I am researching an engineering problem in which I want to model the probability distribution of the number X of independent trials needed to get one success. If the probability of success at each trial is a constant p, the solution clearly reduces to the simple geometric distribution. However, in this particular case, the probability varies with each trial in a way that I can predict.

Sonia Elliott

Sonia Elliott

Answered question

2022-10-20

Geometric distribution with unequal probabilities for trials
I am researching an engineering problem in which I want to model the probability distribution of the number X of independent trials needed to get one success. If the probability of success at each trial is a constant p, the solution clearly reduces to the simple geometric distribution. However, in this particular case, the probability varies with each trial in a way that I can predict.
I feel sure there must already be a body of work on this type of problem that I can refer to, but I have been unable to find it. I have come across the Poisson Binomial distribution which models the probability distribution of the number of successes in a sequence of X independent yes/no experiments with different success probabilities. So I naturally searched for the Poisson Geometric or Geometric Poisson distribution, but this in fact refers to something quite different (a variation on the Poisson rather than Geometric distribution).
Can anyone help with the name of the distribution I am looking for?

Answer & Explanation

lefeuilleton42

lefeuilleton42

Beginner2022-10-21Added 12 answers

Step 1
If the probability of success at trial n is p n and the sequence ( p n ) n 1 is deterministic, then the number X of trials needed to get one success is such that, for every n 1, P ( X = n ) = p n k = 1 n 1 ( 1 p k ) ..
Two remarks. First, it may be the case that one never gets a success, in fact X is almost surely finite if and only if k = 1 ( 1 p k ) = 0..
Step 2
Thus, X is almost surely finite if p n = 1 / ( n + 1 ) for every n 1 but not if p n = 1 / ( n + 1 ) 2 for every n 1. Second, every distribution on the nonnegative integers may be written like that, simply choose p n = P ( X = n X n ) = P ( X = n ) P ( X n ) ..
This proves that this class of distributions has no specific name.

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