If $tau=inf {t ge 0: B(t)= -a text{ or } b}$ is a stopping time of Brownian motion $B_t$, why does...
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The solution to this question states that $E(tau)=E(B_{tau}^2)$. But how do we know this?
probability brownian-motion martingales stopping-times expected-value
asked Dec 14 '18 at 16:44
RasputinRasputin
446211
$endgroup$
add a comment |
$begingroup$
The solution to this question states that $E(tau)=E(B_{tau}^2)$. But how do we know this?
probability brownian-motion martingales stopping-times expected-value
asked Dec 14 '18 at 16:44
RasputinRasputin
446211
$endgroup$
2
$begingroup$
It seems to be explained pretty clearly in the answer. You show that $B_t^2 - t$ is a martingale and then you apply the optional stopping theorem. Can you explain which part of that you are asking about?
$endgroup$
– Nate Eldredge
Dec 14 '18 at 16:57
add a comment |
$begingroup$
The solution to this question states that $E(tau)=E(B_{tau}^2)$. But how do we know this?
probability brownian-motion martingales stopping-times expected-value
asked Dec 14 '18 at 16:44
RasputinRasputin
446211
$endgroup$
The solution to this question states that $E(tau)=E(B_{tau}^2)$. But how do we know this?
probability brownian-motion martingales stopping-times expected-value
probability brownian-motion martingales stopping-times expected-value
asked Dec 14 '18 at 16:44
RasputinRasputin
446211
asked Dec 14 '18 at 16:44
RasputinRasputin
446211
asked Dec 14 '18 at 16:44
RasputinRasputin
446211
asked Dec 14 '18 at 16:44
RasputinRasputin
446211
asked Dec 14 '18 at 16:44
RasputinRasputin
446211
446211
2
$begingroup$
It seems to be explained pretty clearly in the answer. You show that $B_t^2 - t$ is a martingale and then you apply the optional stopping theorem. Can you explain which part of that you are asking about?
$endgroup$
– Nate Eldredge
Dec 14 '18 at 16:57
add a comment |
2
$begingroup$
It seems to be explained pretty clearly in the answer. You show that $B_t^2 - t$ is a martingale and then you apply the optional stopping theorem. Can you explain which part of that you are asking about?
$endgroup$
– Nate Eldredge
Dec 14 '18 at 16:57
2
2
$begingroup$
It seems to be explained pretty clearly in the answer. You show that $B_t^2 - t$ is a martingale and then you apply the optional stopping theorem. Can you explain which part of that you are asking about?
$endgroup$
– Nate Eldredge
Dec 14 '18 at 16:57
$begingroup$
It seems to be explained pretty clearly in the answer. You show that $B_t^2 - t$ is a martingale and then you apply the optional stopping theorem. Can you explain which part of that you are asking about?
$endgroup$
– Nate Eldredge
Dec 14 '18 at 16:57
add a comment |
2 Answers
2
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$begingroup$
In fact, the hint given above only applies to deterministic $t$, not to random time $tau$. The problem here is that $$
E[B^2_tau |tau = t] neq t.$$ You should use the fact that $$
B^2_t - t$$ is a continuous ${mathcal{F}_t}$-martingale. By optional sampling theorem, we have
$$
E[B^2_{tauwedge t} - (tauwedge t)] = 0.
$$ Now, let $tto infty$ and conclude that
$$
E[B^2_{tau}]=E[tau].
$$ (by Lebesgue's dominated convergence and monotone convergence theorem.)
answered Dec 14 '18 at 16:57
SongSong
13.9k633
$endgroup$
add a comment |
$begingroup$
Ohh wait I see, it's because it's a property of standard Brownian motion that $B_t^2=t$ for all $t ge 0$!!
Since $B(t) sim N(0, t)$, so $Var(B_t)=t=E(B_t^2)-E(B_t)^2=E(B_t^2)$.
answered Dec 14 '18 at 16:48
RasputinRasputin
446211
$endgroup$
2
$begingroup$
It's not true that $B_t^2 = t$. It's true that $E[B_t^2] = t$. But this property does not automatically carry over when you replace a constant time $t$ by a stopping time. One has to prove it.
$endgroup$
– Nate Eldredge
Dec 14 '18 at 16:54
add a comment |
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2 Answers
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2 Answers
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active
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votes
$begingroup$
In fact, the hint given above only applies to deterministic $t$, not to random time $tau$. The problem here is that $$
E[B^2_tau |tau = t] neq t.$$ You should use the fact that $$
B^2_t - t$$ is a continuous ${mathcal{F}_t}$-martingale. By optional sampling theorem, we have
$$
E[B^2_{tauwedge t} - (tauwedge t)] = 0.
$$ Now, let $tto infty$ and conclude that
$$
E[B^2_{tau}]=E[tau].
$$ (by Lebesgue's dominated convergence and monotone convergence theorem.)
answered Dec 14 '18 at 16:57
SongSong
13.9k633
$endgroup$
add a comment |
$begingroup$
In fact, the hint given above only applies to deterministic $t$, not to random time $tau$. The problem here is that $$
E[B^2_tau |tau = t] neq t.$$ You should use the fact that $$
B^2_t - t$$ is a continuous ${mathcal{F}_t}$-martingale. By optional sampling theorem, we have
$$
E[B^2_{tauwedge t} - (tauwedge t)] = 0.
$$ Now, let $tto infty$ and conclude that
$$
E[B^2_{tau}]=E[tau].
