Supremum of absolute value of Brownian Motion

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I know that by the reflection principle,
$$
Pleft[sup_0 < s < t B_s > a right] = 2P[B_t> a]
$$
where $B_t$ is a Brownian Motion. But what is $Pleft[sup_0 < s < t |B_s|> a right]$?




brownian-motion




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  • 1




    Hint: $B_t$ has the same distribution as $-B_t$ and $P(sup_0 < s < t |B_s| > a) = P(sup_0 < s < tB_t > a$ or $inf_0 < s < t B_t < -a)$. Use a union bound--for your previous question, you just need to bound this probability, you do not need to know the probability exactly.
    – Chris Janjigian
    Jul 30 at 16:52















up vote
0
down vote

favorite












I know that by the reflection principle,
$$
Pleft[sup_0 < s < t B_s > a right] = 2P[B_t> a]
$$
where $B_t$ is a Brownian Motion. But what is $Pleft[sup_0 < s < t |B_s|> a right]$?




brownian-motion




share|cite|improve this question















  • 1




    Hint: $B_t$ has the same distribution as $-B_t$ and $P(sup_0 < s < t |B_s| > a) = P(sup_0 < s < tB_t > a$ or $inf_0 < s < t B_t < -a)$. Use a union bound--for your previous question, you just need to bound this probability, you do not need to know the probability exactly.
    – Chris Janjigian
    Jul 30 at 16:52













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I know that by the reflection principle,
$$
Pleft[sup_0 < s < t B_s > a right] = 2P[B_t> a]
$$
where $B_t$ is a Brownian Motion. But what is $Pleft[sup_0 < s < t |B_s|> a right]$?




brownian-motion




share|cite|improve this question











I know that by the reflection principle,
$$
Pleft[sup_0 < s < t B_s > a right] = 2P[B_t> a]
$$
where $B_t$ is a Brownian Motion. But what is $Pleft[sup_0 < s < t |B_s|> a right]$?





brownian-motion





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 30 at 16:07









Tohiko

1245




1245







  • 1




    Hint: $B_t$ has the same distribution as $-B_t$ and $P(sup_0 < s < t |B_s| > a) = P(sup_0 < s < tB_t > a$ or $inf_0 < s < t B_t < -a)$. Use a union bound--for your previous question, you just need to bound this probability, you do not need to know the probability exactly.
    – Chris Janjigian
    Jul 30 at 16:52













  • 1




    Hint: $B_t$ has the same distribution as $-B_t$ and $P(sup_0 < s < t |B_s| > a) = P(sup_0 < s < tB_t > a$ or $inf_0 < s < t B_t < -a)$. Use a union bound--for your previous question, you just need to bound this probability, you do not need to know the probability exactly.
    – Chris Janjigian
    Jul 30 at 16:52








1




1




Hint: $B_t$ has the same distribution as $-B_t$ and $P(sup_0 < s < t |B_s| > a) = P(sup_0 < s < tB_t > a$ or $inf_0 < s < t B_t < -a)$. Use a union bound--for your previous question, you just need to bound this probability, you do not need to know the probability exactly.
– Chris Janjigian
Jul 30 at 16:52





Hint: $B_t$ has the same distribution as $-B_t$ and $P(sup_0 < s < t |B_s| > a) = P(sup_0 < s < tB_t > a$ or $inf_0 < s < t B_t < -a)$. Use a union bound--for your previous question, you just need to bound this probability, you do not need to know the probability exactly.
– Chris Janjigian
Jul 30 at 16:52
















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