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】
: 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】
: 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】
: 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】
: 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】
: 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】
: 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】
: 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】
: 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】
: 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
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