0
$\begingroup$

Suppose you have two independent Poisson processes $P_1,P_2$ with rates $\lambda_1,\lambda_2$ respectively.

1. If you merge these two Poisson processes, what is the probability that for the first $n$ outcomes, exactly $k$ of them are from $P_1$?

2. And conditional on the previous events, what is distribution of the time before the next outcome?

For the first problem, my idea is that consider a Poisson process with rate $\lambda_1+\lambda_2,$ for each outcome, there is probability $\frac{\lambda_1}{\lambda_1+\lambda_2}$ that falls into type I and with probability $\frac{\lambda_2}{\lambda_1+\lambda_2}$ it falls into type II. By splitting theorem, I this the number of type I outcomes should equal to outcomes of $P_1.$ Therefore for the first problem, I think the answer is $\binom{n}{k}(\frac{\lambda_1}{\lambda_1+\lambda_2})^k(\frac{\lambda_2}{\lambda_1+\lambda_2})^{n-k}.$

For the second question, I think either we can view it as a renewal process with interarrival time expo$(\lambda_1+\lambda_2)$. Or it is $\min(expo(\lambda_1),expo(\lambda_2))=expo(\lambda_1+\lambda_2).$ I am not sure if my answer is correct.

$\endgroup$
2
$\begingroup$

The answer to your first question is correct: the counting distribution of events of the first type is binomial with parameters $n$ and $p = \lambda_1/(\lambda_1+\lambda_2)$ and your reasoning is proper.

The answer to the second question should become evident by noting that the merged process is equivalent to a process where event type is irrelevant; i.e., it is a Poisson process with rate $\lambda_1 + \lambda_2$, hence the interarrival time is exponentially distributed with this rate parameter. The fact that the minimum order statistic of two independent exponentially distributed random variables is itself exponential (but with rate equal to the sum of individual rates) is a consequence of the memoryless property.

$\endgroup$

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.