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
edited Dec 10 '18 at 15:14
saz
80.5k860125
asked Dec 8 '18 at 16:08
nablanabla
1089
$endgroup$
add a comment |
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
edited Dec 10 '18 at 15:14
saz
80.5k860125
asked Dec 8 '18 at 16:08
nablanabla
1089
$endgroup$
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
add a comment |
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
edited Dec 10 '18 at 15:14
saz
80.5k860125
asked Dec 8 '18 at 16:08
nablanabla
1089
$endgroup$
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
edited Dec 10 '18 at 15:14
saz
80.5k860125
asked Dec 8 '18 at 16:08
nablanabla
1089
edited Dec 10 '18 at 15:14
saz
80.5k860125
asked Dec 8 '18 at 16:08
nablanabla
1089
edited Dec 10 '18 at 15:14
saz
80.5k860125
edited Dec 10 '18 at 15:14
saz
80.5k860125
edited Dec 10 '18 at 15:14
saz
80.5k860125
80.5k860125
asked Dec 8 '18 at 16:08
nablanabla
1089
asked Dec 8 '18 at 16:08
nablanabla
1089
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
add a comment |
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
add a comment |
1 Answer
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3
$begingroup$
Hints:
- Define a stopping time $tau$ by $$tau := inf{t geq 0; X_t = e^{2ab}}.$$ Show that $$mathbb{P}(tau<infty) = mathbb{P}(exists t geq 0: B_t = a+bt). tag{1}$$
- Apply the optional stopping theorem to show that $$mathbb{E}(X_{t wedge tau}) = mathbb{E}(X_0)=1 tag{2}$$ for all $t geq 0$.
- 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).
- 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 $$mathbb{P}(tau<infty) = e^{-2ab}.$$
answered Dec 10 '18 at 15:08
sazsaz
80.5k860125
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
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oldest
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3
$begingroup$
Hints:
- Define a stopping time $tau$ by $$tau := inf{t geq 0; X_t = e^{2ab}}.$$ Show that $$mathbb{P}(tau<infty) = mathbb{P}(exists t geq 0: B_t = a+bt). tag{1}$$
- Apply the optional stopping theorem to show that $$mathbb{E}(X_{t wedge tau}) = mathbb{E}(X_0)=1 tag{2}$$ for all $t geq 0$.
- 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).
- 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 $$mathbb{P}(tau<infty) = e^{-2ab}.$$
answered Dec 10 '18 at 15:08
sazsaz
80.5k860125
$endgroup$
add a comment |
3
$begingroup$
Hints:
- Define a stopping time $tau$ by $$tau := inf{t geq 0; X_t = e^{2ab}}.$$ Show that $$mathbb{P}(tau<infty) = mathbb{P}(exists t geq 0: B_t = a+bt). tag{1}$$
- Apply the optional stopping theorem to show that $$mathbb{E}(X_{t wedge tau}) = mathbb{E}(X_0)=1 tag{2}$$ for all $t geq 0$.
- 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).
- 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 $$mathbb{P}(tau<infty) = e^{-2ab}.$$
answered Dec 10 '18 at 15:08
sazsaz
80.5k860125
$endgroup$
add a comment |
3
3
3
$begingroup$
Hints:
- Define a stopping time $tau$ by $$tau := inf{t geq 0; X_t = e^{2ab}}.$$ Show that $$mathbb{P}(tau<infty) = mathbb{P}(exists t geq 0: B_t = a+bt). tag{1}$$
- Apply the optional stopping theorem to show that $$mathbb{E}(X_{t wedge tau}) = mathbb{E}(X_0)=1 tag{2}$$ for all $t geq 0$.
- 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).
- 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 $$mathbb{P}(tau<infty) = e^{-2ab}.$$
answered Dec 10 '18 at 15:08
sazsaz
80.5k860125
$endgroup$
Hints:
- Define a stopping time $tau$ by $$tau := inf{t geq 0; X_t = e^{2ab}}.$$ Show that $$mathbb{P}(tau<infty) = mathbb{P}(exists t geq 0: B_t = a+bt). tag{1}$$
- Apply the optional stopping theorem to show that $$mathbb{E}(X_{t wedge tau}) = mathbb{E}(X_0)=1 tag{2}$$ for all $t geq 0$.
- 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).
- 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 $$mathbb{P}(tau<infty) = e^{-2ab}.$$
answered Dec 10 '18 at 15:08
sazsaz
80.5k860125
answered Dec 10 '18 at 15:08
sazsaz
80.5k860125
answered Dec 10 '18 at 15:08
sazsaz
80.5k860125
answered Dec 10 '18 at 15:08
sazsaz
80.5k860125
80.5k860125
add a comment |
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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