K*V
发帖数: 192 | 1 为什么 the probability that Bt hits 1 or -1 is 1? |
d********t
发帖数: 9628 | 2 lim_{tau-->Inf.} P(t<=tau) = 1 ?
【在 K*V 的大作中提到】
: 为什么 the probability that Bt hits 1 or -1 is 1?
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l******i
发帖数: 1404 | 3 要用Monotone Convegence Theorem来证E[1_{\tau<\infty}]=1，
where \tau=the first passage time to hit 1 or -1.
1_{\tau<\infty}=1 if \tau<\infty; 1_{\tau<\infty}=0 otherwise.
具体过程：Shreve's Book, Chapter 3, Section 6有详细证明。
请问这是哪家的什么职位的面试题？谢谢楼主共享。 |
h*****n
发帖数: 38 | 4 Sketch of Proof:
Let P_x := probability that S_t doesn't hit -1 and 1 within time [0,1],
starting at x \in [-1,1].
just need to prove P_x <= P_0 < 1, then S_t doesn't hit [-1,1] in [0,n}
would be less than P_0^n and as n-> infinity this probability goes to 0
Now, P_x <=P_0 is proven by considering the distribution of B_1, which is
normal and B_1 is outside of [-1,1] of probability strictly greater than 0.
hence P_0 is strictly less than 1. Done. |
m***w
发帖数: 404 | 5 Use the Reflection Principle (pp 178-179, Intro.to Stochastic Process,
Lawler, 2nd). P(B_s = +1 for some 0 <= s <= t | B_0 = 0) = 2*P(B_s >= 1 for
some some 0 <= s <= t | B_0 = 0), it is easily known that when t->infinity,
2*P(B_s >= 1 for some some 0 <= s <= t | B_0 = 0) -> 1. Same for B_s = -1.
That means that, given infinite time and its starting point 0, B_s will
finally cross +1 or -1 with probability 1. Am I right? |
k*****y
发帖数: 744 | 6 直观点说只要看在T时刻 |B_T| > 1那部分，这时B_T~N(0, sqrt{T})，variance随着T
越来越大，出现在(-1,1)之间的概率就趋向于0了。
【在 K*V 的大作中提到】
: 为什么 the probability that Bt hits 1 or -1 is 1?
|
x******a
发帖数: 6336 | 7 let A= \{ |B_T|<1 \} and B= \{ \sup_{0
P(B) \leq P(A) = P( B_1 \leq 1/\sqrt{T}) \to 0 as T\to \infty. |
R**T
发帖数: 784 | 8 我的理解是stopping time of 1 or -1小于等于stopping time of 1
然后stopping time of 1有probability为1，Shreve的书上证得很详细
【在 K*V 的大作中提到】
: 为什么 the probability that Bt hits 1 or -1 is 1?
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