It started raining today while I was at work. I don’t bring an umbrella to my office unless it is actually raining when I come to work (it wasn’t today). However, I am in luck. I have an umbrella in my office. I left it here the last time when it was raining during my morning commute but had stopped raining by the time I needed to go home.
Those of you who have had a course in stochastic processes will recognize this as the umbrella problem. That is, a professor travels between an office and home. She has U umbrellas in total, and she brings one with her if it is raining. It rains with probability p. We can determine the fraction of time she gets wet using the limiting distribution of a Markov chain.
In the Markov chain, let our state reflect the number of available umbrellas at the professors current location (either home or her office). Then the Markov chain has U+1 states. The transition probability matrix for U = 4 is:
P= [0 0 0 0 1 <-- when there are no umbrellas, all U umbrellas will be at the other location 0 0 0 1-p p 0 0 1-p p 0 <-- with probability p, one umbrella is moved to be with the umbrellas at the other location 0 1-p p 0 0 1-p p 0 0 0] <-- with probability p, one of the U umbrellas will be moved to the other location, and with probability 1-p, there will be no umbrellas at the other location.
Let the limiting distribution be the vector π. Therefore, the proportion of time I get wet is p x π[0 ] (π is the proportion of time I am in a location with no umbrellas).
Since I have U=2 umbrellas, the odds are pretty good that I will get wet on my way to or from my office. I think I lucked out today.
Other umbrella problem links: