Probability on first hitting time of Brownian motion with drift












2












$begingroup$


I am struggling with the following problem:




Let $B$ be a one dimensional Brownian motion and $a,b>0$. Show that $$P[B_t=a + bt text{ for some } tgeq 0] = e^{-2ab}.$$




The following hint is given: Consider the martingale $(X_t)_{tgeq 0} = (exp(2bB_t -2b^2 t))_{t geq 0}$.



I already showed that $(X_t)$ is a martingale but I do not have any idea how I can use this to prove the statement.



Could somebody help me?
Thanks in advance!






probability-theory brownian-motion martingales stopping-times







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    You have to define a suitable stopping time and apply the optional stopping theorem.
    $endgroup$
    – saz
    Dec 8 '18 at 16:24






  • 2




    $begingroup$
    Sub-hint: $(B_t)$ hits the line $B_t=a+bt$ if and only if $(X_t)$ hits...
    $endgroup$
    – Did
    Dec 8 '18 at 16:25










  • $begingroup$
    Ok, so clearly $P[B_t = a+bt text{ for some } tgeq0] = P[X_t = exp(2ba) text{ for some } tgeq0] = P[tau < infty]$ when I define $tau$ to be the hitting time of $exp(2ba)$. But how can I use the optional stopping theorem here? It says that $E(X_tau) = E(X_0)$ if the necessary conditions are fulfilled, and then...?
    $endgroup$
    – nabla
    Dec 9 '18 at 17:21












  • $begingroup$
    See below for further hints. (I didn't get a notification about your comment because you didn't ping me... you have to write @xyz at the beginning of your comment in order to notify the person with username "xyz".)
    $endgroup$
    – saz
    Dec 10 '18 at 15:09
















2












$begingroup$


I am struggling with the following problem:




Let $B$ be a one dimensional Brownian motion and $a,b>0$. Show that $$P[B_t=a + bt text{ for some } tgeq 0] = e^{-2ab}.$$




The following hint is given: Consider the martingale $(X_t)_{tgeq 0} = (exp(2bB_t -2b^2 t))_{t geq 0}$.



I already showed that $(X_t)$ is a martingale but I do not have any idea how I can use this to prove the statement.



Could somebody help me?
Thanks in advance!






probability-theory brownian-motion martingales stopping-times







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    You have to define a suitable stopping time and apply the optional stopping theorem.
    $endgroup$
    – saz
    Dec 8 '18 at 16:24






  • 2




    $begingroup$
    Sub-hint: $(B_t)$ hits the line $B_t=a+bt$ if and only if $(X_t)$ hits...
    $endgroup$
    – Did
    Dec 8 '18 at 16:25










  • $begingroup$
    Ok, so clearly $P[B_t = a+bt text{ for some } tgeq0] = P[X_t = exp(2ba) text{ for some } tgeq0] = P[tau < infty]$ when I define $tau$ to be the hitting time of $exp(2ba)$. But how can I use the optional stopping theorem here? It says that $E(X_tau) = E(X_0)$ if the necessary conditions are fulfilled, and then...?
    $endgroup$
    – nabla
    Dec 9 '18 at 17:21












  • $begingroup$
    See below for further hints. (I didn't get a notification about your comment because you didn't ping me... you have to write @xyz at the beginning of your comment in order to notify the person with username "xyz".)
    $endgroup$
    – saz
    Dec 10 '18 at 15:09














2












2








2


1



$begingroup$


I am struggling with the following problem:




Let $B$ be a one dimensional Brownian motion and $a,b>0$. Show that $$P[B_t=a + bt text{ for some } tgeq 0] = e^{-2ab}.$$




The following hint is given: Consider the martingale $(X_t)_{tgeq 0} = (exp(2bB_t -2b^2 t))_{t geq 0}$.



I already showed that $(X_t)$ is a martingale but I do not have any idea how I can use this to prove the statement.



Could somebody help me?
Thanks in advance!






probability-theory brownian-motion martingales stopping-times







share|cite|improve this question











$endgroup$




I am struggling with the following problem:




Let $B$ be a one dimensional Brownian motion and $a,b>0$. Show that $$P[B_t=a + bt text{ for some } tgeq 0] = e^{-2ab}.$$




The following hint is given: Consider the martingale $(X_t)_{tgeq 0} = (exp(2bB_t -2b^2 t))_{t geq 0}$.



I already showed that $(X_t)$ is a martingale but I do not have any idea how I can use this to prove the statement.



