Two independent Poisson Processes

Question

$X$ and $Y$ are two independent Poisson Processes with rates $\lambda_1, \lambda_2$ respectively. What is the probability that at least two events from $X$ occur before a total of two events from $Y$? A valid example would be $X_1, Y_1, X_2, Y_2...$ since two events from $X$ occurred before a total of 2 events from $Y$.

I think I use complimentary counting to solve this problem. The invalid cases I've noted are $X_1, Y_1, Y_2...$; $Y_2, Y_2...$; $Y_1, X_1, Y_2...$ Am I on the right track or is there a more efficient/cleaner method?

Answer 1

HINT

Derive the distribution of $T_2$, the second event occurrence from a Poisson process and you have to find $\mathbb{P}[T_2<S_2]$ where $T_2$ is 2nd event from the first and $S_2$ is 2nd event from the 2nd process.

Answer 2

Hints:

• Calculate the probabilities that the first event is from $X$, and that the first event is from $Y$
• You only care about the first two or three events
• Use the memoryless property of Poisson processes to find the probabilities of
  • An event from $X$ followed by another event from $X$ without any events from $Y$
  • An event from $X$ followed by an event from $Y$ followed by another event from $X$ without more events from $Y$
  • An event from $Y$ followed by an event from $X$ followed by another event from $X$ without more events from $Y$
• Add up those probabilities

Answer 3

You are on the right track. Considering the combined process as you are doing, you only need to look at the first three arrivals. There are eight possibilities (XXX, XXY, XYX, XYY, YXX, YXY, YYX, YYY), and you can figure out which ones to count. (I think counting the complement is no easier/harder than counting what you actually want, but you have the right idea already.)

Now comes the question of how to compute the probabilities associated with the above eight outcomes. It turns out that the outcomes have the same distribution as three independent coin flips: the probability that an arrival is from $X$ is $\frac{\lambda_1}{\lambda_1+\lambda_2}$, and the probability that an arrival is from $Y$ is $\frac{\lambda_2}{\lambda_1+\lambda_2}$. Summing up the probabilities of the "good" outcomes will give you the answer.

See this page for an explanation of why the arrivals in the combined process behave like coin flips with the above probabilities.