I could simply add up the single means - it just seemed to be too simple! – DaDaDom Jan 11 '13 at 9:47
OK, it works. Here is the Java code: double totalDemand = 0; for (User u : tour.getUsers()) {totalDemand += u.getMean();} double cumulativeProbability = new PoissonDistribution(totalDemand).cumulativeProbability(50); – DaDaDom Jan 11 '13 at 18:32