n*********3 发帖数: 21 | 1 A person stands in the long boat on the lake. Say he is at point 0, the
distance from him to either end of boat is A and B. If this guy follows
Brownian motion, how long will he falls into water. (this question needs
martingale solution.) | t*******e 发帖数: 172 | 2 For all kinds of this porblem:
Theorem:
Suppose there is a start point p in a bounded domain D, then let t be the
stopping time of a brownian motion start from p hit the boundary of D.
Then the expect value of t is u(t). Where u is a fuction such that
Laplace(u)=-2, u=0 on the boundary.
Especially, in a interval [-A,B], the corresponding function u(x)=(x+A)(x-B)
. So the expected value of stopping time is AB.
【在 n*********3 的大作中提到】![](/moin_static193/solenoid/img/up.png) : A person stands in the long boat on the lake. Say he is at point 0, the : distance from him to either end of boat is A and B. If this guy follows : Brownian motion, how long will he falls into water. (this question needs : martingale solution.)
| n****e 发帖数: 629 | 3 This is a nice solution and should be marked.
我前几天胡思乱想也考虑过这道题。事实上有几种情况:
1.如果是random walk(discrete), 那么给定boundary A,B 期望stopping time?
2.如果是biased,往上走概率p,那么给定boundary A,B 期望stopping time?
3.如果改成brownian motion(with/without drift) 那么给定boundary A,B 期望
stopping time?
如果把A->\infty, 就是面试很常见的题了。
都可以用martingale来解。先放在这里大家讨论一下,hehe
B)
【在 t*******e 的大作中提到】![](/moin_static193/solenoid/img/up.png) : For all kinds of this porblem: : Theorem: : Suppose there is a start point p in a bounded domain D, then let t be the : stopping time of a brownian motion start from p hit the boundary of D. : Then the expect value of t is u(t). Where u is a fuction such that : Laplace(u)=-2, u=0 on the boundary. : Especially, in a interval [-A,B], the corresponding function u(x)=(x+A)(x-B) : . So the expected value of stopping time is AB.
| n****e 发帖数: 629 | 4 我来试着解释一下。
这里Laplace(u)多半指的是laplace operator, 二阶导。
就是说,u(t)是u的二次函数。Intuitively, 一个brownian motion走的距离应该是
sqrt(t),所以相对的,stopping time应该是距离的二次函数。
然后把边条件带进去,凑出函数形式。其实前面还应该有个常数要定。这是Steele书里
第一章的做法。
LZ说要用martingale,那么可以考虑(X-A)(X-B)-t 这个函数。X=X(t)是当前位置。证
明这个是一个martigale 然后再用Optional stopping theorem立得。
以上解法仅对没有drift的情况成立。
B)
【在 t*******e 的大作中提到】![](/moin_static193/solenoid/img/up.png) : For all kinds of this porblem: : Theorem: : Suppose there is a start point p in a bounded domain D, then let t be the : stopping time of a brownian motion start from p hit the boundary of D. : Then the expect value of t is u(t). Where u is a fuction such that : Laplace(u)=-2, u=0 on the boundary. : Especially, in a interval [-A,B], the corresponding function u(x)=(x+A)(x-B) : . So the expected value of stopping time is AB.
| z****i 发帖数: 406 | 5 probability of getting A before B: B/(A+B);
probability of getting B before A: A/(A+B);
(gambler ruin problem)
(S_n)^2 - n is a martingale, stopped at A or B is also a martingale;
if stopped at A, E(A^2-n) = 0, therefore E(n) = A^2.
if stopped at B, E(B^2-n) = 0, therefore E(n) = B^2.
So expected time = B/(A+B)*A^2 + A/(A+B)*B^2 = AB.
