- #1
fignewtons
- 28
- 0
Homework Statement
You are playing a game with two bells. Bell A rings according to a homogeneous poisson process at a rate r per hour and Bell B rings once at a time T that is uniformly distributed from 0 to 1 hr (inclusive). You get $1 each time A rings and can quit anytime but if B rings before you quit, you must return all the money you received thus far.
Homework Equations
P[A rings] = rΔt
P[B rings] = Δt/(1-t)
The Attempt at a Solution
At time t, with x in earnings, the optimal stop strategy for this game is continuing if E[W] > 0, where W is net earnings over a small period of time (call it Δt):
E[W] = E[W|A rings]P[A rings] + E[W|B rings]P[B rings] + E[W|no rings]P[no rings] + E[W|many rings]P[many rings]
E[W|A rings] = 1
E[W|B rings] = $-x
E[W|no rings] = $0
and P[many rings] ≈ 0 (since Δt is small time interval)
So E[W] = E[W|A rings]P[A rings] + E[W|B rings]P[B rings]
substituting in the relevant equations...
the strategy is to continue only if x < r(1-t), and stop otherwise.
I'm not sure how to find what's the most amount of money I can win under this strategy. I thought it means I should take the derivative of this winnings strategy but taking the derivative doesn't make sense for the inequality.
Ie. derivative wrt x of x < r(1-t) is d/dx(x-r(1-t)) < 0 which yields 1< 0, doesn't make sense.