$mathbb{P}left {sup_{tin [0,delta]} B_{t} geq delta^{2/3}right} geq 1-delta^alpha$ as $delta$ tends to zero.

1

$begingroup$

$mathbb{P}left {sup_{tin [0,delta]} B_{t} geq delta^{2/3}right} geq 1-delta^alpha$ as $delta$ tends to zero.

Does this type of tail estimate exist in the literature?

By Blumenthal $0-1$ law, the limit would apprach to $1$ as $delta$ tends to zero since

$$mathbb{P}left {sup_{tin [0,delta]} B_{t} geq delta^{2/3}right} geq mathbb{P}left {B_{delta} geq delta^{2/3}right} = mathbb{P}left {frac{B_{delta}}{sqrt{delta}} geq delta^{1/6}right} = 1-Phi(delta^{1/6}) geq 1/10$$
and I was wondering if there is a precise tail estimate.

probability-theory stochastic-processes brownian-motion

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asked Jan 24 at 18:19

XiaoXiao

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1

$begingroup$

$mathbb{P}left {sup_{tin [0,delta]} B_{t} geq delta^{2/3}right} geq 1-delta^alpha$ as $delta$ tends to zero.

Does this type of tail estimate exist in the literature?

By Blumenthal $0-1$ law, the limit would apprach to $1$ as $delta$ tends to zero since

$$mathbb{P}left {sup_{tin [0,delta]} B_{t} geq delta^{2/3}right} geq mathbb{P}left {B_{delta} geq delta^{2/3}right} = mathbb{P}left {frac{B_{delta}}{sqrt{delta}} geq delta^{1/6}right} = 1-Phi(delta^{1/6}) geq 1/10$$
and I was wondering if there is a precise tail estimate.

probability-theory stochastic-processes brownian-motion

share|cite|improve this question

asked Jan 24 at 18:19

XiaoXiao

4,84611636

$endgroup$

add a comment | 

1

1

1

$begingroup$

$mathbb{P}left {sup_{tin [0,delta]} B_{t} geq delta^{2/3}right} geq 1-delta^alpha$ as $delta$ tends to zero.

Does this type of tail estimate exist in the literature?

By Blumenthal $0-1$ law, the limit would apprach to $1$ as $delta$ tends to zero since

$$mathbb{P}left {sup_{tin [0,delta]} B_{t} geq delta^{2/3}right} geq mathbb{P}left {B_{delta} geq delta^{2/3}right} = mathbb{P}left {frac{B_{delta}}{sqrt{delta}} geq delta^{1/6}right} = 1-Phi(delta^{1/6}) geq 1/10$$
and I was wondering if there is a precise tail estimate.

probability-theory stochastic-processes brownian-motion

share|cite|improve this question

asked Jan 24 at 18:19

XiaoXiao

4,84611636

$endgroup$

$mathbb{P}left {sup_{tin [0,delta]} B_{t} geq delta^{2/3}right} geq 1-delta^alpha$ as $delta$ tends to zero.

Does this type of tail estimate exist in the literature?

By Blumenthal $0-1$ law, the limit would apprach to $1$ as $delta$ tends to zero since

$$mathbb{P}left {sup_{tin [0,delta]} B_{t} geq delta^{2/3}right} geq mathbb{P}left {B_{delta} geq delta^{2/3}right} = mathbb{P}left {frac{B_{delta}}{sqrt{delta}} geq delta^{1/6}right} = 1-Phi(delta^{1/6}) geq 1/10$$
and I was wondering if there is a precise tail estimate.

probability-theory stochastic-processes brownian-motion

probability-theory stochastic-processes brownian-motion

share|cite|improve this question

asked Jan 24 at 18:19

XiaoXiao

4,84611636

share|cite|improve this question

asked Jan 24 at 18:19

XiaoXiao

4,84611636

share|cite|improve this question

share|cite|improve this question

asked Jan 24 at 18:19

XiaoXiao

4,84611636

asked Jan 24 at 18:19

XiaoXiao

4,84611636

asked Jan 24 at 18:19

XiaoXiao

4,84611636

4,84611636

add a comment | 

add a comment | 

1 Answer
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$begingroup$

By the reflection principle, it holds that

$$sup_{t in [0,delta]} B_t sim |B_{delta}|,$$

and therefore

$$mathbb{P} left( sup_{t in [0,delta]} B_t geq delta^{2/3} right) = mathbb{P}(|B_{delta}| geq delta^{2/3}).$$

