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Given $N(t)$ a Poisson Process with generation rate $\lambda$ with $t_1<t_2$ and $N_2>N_1$ I'm looking for a way to express the following probability: $$ P[N(t_2)>N_2|N(t_1)<N_1]$$ In other words, what is the probability of counting more than $N_2$ events in time $t_2$ given that at time $t_1$ I had counted less than $N_1$ events? So far I only got to express it in terms of sum of explicit probabilities. $$ =\sum_{k=0}^{N_1-1} P[N(t_1)=k]P[N(t_2-t_1)\geq N_2-k]$$ $$ =\sum_{k=0}^{N_1-1} \frac{(\lambda t_1)^ke^{\lambda t_1}}{k!}\left[ \sum_{j=N_2-k}^{+\infty} \frac{(\lambda (t_2-t_1))^je^{\lambda (t_2-t_1)}}{j!}\right]$$

Is there a way to further simplify this expression? Or another clever way to look at the problem?

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1 Answer 1

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$$P(N(t_2)>N_2|N(t_1)<N_1)=P(N(t_1)<N_1 \land N(t_2)>N_2)/P(N(t_1)<N_1)=\sum_{k=0}^{N_1-1} [P(N(t_1)=k)P(N(t_2) >N_2)]/P(N(t_1)<N_1)= \frac{\sum_{k=0}^{N_1-1} [P(N(t_1)=k)(1-P(N(t_2) \le N_2))]}{P(N(t_1)<N_1)}$$

if it's not for exam then this is (I hope) best form to implement in algorithm, after all not all problems have nice formula as solution

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