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?

  • $\begingroup$ what you need is called Coupon Collector's problem. The mean time till $n$'th event in $O(n \log n)$ $\endgroup$
    – Alex
    Jul 4 '13 at 2:08
  • $\begingroup$ Thank you for the quick response Alex. I did look at the Coupon Collector's problem previously and I decided it did not match my problem exactly. The reason is because while you are searching for coupon i, the success probability is constant for all trials. The probability only changes when coupon i is found. In my problem, the probability changes for each and every trial. $\endgroup$
    – Jason
    Jul 4 '13 at 2:18

If the probability of success at trial $n$ is $p_n$ and the sequence $(p_n)_{n\geqslant1}$ is deterministic, then the number $X$ of trials needed to get one success is such that, for every $n\geqslant1$, $$ P(X=n)=p_n\prod_{k=1}^{n-1}(1-p_k). $$ 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 $$ \prod_{k=1}^{\infty}(1-p_k)=0. $$ Thus, $X$ is almost surely finite if $p_n=1/(n+1)$ for every $n\geqslant1$ but not if $p_n=1/(n+1)^2$ for every $n\geqslant1$. Second, every distribution on the nonnegative integers may be written like that, simply choose $$ p_n=P(X=n\mid X\geqslant n)=\frac{P(X=n)}{P(X\geqslant n)}. $$ This proves that this class of distributions has no specific name.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.