H****h
发帖数: 1037 | 1 某人每天步行上下班。每趟路程上下雨的概率是p(0
他在家里和办公室里共放了n把伞。上下班时，如果没有雨，就不随身带伞；
如果下了雨且手边有伞就打伞，如果没伞就只好挨雨淋了。
求证：无论开始状态如何，单程挨雨淋的概率都将趋于一个定值，并计算之。 | I***e
发帖数: 1136 | 2 I forgot what it was called. But the content says:
If a Markov chain is recurrent and irreducible and have finite
states, then the limiting measure will be the stationary measure
no matter what initial measure you choose. ( Recurrent means with
probability one the Markov chain will revisit its starting point.
Irreducible means you can reach any state from any other state
if you wait long enough. )
It is contained in the Markov chain chapter of Richard Durrett's
book.
Thanks, all.
Icare
【在 H****h 的大作中提到】
: 某人每天步行上下班。每趟路程上下雨的概率是p(0: 他在家里和办公室里共放了n把伞。上下班时，如果没有雨，就不随身带伞；
: 如果下了雨且手边有伞就打伞，如果没伞就只好挨雨淋了。
: 求证：无论开始状态如何，单程挨雨淋的概率都将趋于一个定值，并计算之。
| I***e
发帖数: 1136 | 3 You are right... I mis-configured the markov structure.
So is the right answer equal to:
PROB = p(1-p)/(n+1-p) ?
This structure is a bit more complex than what I imagined.
If we identify situations by (number, localtion) where
number
is the number of umbrellas at office, and location is the
current location
of the guy. Then the stationary distribution is:
Status: (n,home) (n,office) (n-1,home)
(n-1,office) ...
PROB: p1 p2 p2 p2
Status: | I***e
发帖数: 1136 | 4
It seemes that I have to defend my solution a
little more, although the theoretical reasoning
is quite enough... :)
It is wrong to assume that your location and
the location of the umbrella are independent.
Intuitively, when p is close to 1, after a few
steps, your deterministic strategy of carrying the
umbrella whenever rains will guarantee that the
probability of having the umbrella at your current
location is much higher than the counterpart if
p is small. So, as p approaches 1, you are more
【在 H****h 的大作中提到】
: 某人每天步行上下班。每趟路程上下雨的概率是p(0: 他在家里和办公室里共放了n把伞。上下班时，如果没有雨，就不随身带伞；
: 如果下了雨且手边有伞就打伞，如果没伞就只好挨雨淋了。
: 求证：无论开始状态如何，单程挨雨淋的概率都将趋于一个定值，并计算之。
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