s**********i
发帖数: 3 | 1 请各位大神拔刀相助，帮我度过考试难关，感激不尽！
1. I've made appointment to see two students, one at noon(12:00) and the
other at 12:15. The duration of the appointments are independent exponential
with mean 15 minutes. I cannot meet with the second student until I'm done
with the first. What is the expected amount of time the 12:15 student spends
at my office? Assume the students arrive at their appointed times.
2. Consider the following game. There are two clocks sounding alarms. Clock
A’s alarm sounds according to
a homogeneous Poisson process with arrival rate h/hour. Clock B’s alarm
sounds only once at a time T that is
uniformly distributed on [0, 1 hour]. Every time clock A’s alarm sounds I
pay you $1. You can stop playing any
time you want and walk away with your winnings, but if clock B’s alarm
sounds before you stop, the game ends and
you must give back any money I’ve paid you. A stopping strategy is a
function f(n, t) → {stop, continue} where
0 ≤ t ≤ 1 is the elapsed time and n ≥ 0 is how many dollars I’ve paid
you by time t. For example, if play has
not stopped by time t = .5 (1/2 hour) and clock A has sounded 4 times, f(4,
.5) tells you whether to stop at that
moment or continue play.
(a) Determine the optimal stopping strategy.
(b) What is the most money you can win under this strategy?
(c) If h = 1 and you pursue the optimal strategy, what is the expected value
of your winnings?
(d) Challenge: do (c) with h = 2. | F****z
发帖数: 12 | 2 直觉上，第二题的optimal strategy是f(n,t)=continue if and only if Pr(clock A
will sound again before clock B will sound)>n/(n+1)，但我没有证明。 | k*b
发帖数: 15 | 3 1.
E(the amount of time the second student stays in your office)
Prob(the first one is done within 15 min)*15+Prob(the first one spends more
than 15 mins)*(15+E(the amount of time the first one spends after 12:15|the
first one spends more than 15 mins))
=(1-e^{-1})*15+e^{-1}(15+15)=15+15*e^{-1}
exponential
done
spends
Clock
【在 s**********i 的大作中提到】
: 请各位大神拔刀相助，帮我度过考试难关，感激不尽！
: 1. I've made appointment to see two students, one at noon(12:00) and the
: other at 12:15. The duration of the appointments are independent exponential
: with mean 15 minutes. I cannot meet with the second student until I'm done
: with the first. What is the expected amount of time the 12:15 student spends
: at my office? Assume the students arrive at their appointed times.
: 2. Consider the following game. There are two clocks sounding alarms. Clock
: A’s alarm sounds according to
: a homogeneous Poisson process with arrival rate h/hour. Clock B’s alarm
: sounds only once at a time T that is
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