Tunnel-vision
Continuing my recent trend, I will be posting sporadic recently throughout this term, because I’m attempting to dedicate myself to my math. Most posts in the recent future will be the result of my enthusiasm about a particular technique or subject bubbling over so I have to share. Other than that, I shall be incommunicado.
So here’s a parting problem from my Markov Chains course. I like it because it looks daunting, but if you know the ergodic theorem for MCs, it is trivial:
An opera singer is due to perform a long series of concerts. Having a fine artistic temperament, she is liable to pull out each night with probability 1/2. One this has happened, she will not sing again until the promoter convinces her of his high regard. This he does by sending her flowers every day until she returns. Flowers costing
times 10,000 USD,
, bring about a reconciliation with probability
. The promoter stands to make about 7,500 USD from each successful concert. How much should he spend on flowers?
Possibly relevant posts:
- Criminal cupbearers (3/5/2005)
- Continued Fractions (2/1/2004)
- Oldies but goodies (7/3/2005)
Is
the accumulated value of the flowers sent to her since she last stopped singing, or the value of flowers sent to her tonight?
Comment by AnonEcon — 1/30/2007 @ 3:01 am
Each night that she refuses to sing, the promoter buys
worth of flowers.
By the way, nice proof!
Comment by Alex — 1/31/2007 @ 11:36 am
Thanks! This problem is very interesting too. (
?)
I was trying out a variant to teach myself how to solve dynamic programming problems numerically. I assumed that it is the accumulated value of flowers that counts and that the manager discounts future incomes/expenditures at some rate
.
I got a surprising result: the optimum policy for the manager is to just send one shipment of flowers (whose value depends on
) when she stops singing and send nothing after that till she starts singing again.
Do you have any idea why this might be happening? I would have expected the manager to send a smaller bouquet to begin with and then some more flowers if that didn’t work and so on.
Comment by AnonEcon — 1/31/2007 @ 10:36 pm
Yep, I got 0.25. What method did you use?
Hmm, what do you mean by
Comment by Alex — 2/1/2007 @ 12:49 am
For 0.25: I calculated the stationary probabilities for the markov chain when the value of flowers is
, then reasoned that since for most of the periods the transition probabilities would be arbitrary close to the stationary values and therefore any strategy that gave strictly superior per-period payoffs for the stationary probabilities could not do worse than any other strategy over an infinite horizon.
For my variant: there are now two kinds of states. The state S_on when the singer sings and manager gets 0.75 and from which we transition with equal probability to S_on and S_off(0). And the states S_off(K) where the singer is not singing and has already received K worth of flowers. If managers chooses to send I worth of flowers then we transition with probability
to state S_on and with the remaining probability to S_off(K+I). If the manager’s income/expenditure in period t is 
then his goal is to maximize the expected value of
Comment by AnonEcon — 2/1/2007 @ 11:31 am