j******n
发帖数: 91 | 1 Brownian motion start from 0.
T(1) first time reach x = 1 ;
T(-1) first time reach x = -1 ;
T(2) first fime reach x = 2;
P(T(1) < T(-1) < T(2)) = ?
My answer is: P(T(1) < T(-1) < T(2)) = P(T(1) < T(-1) )P(T(-1) < T(2)) = 1/2
*2/3 = 1/3. Is it right | n****1
发帖数: 1136 | 2 好像不对,因为第二个过程的起点是1不是0，要修正坐标
P(T(1) | j******n
发帖数: 91 | | M****i
发帖数: 58 | 4 P(T(1) < T(-1) < T(2))
= P(T(1) < T(-1)) - P(T(1) < T(-1), T(-1) > T(2))
= P(T(1) < T(-1)) - P(T(-1) > T(2))
= 1/2 - 1/3
= 1/6 |
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