Reversal of Brownian motion from first hitting time

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Let $(B_t)_t ge 0$ be a (standard) Brownian motion, and for fixed $C>0$ define the first hitting time
$$T_C := inf t ge 0: B_t = C.$$

I am interested in the reversal of the Brownian motion from the first hitting time $T_C$, i.e. a description of the distribution of the path
$$(B_T_C - s - C)_s le T_C.$$

My question is whether this has the same law as
$$(widetildeB_s)_s le S_-C$$
where $(widetildeB_s)_s ge 0$ is a Brownian motion conditioned to stay non-positive, and $S_-C$ is the last time that the process $(B_s)$ hits $-C$.

My guess above comes from the analogous problem for Brownian motion with negative drift. I was told that the following is true (I would appreciate a reference for that):

Fix $m > 0$ and $C > 0$ as before. Let $T_C$ (resp. $S_-C$) be the first hitting time of $C$ (resp. last hittimg time of $-C$) of the Brownian motion with negative drift $(B_s -ms)_s ge 0$. Then
$$(B_T_C - s + m(T_C - s) - C)_s le T_C oversetd= (widetildeB_s - ms)_s le S_-C$$
where $(widetildeB_s - ms)_s ge 0$ is a Brownian motion with drift $-m$ conditioned to stay non-positive.

As a heuristic I can send $m to 0$ and that would recover my claim above, but this does not constitute a mathematical proof. It would be great if someone could tell me that my claim is correct and point me to references where a proof can be easily found.

probability-theory stochastic-processes brownian-motion stopping-times

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edited Jul 24 at 14:01

asked Jul 24 at 11:28

random_person

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up vote
2
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Let $(B_t)_t ge 0$ be a (standard) Brownian motion, and for fixed $C>0$ define the first hitting time
$$T_C := inf t ge 0: B_t = C.$$

I am interested in the reversal of the Brownian motion from the first hitting time $T_C$, i.e. a description of the distribution of the path
$$(B_T_C - s - C)_s le T_C.$$

My question is whether this has the same law as
$$(widetildeB_s)_s le S_-C$$
where $(widetildeB_s)_s ge 0$ is a Brownian motion conditioned to stay non-positive, and $S_-C$ is the last time that the process $(B_s)$ hits $-C$.

My guess above comes from the analogous problem for Brownian motion with negative drift. I was told that the following is true (I would appreciate a reference for that):

Fix $m > 0$ and $C > 0$ as before. Let $T_C$ (resp. $S_-C$) be the first hitting time of $C$ (resp. last hittimg time of $-C$) of the Brownian motion with negative drift $(B_s -ms)_s ge 0$. Then
$$(B_T_C - s + m(T_C - s) - C)_s le T_C oversetd= (widetildeB_s - ms)_s le S_-C$$
where $(widetildeB_s - ms)_s ge 0$ is a Brownian motion with drift $-m$ conditioned to stay non-positive.

As a heuristic I can send $m to 0$ and that would recover my claim above, but this does not constitute a mathematical proof. It would be great if someone could tell me that my claim is correct and point me to references where a proof can be easily found.

probability-theory stochastic-processes brownian-motion stopping-times

share|cite|improve this question

edited Jul 24 at 14:01

asked Jul 24 at 11:28

random_person

1286

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $(B_t)_t ge 0$ be a (standard) Brownian motion, and for fixed $C>0$ define the first hitting time
$$T_C := inf t ge 0: B_t = C.$$

I am interested in the reversal of the Brownian motion from the first hitting time $T_C$, i.e. a description of the distribution of the path
$$(B_T_C - s - C)_s le T_C.$$

My question is whether this has the same law as
$$(widetildeB_s)_s le S_-C$$
where $(widetildeB_s)_s ge 0$ is a Brownian motion conditioned to stay non-positive, and $S_-C$ is the last time that the process $(B_s)$ hits $-C$.

My guess above comes from the analogous problem for Brownian motion with negative drift. I was told that the following is true (I would appreciate a reference for that):

Fix $m > 0$ and $C > 0$ as before. Let $T_C$ (resp. $S_-C$) be the first hitting time of $C$ (resp. last hittimg time of $-C$) of the Brownian motion with negative drift $(B_s -ms)_s ge 0$. Then
$$(B_T_C - s + m(T_C - s) - C)_s le T_C oversetd= (widetildeB_s - ms)_s le S_-C$$
where $(widetildeB_s - ms)_s ge 0$ is a Brownian motion with drift $-m$ conditioned to stay non-positive.

As a heuristic I can send $m to 0$ and that would recover my claim above, but this does not constitute a mathematical proof. It would be great if someone could tell me that my claim is correct and point me to references where a proof can be easily found.

probability-theory stochastic-processes brownian-motion stopping-times

share|cite|improve this question

edited Jul 24 at 14:01

asked Jul 24 at 11:28

random_person

1286

Let $(B_t)_t ge 0$ be a (standard) Brownian motion, and for fixed $C>0$ define the first hitting time
$$T_C := inf t ge 0: B_t = C.$$

I am interested in the reversal of the Brownian motion from the first hitting time $T_C$, i.e. a description of the distribution of the path
$$(B_T_C - s - C)_s le T_C.$$

My question is whether this has the same law as
$$(widetildeB_s)_s le S_-C$$
where $(widetildeB_s)_s ge 0$ is a Brownian motion conditioned to stay non-positive, and $S_-C$ is the last time that the process $(B_s)$ hits $-C$.

My guess above comes from the analogous problem for Brownian motion with negative drift. I was told that the following is true (I would appreciate a reference for that):

Fix $m > 0$ and $C > 0$ as before. Let $T_C$ (resp. $S_-C$) be the first hitting time of $C$ (resp. last hittimg time of $-C$) of the Brownian motion with negative drift $(B_s -ms)_s ge 0$. Then
$$(B_T_C - s + m(T_C - s) - C)_s le T_C oversetd= (widetildeB_s - ms)_s le S_-C$$
where $(widetildeB_s - ms)_s ge 0$ is a Brownian motion with drift $-m$ conditioned to stay non-positive.

As a heuristic I can send $m to 0$ and that would recover my claim above, but this does not constitute a mathematical proof. It would be great if someone could tell me that my claim is correct and point me to references where a proof can be easily found.

probability-theory stochastic-processes brownian-motion stopping-times

share|cite|improve this question

edited Jul 24 at 14:01

asked Jul 24 at 11:28

random_person

1286

share|cite|improve this question

share|cite|improve this question

edited Jul 24 at 14:01

edited Jul 24 at 14:01

edited Jul 24 at 14:01

asked Jul 24 at 11:28

random_person

1286

asked Jul 24 at 11:28

random_person

1286

asked Jul 24 at 11:28

random_person

1286

1286

add a comment | 

add a comment | 

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