Proving that stopping time is finite a.s.












0












$begingroup$



Let $$tau_{a} = inf{t>0 : W_{t} + at = 5}.$$
Prove that $mathbb{P}(tau_{a}<infty) = 1$ for $age0.$




My solution:



We know that $W_{0} +a*0 < 5$. Furthermore, because $W_{t} sim sqrt{2tlnlnt}$, we can say that $W_{t} + at xrightarrow{t rightarrowinfty}infty$. And that is why $mathbb{P}(tau_{a}<infty) = 1.$



My question is whether it can be solved like this. I'm not sure about using $W_{t} sim sqrt{2tlnlnt}$.






brownian-motion stopping-times







share|cite|improve this question











$endgroup$












  • $begingroup$
    Well, it's not true that $W_t approx sqrt{2tlog(log(t))}$, that holds only for the limes superior, assuming that $W_t$ is standard Brownian motion, which you should explain in your question.
    $endgroup$
    – user159517
    Jan 8 at 22:51












  • $begingroup$
    Also, the title of your question asks to show that the stopping time is bounded, while in the body of your question you ask to show that the stopping time is almost surely finite.
    $endgroup$
    – user159517
    Jan 8 at 22:55
















0












$begingroup$



Let $$tau_{a} = inf{t>0 : W_{t} + at = 5}.$$
Prove that $mathbb{P}(tau_{a}<infty) = 1$ for $age0.$




My solution:



We know that $W_{0} +a*0 < 5$. Furthermore, because $W_{t} sim sqrt{2tlnlnt}$, we can say that $W_{t} + at xrightarrow{t rightarrowinfty}infty$. And that is why $mathbb{P}(tau_{a}<infty) = 1.$



My question is whether it can be solved like this. I'm not sure about using $W_{t} sim sqrt{2tlnlnt}$.






brownian-motion stopping-times







share|cite|improve this question











$endgroup$












  • $begingroup$
    Well, it's not true that $W_t approx sqrt{2tlog(log(t))}$, that holds only for the limes superior, assuming that $W_t$ is standard Brownian motion, which you should explain in your question.
    $endgroup$
    – user159517
    Jan 8 at 22:51












  • $begingroup$
    Also, the title of your question asks to show that the stopping time is bounded, while in the body of your question you ask to show that the stopping time is almost surely finite.
    $endgroup$
    – user159517
    Jan 8 at 22:55














0












0








0





$begingroup$



Let $$tau_{a} = inf{t>0 : W_{t} + at = 5}.$$
Prove that $mathbb{P}(tau_{a}<infty) = 1$ for $age0.$




My solution:



We know that $W_{0} +a*0 < 5$. Furthermore, because $W_{t} sim sqrt{2tlnlnt}$, we can say that $W_{t} + at xrightarrow{t rightarrowinfty}infty$. And that is why $mathbb{P}(tau_{a}<infty) = 1.$



My question is whether it can be solved like this. I'm not sure about using $W_{t} sim sqrt{2tlnlnt}$.






brownian-motion stopping-times







share|cite|improve this question











$endgroup$





Let $$tau_{a} = inf{t>0 : W_{t} + at = 5}.$$
Prove that $mathbb{P}(tau_{a}<infty) = 1$ for $age0.$




My solution:



We know that $W_{0} +a*0 < 5$. Furthermore, because $W_{t} sim sqrt{2tlnlnt}$, we can say that $W_{t} + at xrightarrow{t rightarrowinfty}infty$. And that is why $mathbb{P}(tau_{a}<infty) = 1.$



My question is whether it can be solved like this. I'm not sure about using $W_{t} sim sqrt{2tlnlnt}$.






brownian-motion stopping-times




brownian-motion stopping-times






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 23:00







Guesttt

















asked Jan 8 at 21:03









GuestttGuesttt

568




568












  • $begingroup$
    Well, it's not true that $W_t approx sqrt{2tlog(log(t))}$, that holds only for the limes superior, assuming that $W_t$ is standard Brownian motion, which you should explain in your question.
    $endgroup$
    – user159517
    Jan 8 at 22:51












  • $begingroup$
    Also, the title of your question asks to show that the stopping time is bounded, while in the body of your question you ask to show that the stopping time is almost surely finite.
    $endgroup$
    – user159517
    Jan 8 at 22:55


















  • $begingroup$
    Well, it's not true that $W_t approx sqrt{2tlog(log(t))}$, that holds only for the limes superior, assuming that $W_t$ is standard Brownian motion, which you should explain in your question.
    $endgroup$
    – user159517
    Jan 8 at 22:51












