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In the Poisson binomial distribution each trial is either yes (1) or no (0). Is there a distribution where each trial is either yes (N) or no (0)?

I'd like to model a situation where I have, for example, 10 workers. Each worker can produce $N$ units of good in one day or fail with known probability and produce nothing. $N$ can be different for each worker. I need to know distribution of goods after a day of work. Poisson binomial distribution gives me an answer in case each worker produces 1 good.

kjetil b halvorsen
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Boris Mikhaylov
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    You seems to have count data, so you can just use the poisson distribution, or the negative binomial? – kjetil b halvorsen Aug 01 '17 at 14:24
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    With $N$ different for each worker, you will just have to work out the possibilities by considering all possible combinations. There really isn't any other way to do it. If multiple workers have the same $N$, you can use a binomial distribution for the count of such workers who produce $N$, then multiply that result by $N$ to get the count of goods produced; this gives you the distribution for that subpopulation, saving some time. – jbowman Oct 08 '17 at 23:52

2 Answers2

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This is a Generalized Poisson Binomial distribution. Have a look at R package PoissonBinomial (vignete) implementing a number of new high-performance algorithms for this distribution.

VitoshKa
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First, you probably don't want to use a Poisson distribution, as it describes a random variable that can take on any of an infinite collection of integers. Stick with a Binomial(1,p), otherwise known as a Bernoulli(p).

If $X \sim \text{Bernoulli}(p)$, and $Y = NX$, then the pmf of $Y$ would be $$ p_Y(y) = p_X([y/N]) = p^{y/N}(1-p)^{1-y/N}. $$ For an individual worker, you can simply scale a Bernoulli rv to get what you want.

For adding the outcomes of multiple, independent workers, it is helpful to look at MGFs. The MGF from above is $$ M_Y(t) = E[e^{tY}] = E[e^{tNX}] = M_X(tN) = (1-p+pe^{tN}). $$

If you have several ($M$) workers, all with the same $N$, then the MGF of the sum is $$ (1-p+pe^{tN})^M. $$ This is the MGF of the random variable $\sum_{i=1}^M N X_i$. Add together $M$ Bernoulli random variables, and then multiply the result by $N$.

As @jbowman was saying, if you have a different $N_i$ for each worker, then you will not get the above scaled Bernoulli distribution, and you will have to calculate probabilities by hand.

Taylor
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