Independent Poisson Processes

1. Feb 4, 2008

circa415

I'm not sure how to get started with this:

Suppose you have two independent poisson processes, X ~ Po(Lambda1) Y ~Po (Lambda2). How would you figure out the probability of one coming before the other?

Thanks

2. Feb 4, 2008

EnumaElish

What are you modeling, arrival times or number of arrivals? Poisson is generally used for the latter.

In problems involving a comparison between two random variables (e.g. X < Y) one usually needs to define a new random variable, say Z = X - Y, then calculate Pr{Z < 0} from the joint distribution of X and Y.

3. Feb 4, 2008

circa415

thanks for the tip

I'm modeling arrival times for a poisson process. Ex: the time it takes for an insurance company to receive n claims. For instance, what is the probability that One insurance company receives n claims before another insurance company receives n claims

Last edited: Feb 4, 2008
4. Feb 4, 2008

EnumaElish

AFAIK the Poisson distribution describes the number of arrivals in a fixed time frame. This makes me think that you may want to reformulate your question as the number of arrivals in the first process being less (or greater) than those in the second process within a given time period.

5. Feb 4, 2008

gel

A Po(Lambda) process models a process whose jumps (arrivals) occur at rate Lambda - probability of a jump in time dt is Lambda * dt to highest order in dt (http://en.wikipedia.org/wiki/Poisson_process" [Broken].) You could use the fact that the distance between these jumps are independent and exponentially distributed. So the time to the n'th jump (n'th claim) is the sum of n independent exponential r.v.s, which is a chi(2n)-squared distribution.

You could also work out the result for n = 1, and then extend to larger n. If t(n) is the time of the n'th arrival of X & Y combined, then look at the process Z(n) = X(t(n))-Y(t(n)). You can use the memorylessness property to show that this will be a random walk, and is given by the binomial distribution for each n. To say that the first n arrivals of X occur before the first n arrivals of Y is equivalent to Z(2n-1)>=0, which you can then get from the binomial distribution.

Do you know the result for n=1?

Last edited by a moderator: May 3, 2017