  • $begingroup$
    Also, the title of your question asks to show that the stopping time is bounded, while in the body of your question you ask to show that the stopping time is almost surely finite.
    $endgroup$
    – user159517
    Jan 8 at 22:55
















$begingroup$
Well, it's not true that $W_t approx sqrt{2tlog(log(t))}$, that holds only for the limes superior, assuming that $W_t$ is standard Brownian motion, which you should explain in your question.
$endgroup$
– user159517
Jan 8 at 22:51






$begingroup$
Well, it's not true that $W_t approx sqrt{2tlog(log(t))}$, that holds only for the limes superior, assuming that $W_t$ is standard Brownian motion, which you should explain in your question.
$endgroup$
– user159517
Jan 8 at 22:51














$begingroup$
Also, the title of your question asks to show that the stopping time is bounded, while in the body of your question you ask to show that the stopping time is almost surely finite.
$endgroup$
– user159517
Jan 8 at 22:55




$begingroup$
Also, the title of your question asks to show that the stopping time is bounded, while in the body of your question you ask to show that the stopping time is almost surely finite.
$endgroup$
– user159517
Jan 8 at 22:55










2 Answers
2






active

oldest

votes


















0












$begingroup$

LIL says $lim sup _{t to infty} frac {W_t} {sqrt {2tln, ln, t}}=1$ almost surely. This implies that $lim sup_{t to infty} W_t= infty$ almost surely. Hence, $lim sup_{t to infty} (W_t+at)= infty$ almost surely. From this and IVP you get $P{tau_a <infty}=1$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This is not wrong, but using LIL to show that $limsup_{t to infty} W_t = +infty$ is really overkill.
    $endgroup$
    – Nate Eldredge
    Jan 9 at 0:08










  • $begingroup$
    @NateEldredge I agree. But my intention was to correct a mistake made by OP.
    $endgroup$
    – Kavi Rama Murthy
    Jan 9 at 0:35



















0












$begingroup$

Brownian motion is recurrent, so almost surely there exists $t$ with $B_t ge 5$. That is, for almost every $omega$, there is $t < infty$, depending on $omega$, such that $B_t(omega) ge 5$. Since $a ge 0$, we have $B_t(omega) + at ge B_t(omega) ge 5$. So by the intermediate value theorem, there is $s le t$ with $B_t(omega) + at =5$. Then $tau_a(omega) le s$. So $tau_a(omega) < infty$, and this is true for almost every $omega$.



Intuitively, $B_t$ is a.s. going to cross the level 5, and with the positive drift, $B_t + at$ is going to cross it even sooner.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    LIL says $lim sup _{t to infty} frac {W_t} {sqrt {2tln, ln, t}}=1$ almost surely. This implies that $lim sup_{t to infty} W_t= infty$ almost surely. Hence, $lim sup_{t to infty} (W_t+at)= infty$ almost surely. From this and IVP you get $P{tau_a <infty}=1$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      This is not wrong, but using LIL to show that $limsup_{t to infty} W_t = +infty$ is really overkill.
      $endgroup$
      – Nate Eldredge
      Jan 9 at 0:08










    • $begingroup$
      @NateEldredge I agree. But my intention was to correct a mistake made by OP.
      $endgroup$
      – Kavi Rama Murthy
      Jan 9 at 0:35
















    0












    $begingroup$

    LIL says $lim sup _{t to infty} frac {W_t} {sqrt {2tln, ln, t}}=1$ almost surely. This implies that $lim sup_{t to infty} W_t= infty$ almost surely. Hence, $lim sup_{t to infty} (W_t+at)= infty$ almost surely. From this and IVP you get $P{tau_a <infty}=1$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      This is not wrong, but using LIL to show that $limsup_{t to infty} W_t = +infty$ is really overkill.
      $endgroup$
      – Nate Eldredge
      Jan 9 at 0:08










    • $begingroup$
      @NateEldredge I agree. But my intention was to correct a mistake made by OP.
      $endgroup$
      – Kavi Rama Murthy
      Jan 9 at 0:35














    0












    0








    0





    $begingroup$

    LIL says $lim sup _{t to infty} frac {W_t} {sqrt {2tln, ln, t}}=1$ almost surely. This implies that $lim sup_{t to infty} W_t= infty$ almost surely. Hence, $lim sup_{t to infty} (W_t+at)= infty$ almost surely. From this and IVP you get $P{tau_a <infty}=1$.






    share|cite|improve this answer









    $endgroup$



    LIL says $lim sup _{t to infty} frac {W_t} {sqrt {2tln, ln, t}}=1$ almost surely. This implies that $lim sup_{t to infty} W_t= infty$ almost surely. Hence, $lim sup_{t to infty} (W_t+at)= infty$ almost surely. From this and IVP you get $P{tau_a <infty}=1$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 8 at 23:56









