n**********E
发帖数: 157 | 1 给一个centered unit circle C_1，一个no drift的BM从（x,y）点开始，求circle的
hitting time distribution，不是求expected hitting time。
谢谢 |
k*****n
发帖数: 117 | 2 Xt and Yt are jointly normal and independent
So you can work out distribution of Dt = Xt^2 + Yt^2
Now you can follow the same argument of calculating hitting time
distribution of a Brownian motion
P(T < t ) = 2P( Dt > 1 ) = ... |
n****e
发帖数: 629 | 3 趁没人看见赶紧删了吧:D
【在 k*****n 的大作中提到】
: Xt and Yt are jointly normal and independent
: So you can work out distribution of Dt = Xt^2 + Yt^2
: Now you can follow the same argument of calculating hitting time
: distribution of a Brownian motion
: P(T < t ) = 2P( Dt > 1 ) = ...
|
k*****n
发帖数: 117 | 4 why?
【在 n****e 的大作中提到】
: 趁没人看见赶紧删了吧:D
|
k*****n
发帖数: 117 | 5 hmm.. i see your point now.
【在 k*****n 的大作中提到】
: why?
|
n*********e
发帖数: 9 | 6 这开头挺好的啊
这样可以算出来从中心到半径为r的圆的hitting time的分布，记为F_r(t)
那么从一个距离中心z = sqrt(x^2+y^2)的点到半径为1的圆的hitting time D_z(t)，
应该满足
F_1（t) = int_0^t D_z(t-x) f_r(x) dx
对t求导有 D_z(t-x) = f_1(t)/f_r(x)
这是点在圆内的情况。
【在 k*****n 的大作中提到】
: Xt and Yt are jointly normal and independent
: So you can work out distribution of Dt = Xt^2 + Yt^2
: Now you can follow the same argument of calculating hitting time
: distribution of a Brownian motion
: P(T < t ) = 2P( Dt > 1 ) = ...
|