$$ (by Lebesgue's dominated convergence and monotone convergence theorem.)
answered Dec 14 '18 at 16:57
SongSong
13.9k633
$endgroup$
add a comment |
$begingroup$
In fact, the hint given above only applies to deterministic $t$, not to random time $tau$. The problem here is that $$
E[B^2_tau |tau = t] neq t.$$ You should use the fact that $$
B^2_t - t$$ is a continuous ${mathcal{F}_t}$-martingale. By optional sampling theorem, we have
$$
E[B^2_{tauwedge t} - (tauwedge t)] = 0.
$$ Now, let $tto infty$ and conclude that
$$
E[B^2_{tau}]=E[tau].
$$ (by Lebesgue's dominated convergence and monotone convergence theorem.)
answered Dec 14 '18 at 16:57
SongSong
13.9k633
$endgroup$
In fact, the hint given above only applies to deterministic $t$, not to random time $tau$. The problem here is that $$
E[B^2_tau |tau = t] neq t.$$ You should use the fact that $$
B^2_t - t$$ is a continuous ${mathcal{F}_t}$-martingale. By optional sampling theorem, we have
$$
E[B^2_{tauwedge t} - (tauwedge t)] = 0.
$$ Now, let $tto infty$ and conclude that
$$
E[B^2_{tau}]=E[tau].
$$ (by Lebesgue's dominated convergence and monotone convergence theorem.)
answered Dec 14 '18 at 16:57
SongSong
13.9k633
edited Dec 14 '18 at 17:09
answered Dec 14 '18 at 16:57
SongSong
13.9k633
answered Dec 14 '18 at 16:57
SongSong
13.9k633
answered Dec 14 '18 at 16:57
SongSong
13.9k633
13.9k633
add a comment |
add a comment |
$begingroup$
Ohh wait I see, it's because it's a property of standard Brownian motion that $B_t^2=t$ for all $t ge 0$!!
Since $B(t) sim N(0, t)$, so $Var(B_t)=t=E(B_t^2)-E(B_t)^2=E(B_t^2)$.
answered Dec 14 '18 at 16:48
RasputinRasputin
446211
$endgroup$
2
$begingroup$
It's not true that $B_t^2 = t$. It's true that $E[B_t^2] = t$. But this property does not automatically carry over when you replace a constant time $t$ by a stopping time. One has to prove it.
$endgroup$
– Nate Eldredge
Dec 14 '18 at 16:54
add a comment |
$begingroup$
Ohh wait I see, it's because it's a property of standard Brownian motion that $B_t^2=t$ for all $t ge 0$!!
Since $B(t) sim N(0, t)$, so $Var(B_t)=t=E(B_t^2)-E(B_t)^2=E(B_t^2)$.
answered Dec 14 '18 at 16:48
RasputinRasputin
446211
$endgroup$
2
$begingroup$
It's not true that $B_t^2 = t$. It's true that $E[B_t^2] = t$. But this property does not automatically carry over when you replace a constant time $t$ by a stopping time. One has to prove it.
$endgroup$
– Nate Eldredge
Dec 14 '18 at 16:54
add a comment |
$begingroup$
Ohh wait I see, it's because it's a property of standard Brownian motion that $B_t^2=t$ for all $t ge 0$!!
Since $B(t) sim N(0, t)$, so $Var(B_t)=t=E(B_t^2)-E(B_t)^2=E(B_t^2)$.
answered Dec 14 '18 at 16:48
RasputinRasputin
446211
$endgroup$
Ohh wait I see, it's because it's a property of standard Brownian motion that $B_t^2=t$ for all $t ge 0$!!
Since $B(t) sim N(0, t)$, so $Var(B_t)=t=E(B_t^2)-E(B_t)^2=E(B_t^2)$.
answered Dec 14 '18 at 16:48
RasputinRasputin
446211
answered Dec 14 '18 at 16:48
RasputinRasputin
446211
answered Dec 14 '18 at 16:48
RasputinRasputin
446211
answered Dec 14 '18 at 16:48
RasputinRasputin
446211
446211
2
$begingroup$
It's not true that $B_t^2 = t$. It's true that $E[B_t^2] = t$. But this property does not automatically carry over when you replace a constant time $t$ by a stopping time. One has to prove it.
$endgroup$
– Nate Eldredge
Dec 14 '18 at 16:54
add a comment |
2
$begingroup$
It's not true that $B_t^2 = t$. It's true that $E[B_t^2] = t$. But this property does not automatically carry over when you replace a constant time $t$ by a stopping time. One has to prove it.
$endgroup$
– Nate Eldredge
Dec 14 '18 at 16:54
2
2
$begingroup$
It's not true that $B_t^2 = t$. It's true that $E[B_t^2] = t$. But this property does not automatically carry over when you replace a constant time $t$ by a stopping time. One has to prove it.
$endgroup$
– Nate Eldredge
Dec 14 '18 at 16:54
$begingroup$
It's not true that $B_t^2 = t$. It's true that $E[B_t^2] = t$. But this property does not automatically carry over when you replace a constant time $t$ by a stopping time. One has to prove it.
$endgroup$
– Nate Eldredge
Dec 14 '18 at 16:54
add a comment |
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$begingroup$
It seems to be explained pretty clearly in the answer. You show that $B_t^2 - t$ is a martingale and then you apply the optional stopping theorem. Can you explain which part of that you are asking about?
$endgroup$
– Nate Eldredge
Dec 14 '18 at 16:57