As $|B_{delta}| sim sqrt{delta} |B_1|$ this gives

$$mathbb{P} left( sup_{t in [0,delta]} B_t geq delta^{2/3} right) =mathbb{P}(|B_1|geq delta^{4/3}) = 1-mathbb{P}(|B_1|< delta^{4/3}).$$

Noting that

$$mathbb{P}(|B_1| < delta^{4/3}) = frac{1}{sqrt{2pi}} int_{-delta^{4/3}}^{delta^{4/3}} exp left(- frac{y^2}{2} right) , dy$$

we find that

$$mathbb{P}(|B_1| < delta^{4/3}) approx sqrt{frac{2}{pi}} exp left(- frac{delta^{8/3}}{2} right) delta^{4/3} $$

for small $delta>0$. This means that

$$mathbb{P} left( sup_{t in [0,delta]} B_t geq delta^{2/3} right) approx 1- sqrt{frac{2}{pi}} underbrace{exp left(- frac{delta^{8/3}}{2} right)}_{approx 1} delta^{4/3}$$

for small $delta>0$.

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edited Jan 25 at 11:34

answered Jan 24 at 18:44

sazsaz

81.8k861129

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add a comment | 

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1 Answer
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1 Answer
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1

$begingroup$

By the reflection principle, it holds that

$$sup_{t in [0,delta]} B_t sim |B_{delta}|,$$

and therefore

$$mathbb{P} left( sup_{t in [0,delta]} B_t geq delta^{2/3} right) = mathbb{P}(|B_{delta}| geq delta^{2/3}).$$

As $|B_{delta}| sim sqrt{delta} |B_1|$ this gives

$$mathbb{P} left( sup_{t in [0,delta]} B_t geq delta^{2/3} right) =mathbb{P}(|B_1|geq delta^{4/3}) = 1-mathbb{P}(|B_1|< delta^{4/3}).$$

Noting that

$$mathbb{P}(|B_1| < delta^{4/3}) = frac{1}{sqrt{2pi}} int_{-delta^{4/3}}^{delta^{4/3}} exp left(- frac{y^2}{2} right) , dy$$

we find that

$$mathbb{P}(|B_1| < delta^{4/3}) approx sqrt{frac{2}{pi}} exp left(- frac{delta^{8/3}}{2} right) delta^{4/3} $$

for small $delta>0$. This means that

$$mathbb{P} left( sup_{t in [0,delta]} B_t geq delta^{2/3} right) approx 1- sqrt{frac{2}{pi}} underbrace{exp left(- frac{delta^{8/3}}{2} right)}_{approx 1} delta^{4/3}$$

for small $delta>0$.

share|cite|improve this answer

edited Jan 25 at 11:34

answered Jan 24 at 18:44

sazsaz

81.8k861129

$endgroup$

add a comment | 

1

$begingroup$

By the reflection principle, it holds that

$$sup_{t in [0,delta]} B_t sim |B_{delta}|,$$

and therefore

$$mathbb{P} left( sup_{t in [0,delta]} B_t geq delta^{2/3} right) = mathbb{P}(|B_{delta}| geq delta^{2/3}).$$

As $|B_{delta}| sim sqrt{delta} |B_1|$ this gives

$$mathbb{P} left( sup_{t in [0,delta]} B_t geq delta^{2/3} right) =mathbb{P}(|B_1|geq delta^{4/3}) = 1-mathbb{P}(|B_1|< delta^{4/3}).$$