Don't know how to construct a martingale if p is not 1/2. Anybody give a
hint please? | d*j 发帖数: 13780 | 6 看绿皮书。。。。
【在 z****i 的大作中提到】![](/moin_static193/solenoid/img/up.png) : probability of getting A before B: B/(A+B); : probability of getting B before A: A/(A+B); : (gambler ruin problem) : (S_n)^2 - n is a martingale, stopped at A or B is also a martingale; : if stopped at A, E(A^2-n) = 0, therefore E(n) = A^2. : if stopped at B, E(B^2-n) = 0, therefore E(n) = B^2. : So expected time = B/(A+B)*A^2 + A/(A+B)*B^2 = AB. : Don't know how to construct a martingale if p is not 1/2. Anybody give a : hint please?
| z****i 发帖数: 406 | 7 哪本?Zhou 那本?
【在 d*j 的大作中提到】![](/moin_static193/solenoid/img/up.png) : 看绿皮书。。。。
| s*******s 发帖数: 1568 | 8 I got a explanation,
Use black-shoes PDE, and let dS(t)=dW(t)
thus the B-S PDE can be written as
df/dt + 1/2df/ds^2 = 0, with f(A)=f(B)=0,
f is stopping time, thus df/dt =1
this actually is the connection between the brownian motion and la-place
equation
【在 n****e 的大作中提到】![](/moin_static193/solenoid/img/up.png) : 我来试着解释一下。 : 这里Laplace(u)多半指的是laplace operator, 二阶导。 : 就是说,u(t)是u的二次函数。Intuitively, 一个brownian motion走的距离应该是 : sqrt(t),所以相对的,stopping time应该是距离的二次函数。 : 然后把边条件带进去,凑出函数形式。其实前面还应该有个常数要定。这是Steele书里 : 第一章的做法。 : LZ说要用martingale,那么可以考虑(X-A)(X-B)-t 这个函数。X=X(t)是当前位置。证 : 明这个是一个martigale 然后再用Optional stopping theorem立得。 : 以上解法仅对没有drift的情况成立。 :
| d*j 发帖数: 13780 | 9 so niu X
【在 s*******s 的大作中提到】![](/moin_static193/solenoid/img/up.png) : I got a explanation, : Use black-shoes PDE, and let dS(t)=dW(t) : thus the B-S PDE can be written as : df/dt + 1/2df/ds^2 = 0, with f(A)=f(B)=0, : f is stopping time, thus df/dt =1 : this actually is the connection between the brownian motion and la-place : equation
| d*j 发帖数: 13780 | 10 哦, 你是说 expected stopping time
上面没有说, 但是应该同一个方法
【在 z****i 的大作中提到】![](/moin_static193/solenoid/img/up.png) : 哪本?Zhou 那本?
| w**********y 发帖数: 1691 | 11 sorry..有错..第二个martingale好像不对..跟老板meet..回来再检查..
E(S_1)=p-(1-p)=2p-1
Define X_n=S_n-(2p-1)n
1.X_n is martingale
2.x_n^2-n is martingale
3. exp(aX_n-0.5a^2*n) is martingale
Part a:
From 3, exp[a(S_n-(2p-1)n)-0.5a^2*n] is martingale
let a=-2(2p-1)
then exp(-2(2p-1)S_n) is martingale
exp(-2(2p-1)S_tau) is martingale
E(exp(-2(2p-1)S_tau))=1
so P(s_tau=A)*exp[-2(2p-1)A]+(1-P)*exp[-2(2p-1)B]=1
solution: P(s hit A first)=(exp[-2(2p-1)B]-1)/(exp[-2(2p-1)B]-exp[-2(2p-1)A])
Part b:
From the linear weighted | n****8 发帖数: 1716 | 12 The guy is an idiot by following Brownian motion. | z****i 发帖数: 406 | 13 part a 里面的tau还是定义为S第1次到A或者B的时间吧?怎么证明它对于X也是停时呢?
【在 w**********y 的大作中提到】![](/moin_static193/solenoid/img/up.png) : sorry..有错..第二个martingale好像不对..跟老板meet..回来再检查.. : E(S_1)=p-(1-p)=2p-1 : Define X_n=S_n-(2p-1)n : 1.X_n is martingale : 2.x_n^2-n is martingale : 3. exp(aX_n-0.5a^2*n) is martingale : Part a: : From 3, exp[a(S_n-(2p-1)n)-0.5a^2*n] is martingale : let a=-2(2p-1) : then exp(-2(2p-1)S_n) is martingale
|
|