Mixing
Proving that stopping time is finite a.s.
$begingroup$
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
asked Jan 8 at 21:03
GuestttGuesttt
538
$endgroup$
add a comment |
$begingroup$
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
asked Jan 8 at 21:03
GuestttGuesttt
538
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 8 at 21:03
GuestttGuesttt
538
$endgroup$
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
asked Jan 8 at 21:03
GuestttGuesttt
538
asked Jan 8 at 21:03
GuestttGuesttt
538
edited Jan 8 at 23:00
Guesttt
asked Jan 8 at 21:03
GuestttGuesttt
538
asked Jan 8 at 21:03
GuestttGuesttt
538
asked Jan 8 at 21:03
GuestttGuesttt
538
538
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $P{tau_a <infty}=1$.
answered Jan 8 at 23:56
Kavi Rama MurthyKavi Rama Murthy
57.3k42160
$endgroup$
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
add a comment |
$begingroup$
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.
answered Jan 9 at 0:08
Nate EldredgeNate Eldredge
63k682171
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
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$begingroup$
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 $P{tau_a <infty}=1$.
answered Jan 8 at 23:56
Kavi Rama MurthyKavi Rama Murthy
57.3k42160
$endgroup$
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
add a comment |
$begingroup$
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 $P{tau_a <infty}=1$.
answered Jan 8 at 23:56
Kavi Rama MurthyKavi Rama Murthy
57.3k42160
$endgroup$
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
add a comment |
$begingroup$
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 $P{tau_a <infty}=1$.
answered Jan 8 at 23:56
Kavi Rama MurthyKavi Rama Murthy
57.3k42160
$endgroup$
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 $P{tau_a <infty}=1$.
answered Jan 8 at 23:56
Kavi Rama MurthyKavi Rama Murthy
57.3k42160
answered Jan 8 at 23:56
Kavi Rama MurthyKavi Rama Murthy
57.3k42160
answered Jan 8 at 23:56
Kavi Rama MurthyKavi Rama Murthy
57.3k42160
answered Jan 8 at 23:56
Kavi Rama MurthyKavi Rama Murthy
57.3k42160
57.3k42160
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
add a comment |
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
add a comment |
$begingroup$
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.
answered Jan 9 at 0:08
Nate EldredgeNate Eldredge
63k682171
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 9 at 0:08
Nate EldredgeNate Eldredge
63k682171
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 9 at 0:08
Nate EldredgeNate Eldredge
63k682171
$endgroup$
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.
answered Jan 9 at 0:08
Nate EldredgeNate Eldredge
63k682171
answered Jan 9 at 0:08
Nate EldredgeNate Eldredge
63k682171
answered Jan 9 at 0:08
Nate EldredgeNate Eldredge
63k682171
answered Jan 9 at 0:08
Nate EldredgeNate Eldredge
63k682171
63k682171
add a comment |
add a comment |
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$begingroup$
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