Noting that

$$mathbb{P}(|B_1| < delta^{4/3}) = frac{1}{sqrt{2pi}} int_{-delta^{4/3}}^{delta^{4/3}} exp left(- frac{y^2}{2} right) , dy$$

we find that

$$mathbb{P}(|B_1| < delta^{4/3}) approx sqrt{frac{2}{pi}} exp left(- frac{delta^{8/3}}{2} right) delta^{4/3} $$

for small $delta>0$. This means that

$$mathbb{P} left( sup_{t in [0,delta]} B_t geq delta^{2/3} right) approx 1- sqrt{frac{2}{pi}} underbrace{exp left(- frac{delta^{8/3}}{2} right)}_{approx 1} delta^{4/3}$$

for small $delta>0$.

share|cite|improve this answer

edited Jan 25 at 11:34

answered Jan 24 at 18:44

sazsaz

81.8k861129

$endgroup$

add a comment | 

1

1

1

$begingroup$

By the reflection principle, it holds that

$$sup_{t in [0,delta]} B_t sim |B_{delta}|,$$

and therefore

$$mathbb{P} left( sup_{t in [0,delta]} B_t geq delta^{2/3} right) = mathbb{P}(|B_{delta}| geq delta^{2/3}).$$

As $|B_{delta}| sim sqrt{delta} |B_1|$ this gives

$$mathbb{P} left( sup_{t in [0,delta]} B_t geq delta^{2/3} right) =mathbb{P}(|B_1|geq delta^{4/3}) = 1-mathbb{P}(|B_1|< delta^{4/3}).$$

Noting that

$$mathbb{P}(|B_1| < delta^{4/3}) = frac{1}{sqrt{2pi}} int_{-delta^{4/3}}^{delta^{4/3}} exp left(- frac{y^2}{2} right) , dy$$

we find that

$$mathbb{P}(|B_1| < delta^{4/3}) approx sqrt{frac{2}{pi}} exp left(- frac{delta^{8/3}}{2} right) delta^{4/3} $$

for small $delta>0$. This means that

$$mathbb{P} left( sup_{t in [0,delta]} B_t geq delta^{2/3} right) approx 1- sqrt{frac{2}{pi}} underbrace{exp left(- frac{delta^{8/3}}{2} right)}_{approx 1} delta^{4/3}$$

for small $delta>0$.

share|cite|improve this answer

edited Jan 25 at 11:34

answered Jan 24 at 18:44

sazsaz

81.8k861129

$endgroup$

By the reflection principle, it holds that

$$sup_{t in [0,delta]} B_t sim |B_{delta}|,$$

and therefore

$$mathbb{P} left( sup_{t in [0,delta]} B_t geq delta^{2/3} right) = mathbb{P}(|B_{delta}| geq delta^{2/3}).$$

As $|B_{delta}| sim sqrt{delta} |B_1|$ this gives

$$mathbb{P} left( sup_{t in [0,delta]} B_t geq delta^{2/3} right) =mathbb{P}(|B_1|geq delta^{4/3}) = 1-mathbb{P}(|B_1|< delta^{4/3}).$$

Noting that

$$mathbb{P}(|B_1| < delta^{4/3}) = frac{1}{sqrt{2pi}} int_{-delta^{4/3}}^{delta^{4/3}} exp left(- frac{y^2}{2} right) , dy$$

we find that

$$mathbb{P}(|B_1| < delta^{4/3}) approx sqrt{frac{2}{pi}} exp left(- frac{delta^{8/3}}{2} right) delta^{4/3} $$

for small $delta>0$. This means that

$$mathbb{P} left( sup_{t in [0,delta]} B_t geq delta^{2/3} right) approx 1- sqrt{frac{2}{pi}} underbrace{exp left(- frac{delta^{8/3}}{2} right)}_{approx 1} delta^{4/3}$$

for small $delta>0$.

share|cite|improve this answer

edited Jan 25 at 11:34

answered Jan 24 at 18:44

sazsaz

81.8k861129

share|cite|improve this answer

share|cite|improve this answer

edited Jan 25 at 11:34

edited Jan 25 at 11:34

edited Jan 25 at 11:34

answered Jan 24 at 18:44

sazsaz

81.8k861129

answered Jan 24 at 18:44

sazsaz

81.8k861129

answered Jan 24 at 18:44

sazsaz

81.8k861129

81.8k861129

add a comment | 

add a comment | 

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