P****l 发帖数: 156 | 1 求,QAUNT版,各位大牛~~~~
给个解法~~~~ |
A*****s 发帖数: 13748 | 2 再现小女子求助帖 lol
【在 P****l 的大作中提到】 : 求,QAUNT版,各位大牛~~~~ : 给个解法~~~~
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V****u 发帖数: 73 | 3 too simple...
【在 P****l 的大作中提到】 : 求,QAUNT版,各位大牛~~~~ : 给个解法~~~~
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k*******d 发帖数: 1340 | 4 不觉得simple啊,完整写出来得挺复杂的,求大牛们的简便做法
定义stopping time \tau = the first time B(t) = a+(a-b)t, 要求的就是P(\tau<\
infty).
Racall that M(t) = exp(\theta B(t) - 1/2 * \theta^2 t) is a martingale,
M(the min of \tau and t) is also a martingale, then
E[M(the min of \tau and t)] = E[M(0)] = 1; i.e.
E[exp(\theta B(min of \tau and t) - 1/2 \theta^2(min of \tau and t))] =
用indicator function 把它拆开
E[exp(\theta B(\tau) - 1/2 \theta^2 \tau) * Indicator function(\tau<=t)]
+ E[exp(\thata B(t) - 1/2 \theta^2 t) * Indicator |
n****e 发帖数: 629 | 5 其实就是一个带drift的brownian hit boundary的概率的问题。
标准做法是写到指数上去,搞成martingale
这里讨论过很多次了吧。
【在 k*******d 的大作中提到】 : 不觉得simple啊,完整写出来得挺复杂的,求大牛们的简便做法 : 定义stopping time \tau = the first time B(t) = a+(a-b)t, 要求的就是P(\tau<\ : infty). : Racall that M(t) = exp(\theta B(t) - 1/2 * \theta^2 t) is a martingale, : M(the min of \tau and t) is also a martingale, then : E[M(the min of \tau and t)] = E[M(0)] = 1; i.e. : E[exp(\theta B(min of \tau and t) - 1/2 \theta^2(min of \tau and t))] = : 用indicator function 把它拆开 : E[exp(\theta B(\tau) - 1/2 \theta^2 \tau) * Indicator function(\tau<=t)] : + E[exp(\thata B(t) - 1/2 \theta^2 t) * Indicator
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b*****t 发帖数: 10 | 6 最后那步怎么证的?
我觉得 可以 直接 让 lambda=2a 就把 tau_a 消去了吧 |
f*****s 发帖数: 141 | 7 also too naive?
【在 V****u 的大作中提到】 : too simple...
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P****l 发帖数: 156 | 8 谢谢牛人~~~我去试试~~~~
是作业题哈~~~~竟然被看出来了
【在 k*******d 的大作中提到】 : 不觉得simple啊,完整写出来得挺复杂的,求大牛们的简便做法 : 定义stopping time \tau = the first time B(t) = a+(a-b)t, 要求的就是P(\tau<\ : infty). : Racall that M(t) = exp(\theta B(t) - 1/2 * \theta^2 t) is a martingale, : M(the min of \tau and t) is also a martingale, then : E[M(the min of \tau and t)] = E[M(0)] = 1; i.e. : E[exp(\theta B(min of \tau and t) - 1/2 \theta^2(min of \tau and t))] = : 用indicator function 把它拆开 : E[exp(\theta B(\tau) - 1/2 \theta^2 \tau) * Indicator function(\tau<=t)] : + E[exp(\thata B(t) - 1/2 \theta^2 t) * Indicator
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