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If we have two Poisson process $A$ and $B$ with arrival rate $\lambda_{1}$ and $\lambda_{2}$. Now I have question for calculating the probability of an arrival e.g., from process A in the case of merged Poisson process. Which of the approach we should use.

Considering binomial distribution, given an arrival, the probability that this arrival belongs to process A is $\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}$

Or should we use the traditional Poisson pmf to compute the probability of an arrival from process A, such as using, $$p_{m}=\frac{(\lambda_{1}t)^{m}.e^{-\lambda_{1}t}}{m!}$$ and we can put m=1 to get the probability of an arrival from process A using only $\lambda_{1}$ in the equation

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closed as unclear what you're asking by Did, Sahiba Arora, Namaste, Leucippus, user91500 Aug 2 '17 at 8:55

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  • $\begingroup$ Your second approach is unclear. Care to expand on it? $\endgroup$ – Did Aug 1 '17 at 12:53
  • $\begingroup$ Since a standard Poisson pmf is $$p_{m}=\frac{(\lambda t)^{m}.e^{-\lambda t}}{m!}$$ where $m$ goes form $0,1,2,3...................$ I used this approach for Process A as Process A has an arrival rate of $\lambda_{1}$ and if I consider a time interval of $t$ for both processes (as observing time interval).. $\endgroup$ – Hallian1990 Aug 1 '17 at 13:05
  • $\begingroup$ Sorry but your explanations do not make much sense to me (and recalling the definition of the Poisson distribution was not exactly useful in this perspective, sorry to say). $\endgroup$ – Did Aug 1 '17 at 21:57
  • $\begingroup$ No, my question is whether the second way is right or wrong? the objective is to find the probability of an arrival from process A. The first approach is right I know considering the binomial distribution. Since, the Process A is a poisson process ,independent of process B, and for an interval $t$, why can't we use the definition of Poisson process to compute the probability of an arrival from Process A? $\endgroup$ – Hallian1990 Aug 1 '17 at 22:58
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If we have two Poisson process $A$ and $B$ with arrival rate $\lambda_{1}$ and $\lambda_{2}$. Now I have question for calculating the probability of an arrival e.g., from process A in the case of merged Poisson process. Which of the approach we should use.

If you have two independent Poisson processes, then the probabiliy that any particular arrival in the merged process was generated by process $A$ is indeed $\lambda_1/(\lambda_1+\lambda_2)$ .

A Poisson process consists of point events (arrivals) that occur over an interval at a constant average rate but each independently of any other arrival in the process.

Thus in the merged process you have arrivals that occur at a constant joint rate $(\lambda_1+\lambda_2)$ and each independently generated by one from the two independent Possion processes.


Your aproach seems to be asking the condtional probability that, in a very small interval, we get one arrival from process A and no arrival from process B, given that we get only one arrival from the merged process in that very small interval.   Well...sure.

$$\lim\limits_{\Delta t\to 0^+}\cfrac{~\cfrac{(\lambda_1\Delta t)^1e^{-\lambda_1\Delta t}}{1!}\cdot\cfrac{(\lambda_2\Delta t)^0e^{-\lambda_2\Delta t}~}{0!}}{\cfrac{((\lambda_1+\lambda_2)\Delta t)^1e^{-(\lambda_1+\lambda_2)\Delta t}}{1!}}~=~\dfrac{\lambda_1}{\lambda_1+\lambda_2}$$


[Of course, to use this approach you have to first establish that the count of arrivals in the joint process follows a Poisson distribution with the combined average rate.]

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  • $\begingroup$ Many thanks @Graham Kemp. got it. $\endgroup$ – Hallian1990 Aug 2 '17 at 12:18

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