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









    $endgroup$















      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









      $endgroup$





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













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 24 at 18:19









      XiaoXiao

      4,84611636




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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 $delta>0$.






          share|cite|improve this 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











            $endgroup$


















              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











              $endgroup$
















                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











                $endgroup$



                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














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 25 at 11:34

























                answered Jan 24 at 18:44









                sazsaz

                81.8k861129




                81.8k861129






























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