Thread: Poisson arrivals, departures via exponential and uniform distributions

Question: "Customers arrive at a sporting goods store at times of a Poisson process with rate 10 per hour. 60% of the customers are men and 40% are women. Women spend an amount of time shopping that is uniformly distributed on [0, 30] minutes, while men spend an exponentially distributed amount of time with mean 30 minutes. Suppose the store has been open for a long time. Find the joint distribution of X = the number of men and Y = the number of women shopping."

This doesn't make sense to me. If the rate of arrival of men is 6 per hour but the rate of departure for men is 2 per hour (mean 1/2 hour per visit = 2 visit/hour), wouldn't the long-run distribution of men explode?

Could someone please give me a hint on how to construct this joint distribution? Thanks a bunch!

Using decomposition of Poisson processes, the arrival of men is a Poisson process with rate 6 per hour while the arrival of women is Poisson with rate 4 per hour and the two arrivals are independent of each other.

I'm assuming that the store can accommodate an infinite amount of customers and so for the men customers, it's equivalent to an queueing system while for women, it's . All the customers are served upon entering the store and so the long-run distribution of men will not explode

Additionally, it's stated that the store has been opened for a long time and so what we need here is the equilibrium distribution. For , the equilibrium distribution is Poisson with parameter a = mean arrival rate * mean service time. For , due to the insensitivity property, the equilibrium distribution is also Poisson with the same parameter.

p/s: Ithaca? Are you from Cornell?

I got it - thanks a bunch!

And yes I am. How did you guess? I could be from IC, you know