Could somebody help me?
Thanks in advance!






probability-theory brownian-motion martingales stopping-times




probability-theory brownian-motion martingales stopping-times






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 15:14









saz

79.8k860124




79.8k860124










asked Dec 8 '18 at 16:08









nablanabla

1089




1089








  • 3




    $begingroup$
    You have to define a suitable stopping time and apply the optional stopping theorem.
    $endgroup$
    – saz
    Dec 8 '18 at 16:24






  • 2




    $begingroup$
    Sub-hint: $(B_t)$ hits the line $B_t=a+bt$ if and only if $(X_t)$ hits...
    $endgroup$
    – Did
    Dec 8 '18 at 16:25










  • $begingroup$
    Ok, so clearly $P[B_t = a+bt text{ for some } tgeq0] = P[X_t = exp(2ba) text{ for some } tgeq0] = P[tau < infty]$ when I define $tau$ to be the hitting time of $exp(2ba)$. But how can I use the optional stopping theorem here? It says that $E(X_tau) = E(X_0)$ if the necessary conditions are fulfilled, and then...?
    $endgroup$
    – nabla
    Dec 9 '18 at 17:21












  • $begingroup$
    See below for further hints. (I didn't get a notification about your comment because you didn't ping me... you have to write @xyz at the beginning of your comment in order to notify the person with username "xyz".)
    $endgroup$
    – saz
    Dec 10 '18 at 15:09














  • 3




    $begingroup$
    You have to define a suitable stopping time and apply the optional stopping theorem.
    $endgroup$
    – saz
    Dec 8 '18 at 16:24






  • 2




    $begingroup$
    Sub-hint: $(B_t)$ hits the line $B_t=a+bt$ if and only if $(X_t)$ hits...
    $endgroup$
    – Did
    Dec 8 '18 at 16:25










  • $begingroup$
    Ok, so clearly $P[B_t = a+bt text{ for some } tgeq0] = P[X_t = exp(2ba) text{ for some } tgeq0] = P[tau < infty]$ when I define $tau$ to be the hitting time of $exp(2ba)$. But how can I use the optional stopping theorem here? It says that $E(X_tau) = E(X_0)$ if the necessary conditions are fulfilled, and then...?
    $endgroup$
    – nabla
    Dec 9 '18 at 17:21












  • $begingroup$
    See below for further hints. (I didn't get a notification about your comment because you didn't ping me... you have to write @xyz at the beginning of your comment in order to notify the person with username "xyz".)
    $endgroup$
    – saz
    Dec 10 '18 at 15:09








3




3




$begingroup$
You have to define a suitable stopping time and apply the optional stopping theorem.
$endgroup$
– saz
Dec 8 '18 at 16:24




$begingroup$
You have to define a suitable stopping time and apply the optional stopping theorem.
$endgroup$
– saz
Dec 8 '18 at 16:24




2




2




$begingroup$
Sub-hint: $(B_t)$ hits the line $B_t=a+bt$ if and only if $(X_t)$ hits...
$endgroup$
– Did
Dec 8 '18 at 16:25




$begingroup$
Sub-hint: $(B_t)$ hits the line $B_t=a+bt$ if and only if $(X_t)$ hits...
$endgroup$
– Did
Dec 8 '18 at 16:25












$begingroup$
Ok, so clearly $P[B_t = a+bt text{ for some } tgeq0] = P[X_t = exp(2ba) text{ for some } tgeq0] = P[tau < infty]$ when I define $tau$ to be the hitting time of $exp(2ba)$. But how can I use the optional stopping theorem here? It says that $E(X_tau) = E(X_0)$ if the necessary conditions are fulfilled, and then...?
$endgroup$
– nabla
Dec 9 '18 at 17:21






$begingroup$
Ok, so clearly $P[B_t = a+bt text{ for some } tgeq0] = P[X_t = exp(2ba) text{ for some } tgeq0] = P[tau < infty]$ when I define $tau$ to be the hitting time of $exp(2ba)$. But how can I use the optional stopping theorem here? It says that $E(X_tau) = E(X_0)$ if the necessary conditions are fulfilled, and then...?
$endgroup$
– nabla
Dec 9 '18 at 17:21














$begingroup$
See below for further hints. (I didn't get a notification about your comment because you didn't ping me... you have to write @xyz at the beginning of your comment in order to notify the person with username "xyz".)
$endgroup$
– saz
Dec 10 '18 at 15:09




$begingroup$
See below for further hints. (I didn't get a notification about your comment because you didn't ping me... you have to write @xyz at the beginning of your comment in order to notify the person with username "xyz".)
$endgroup$
– saz
Dec 10 '18 at 15:09










1 Answer
1






active

oldest

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3












$begingroup$

Hints:




  1. Define a stopping time $tau$ by $$tau := inf{t geq 0; X_t = e^{2ab}}.$$ Show that $$mathbb{P}(tau<infty) = mathbb{P}(exists t geq 0: B_t = a+bt). tag{1}$$

  2. Apply the optional stopping theorem to show that $$mathbb{E}(X_{t wedge tau}) = mathbb{E}(X_0)=1 tag{2}$$ for all $t geq 0$.

  3. Show that $$lim_{t to infty} X_{t wedge tau}(omega)=e^{2ab} quad text{for $omega in {tau<infty}$}$$ and $$lim_{t to infty} X_{t wedge tau}(omega)=0 quad text{for $omega in {tau=infty}$}$$ (use $lim_{t to infty} B_t/t=0$ almost surely).

