l**********r
发帖数: 12 | 1 how to calculate the distribution of the time of reaching minimum of a
Brownian Motion Bt? 0<=t<=1
I guess that its pdf looks like
- symmetric around 1/2
- higher at 0, 1 and lower at 1/2 | y***s
发帖数: 23 | 2 As B_t and -B_t have the same distribution, it's
enough to consider the max.
Pr(max B_t > a)=Pr(min B_t > a)=exp(-2a^2).
Take derivative to get the pdf.
reference
3.3' in
http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view= | l**********r
发帖数: 12 | 3 sorry I don't understand
could you explain a bit how to connect argmin{Bt} with 3.3' in the paper?
Thanks
【在 y***s 的大作中提到】
: As B_t and -B_t have the same distribution, it's
: enough to consider the max.
: Pr(max B_t > a)=Pr(min B_t > a)=exp(-2a^2).
: Take derivative to get the pdf.
: reference
: 3.3' in
: http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=
| y***s
发帖数: 23 | 4 Pr(min{B_t,0a)=Pr(min{-B_t,0a)(B_t and -B_t has the same
distribution)
=Pr(-max{B_t,0a)
=Pr(max{B_t,0
=1-Pr(max{B_t,0-a)
1-exp(-2a^2) by 3.3' in the paper
The pdf of distribution of argmin{ Bt, t in [0, 1] } is
(-4a)exp(-2a^2)
where a is negative
by taking derivative of 1-[1-exp(-2a^2)] with respect to a. | k*******d
发帖数: 1340 | 5 My understanding is argmin{Bt, t in [0,1]} is the argument t, so we are
asked to find the distribution of t_min where B(t_min) is the mininum of B(t
) when t in [0,1]. It seems to me you are finding the pdf of the minimum
value.
【在 y***s 的大作中提到】
: Pr(min{B_t,0a)=Pr(min{-B_t,0a)(B_t and -B_t has the same
: distribution)
: =Pr(-max{B_t,0a)
: =Pr(max{B_t,0: =1-Pr(max{B_t,0-a)
: 1-exp(-2a^2) by 3.3' in the paper
: The pdf of distribution of argmin{ Bt, t in [0, 1] } is
: (-4a)exp(-2a^2)
: where a is negative
: by taking derivative of 1-[1-exp(-2a^2)] with respect to a.
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