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Just a quick question regarding two Poisson Processes:

Let $X_t$ and $Y_t$ be two independent Poisson Processes with rate parameters $\lambda_1$ and $\lambda_2$, respectively, measuring the number of customers arriving in stores 1 and 2, respectively. What is the probability that a customer arrives in store 1 before any customers arrive in store 2?

My approach to this problem thus far has been to consider all possible times where store 1 could have a customer arrive, but that gets into dealing with infinity and I'm not so sure that's correct. Mathematically, I'm thinking I should calculate

$$P(X_1 = 1)P(Y_1 = 0) + P(X_2 = 1|X_1 = 0)P(Y_2 = 0) + P(X_3 = 1|X_2 = 0)P(Y_3 = 0) + ...+ P(X_n = 1|X_{n-1} = 0)P(Y_n = 0).$$

Is there an easier approach than the one I am taking? Is the approach I'm taking even correct?

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  • $\begingroup$ Do you know the relationship between Poisson distribution and exponential distribution? $\endgroup$ – André Nicolas Oct 10 '12 at 6:05
  • $\begingroup$ I know that the expected value of the waiting time is $1/{\lambda}$. Is that what you're getting at? $\endgroup$ – jrad Oct 10 '12 at 6:06
  • $\begingroup$ More. That the waiting time is exponentially distributed. $\endgroup$ – André Nicolas Oct 10 '12 at 6:07
  • $\begingroup$ Sorry, but I'm not really seeing where this is going. Care to elaborate a bit? $\endgroup$ – jrad Oct 10 '12 at 6:13
  • $\begingroup$ Should I give an answer? $\endgroup$ – André Nicolas Oct 10 '12 at 6:17
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Let the random variables $X$ and $Y$ denote the respective waiting times until the first customer. These two random variables have exponential distribution with parameters say $\lambda$ and $\mu$. We want the probability that $X\lt Y$.

By independence, the joint density is the product of the individual densities, so $$\Pr(X\lt Y)=\int_{y=0}^\infty \mu e^{-\mu y}\left(\int_{x=0}^y \lambda e^{-\lambda x}\,dx\right)\,dy.$$ The integrations are not difficult.

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  • $\begingroup$ It's no question why you have so much reputation. Thank you. $\endgroup$ – jrad Oct 10 '12 at 6:23

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