  4. By Step 3, we have $$lim_{t to infty} X_{t wedge tau} = e^{2ab} cdot 1_{{tau<infty}}.$$ Use the fact that $|X_{t wedge tau}| leq e^{2ab}$ to conclude from $(2)$ and the dominated convergence theorem that $$mathbb{P}(tau<infty) = e^{-2ab}.$$






share|cite|improve this answer









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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Hints:




    1. Define a stopping time $tau$ by $$tau := inf{t geq 0; X_t = e^{2ab}}.$$ Show that $$mathbb{P}(tau<infty) = mathbb{P}(exists t geq 0: B_t = a+bt). tag{1}$$

    2. Apply the optional stopping theorem to show that $$mathbb{E}(X_{t wedge tau}) = mathbb{E}(X_0)=1 tag{2}$$ for all $t geq 0$.

    3. Show that $$lim_{t to infty} X_{t wedge tau}(omega)=e^{2ab} quad text{for $omega in {tau<infty}$}$$ and $$lim_{t to infty} X_{t wedge tau}(omega)=0 quad text{for $omega in {tau=infty}$}$$ (use $lim_{t to infty} B_t/t=0$ almost surely).

    4. By Step 3, we have $$lim_{t to infty} X_{t wedge tau} = e^{2ab} cdot 1_{{tau<infty}}.$$ Use the fact that $|X_{t wedge tau}| leq e^{2ab}$ to conclude from $(2)$ and the dominated convergence theorem that $$mathbb{P}(tau<infty) = e^{-2ab}.$$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Hints:




      1. Define a stopping time $tau$ by $$tau := inf{t geq 0; X_t = e^{2ab}}.$$ Show that $$mathbb{P}(tau<infty) = mathbb{P}(exists t geq 0: B_t = a+bt). tag{1}$$

      2. Apply the optional stopping theorem to show that $$mathbb{E}(X_{t wedge tau}) = mathbb{E}(X_0)=1 tag{2}$$ for all $t geq 0$.

      3. Show that $$lim_{t to infty} X_{t wedge tau}(omega)=e^{2ab} quad text{for $omega in {tau<infty}$}$$ and $$lim_{t to infty} X_{t wedge tau}(omega)=0 quad text{for $omega in {tau=infty}$}$$ (use $lim_{t to infty} B_t/t=0$ almost surely).

      4. By Step 3, we have $$lim_{t to infty} X_{t wedge tau} = e^{2ab} cdot 1_{{tau<infty}}.$$ Use the fact that $|X_{t wedge tau}| leq e^{2ab}$ to conclude from $(2)$ and the dominated convergence theorem that $$mathbb{P}(tau<infty) = e^{-2ab}.$$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Hints:




        1. Define a stopping time $tau$ by $$tau := inf{t geq 0; X_t = e^{2ab}}.$$ Show that $$mathbb{P}(tau<infty) = mathbb{P}(exists t geq 0: B_t = a+bt). tag{1}$$

        2. Apply the optional stopping theorem to show that $$mathbb{E}(X_{t wedge tau}) = mathbb{E}(X_0)=1 tag{2}$$ for all $t geq 0$.

        3. Show that $$lim_{t to infty} X_{t wedge tau}(omega)=e^{2ab} quad text{for $omega in {tau<infty}$}$$ and $$lim_{t to infty} X_{t wedge tau}(omega)=0 quad text{for $omega in {tau=infty}$}$$ (use $lim_{t to infty} B_t/t=0$ almost surely).

        4. By Step 3, we have $$lim_{t to infty} X_{t wedge tau} = e^{2ab} cdot 1_{{tau<infty}}.$$ Use the fact that $|X_{t wedge tau}| leq e^{2ab}$ to conclude from $(2)$ and the dominated convergence theorem that $$mathbb{P}(tau<infty) = e^{-2ab}.$$






        share|cite|improve this answer









        $endgroup$



        Hints:




        1. Define a stopping time $tau$ by $$tau := inf{t geq 0; X_t = e^{2ab}}.$$ Show that $$mathbb{P}(tau<infty) = mathbb{P}(exists t geq 0: B_t = a+bt). tag{1}$$

        2. Apply the optional stopping theorem to show that $$mathbb{E}(X_{t wedge tau}) = mathbb{E}(X_0)=1 tag{2}$$ for all $t geq 0$.

        3. Show that $$lim_{t to infty} X_{t wedge tau}(omega)=e^{2ab} quad text{for $omega in {tau<infty}$}$$ and $$lim_{t to infty} X_{t wedge tau}(omega)=0 quad text{for $omega in {tau=infty}$}$$ (use $lim_{t to infty} B_t/t=0$ almost surely).

        4. By Step 3, we have $$lim_{t to infty} X_{t wedge tau} = e^{2ab} cdot 1_{{tau<infty}}.$$ Use the fact that $|X_{t wedge tau}| leq e^{2ab}$ to conclude from $(2)$ and the dominated convergence theorem that $$mathbb{P}(tau<infty) = e^{-2ab}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 15:08









        sazsaz

        79.8k860124




        79.8k860124






























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