    Kavi Rama MurthyKavi Rama Murthy

    53.8k32055




    53.8k32055








    • 1




      $begingroup$
      This is not wrong, but using LIL to show that $limsup_{t to infty} W_t = +infty$ is really overkill.
      $endgroup$
      – Nate Eldredge
      Jan 9 at 0:08










    • $begingroup$
      @NateEldredge I agree. But my intention was to correct a mistake made by OP.
      $endgroup$
      – Kavi Rama Murthy
      Jan 9 at 0:35














    • 1




      $begingroup$
      This is not wrong, but using LIL to show that $limsup_{t to infty} W_t = +infty$ is really overkill.
      $endgroup$
      – Nate Eldredge
      Jan 9 at 0:08










    • $begingroup$
      @NateEldredge I agree. But my intention was to correct a mistake made by OP.
      $endgroup$
      – Kavi Rama Murthy
      Jan 9 at 0:35








    1




    1




    $begingroup$
    This is not wrong, but using LIL to show that $limsup_{t to infty} W_t = +infty$ is really overkill.
    $endgroup$
    – Nate Eldredge
    Jan 9 at 0:08




    $begingroup$
    This is not wrong, but using LIL to show that $limsup_{t to infty} W_t = +infty$ is really overkill.
    $endgroup$
    – Nate Eldredge
    Jan 9 at 0:08












    $begingroup$
    @NateEldredge I agree. But my intention was to correct a mistake made by OP.
    $endgroup$
    – Kavi Rama Murthy
    Jan 9 at 0:35




    $begingroup$
    @NateEldredge I agree. But my intention was to correct a mistake made by OP.
    $endgroup$
    – Kavi Rama Murthy
    Jan 9 at 0:35











    0












    $begingroup$

    Brownian motion is recurrent, so almost surely there exists $t$ with $B_t ge 5$. That is, for almost every $omega$, there is $t < infty$, depending on $omega$, such that $B_t(omega) ge 5$. Since $a ge 0$, we have $B_t(omega) + at ge B_t(omega) ge 5$. So by the intermediate value theorem, there is $s le t$ with $B_t(omega) + at =5$. Then $tau_a(omega) le s$. So $tau_a(omega) < infty$, and this is true for almost every $omega$.



    Intuitively, $B_t$ is a.s. going to cross the level 5, and with the positive drift, $B_t + at$ is going to cross it even sooner.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Brownian motion is recurrent, so almost surely there exists $t$ with $B_t ge 5$. That is, for almost every $omega$, there is $t < infty$, depending on $omega$, such that $B_t(omega) ge 5$. Since $a ge 0$, we have $B_t(omega) + at ge B_t(omega) ge 5$. So by the intermediate value theorem, there is $s le t$ with $B_t(omega) + at =5$. Then $tau_a(omega) le s$. So $tau_a(omega) < infty$, and this is true for almost every $omega$.



      Intuitively, $B_t$ is a.s. going to cross the level 5, and with the positive drift, $B_t + at$ is going to cross it even sooner.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Brownian motion is recurrent, so almost surely there exists $t$ with $B_t ge 5$. That is, for almost every $omega$, there is $t < infty$, depending on $omega$, such that $B_t(omega) ge 5$. Since $a ge 0$, we have $B_t(omega) + at ge B_t(omega) ge 5$. So by the intermediate value theorem, there is $s le t$ with $B_t(omega) + at =5$. Then $tau_a(omega) le s$. So $tau_a(omega) < infty$, and this is true for almost every $omega$.



        Intuitively, $B_t$ is a.s. going to cross the level 5, and with the positive drift, $B_t + at$ is going to cross it even sooner.






        share|cite|improve this answer









        $endgroup$



        Brownian motion is recurrent, so almost surely there exists $t$ with $B_t ge 5$. That is, for almost every $omega$, there is $t < infty$, depending on $omega$, such that $B_t(omega) ge 5$. Since $a ge 0$, we have $B_t(omega) + at ge B_t(omega) ge 5$. So by the intermediate value theorem, there is $s le t$ with $B_t(omega) + at =5$. Then $tau_a(omega) le s$. So $tau_a(omega) < infty$, and this is true for almost every $omega$.



        Intuitively, $B_t$ is a.s. going to cross the level 5, and with the positive drift, $B_t + at$ is going to cross it even sooner.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 0:08









        Nate EldredgeNate Eldredge

        62.6k682170




        62.6k682170






























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