I am trying to work out a research problem I recently faced. I have a group of Poisson random variables and I want to find the distribution of the first sample that is equal to a specific number. In math terms I have

Given Poisson random variables $S_{1},S_{2}...S_{b}..$ that are generated from rate parameters $\beta_{1},\beta_{2}...\beta_{b}..$

Let $A=min_{b} \big( {S_{b}=s} \big) $

What is the probability of $\beta_{A}$ i.e $P(\beta_{A}=z)$


1 Answer 1


Given $z$, let $\mathcal{B}_z$ be the set of indices $b$ for which $\beta_b = z$. Then $\beta_A = z$ precisely when the first Poisson variable having the value $s$ has an index belonging to the set $\mathcal{B}_z$, that is,

$$ P(\beta_A = z) = P(A \in \mathcal{B}_z) = \sum_{b \in \mathcal{B}_z} P(A = b),$$

where the second equality holds as the events $\{ A = b \}$ are mutually exclusive for $b \in \mathcal{B}_z$. Then, $A = b$ if and only if the first $b - 1$ Poisson variables do not equal $s$ but the $b$th one does, $P(A = b) = P((S_1 \neq s) \cap \cdots \cap (S_{b-1} \neq s) \cap (S_b = s)) $. If you are willing to assume that the Poisson-variates are independent this can be further factorized and the the desired probability gets the form

\begin{align} P(\beta_A = z) &= \sum_{b \in \mathcal{B}_z} \left( \prod_{i=1}^{b-1} P(S_i \neq s) \right) P(S_b = s) \\ &= \sum_{b \in \mathcal{B}_z} \left( \prod_{i=1}^{b-1} \left( \sum_{j \neq s}^\infty \frac{\beta_i^j}{j!} e^{-\beta_i} \right) \right) \frac{\beta_b^s}{s!} e^{-\beta_b}, \end{align}

which is actually defined for all $z \in \mathbb{R}$ as if $\mathcal{B}_z = \emptyset$ we end up with an empty sum which equals zero.

  • $\begingroup$ thank you. I think you can consider that $A$ is a first succes probability and treat it as a geometric diustrubution , then the calculations will be simplified. Please check if that is possibel and I will accept the answer afterwards $\endgroup$
    – Wis
    Dec 9, 2016 at 5:08
  • $\begingroup$ @raw5 Geometric distribution assumes that the probability of success is the same for each replication and that is not the case here, unless the rate parameters are all equal (in which case the distribution of $\beta_A$ is degenerate). $\endgroup$
    – J. Virta
    Dec 9, 2016 at 18:41

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