Mixing
Expected value of Brownian motion at a time decided by a rate one Poisson process.
$begingroup$
Situation:
We have that ${W_{t},tgeq 0}$ is a Brownian motion and
${N_{t},tgeq 0}$ is a Poisson process such that $N_{t}$ follows a
Poisson distribution with parameter $t$. This process is independent
from our Brownian motion.
Question: a) Show that $mathbb{E}[W^{2}_{N_{t}}]=t$
b) For $ageq 0$ we define the stopping time: $tau=inf{tgeq 0:|W_{t}|>a}$. Show that this stopping time is finite.
My attempt: a) Keeping in mind that Brownian motion has independent normally distributed increments, I rewrite the problem into:
$mathbb{E}[W^{2}_{N_{t}}]=mathbb{E}[(W_{N_{t}}-W_{s}+W_{s})^{2}]$ where $0leq s<N_{t}$.
Using the Brownian motion properties a few more times, this reduces to $mathbb{E}[W^{2}_{N_{t}}]=N_{t}$, which looks like the desired answer but obviously is not. I thought of doing the same trick but with $N_{s}$ instead of $s$, but again no luck. This is a trick I learned during lectures but I'm starting to doubt if it works here aswell.
b) Intuitively, this makes obvious sense. Eventually we will cross hit $a$ and thus the stopping is finite. However, I struggle to make this a solid proof. Perhaps I could use the Law of Large Numbers but I'm not sure if it is applicable here.
Any help is appreciated.
probability-theory brownian-motion stopping-times poisson-process expected-value
edited Jan 1 at 8:49
saz
78.6k858123
asked Dec 31 '18 at 17:23
S. CrimS. Crim
15912
$endgroup$
add a comment |
$begingroup$
Situation:
We have that ${W_{t},tgeq 0}$ is a Brownian motion and
${N_{t},tgeq 0}$ is a Poisson process such that $N_{t}$ follows a
Poisson distribution with parameter $t$. This process is independent
from our Brownian motion.
Question: a) Show that $mathbb{E}[W^{2}_{N_{t}}]=t$
b) For $ageq 0$ we define the stopping time: $tau=inf{tgeq 0:|W_{t}|>a}$. Show that this stopping time is finite.
My attempt: a) Keeping in mind that Brownian motion has independent normally distributed increments, I rewrite the problem into:
$mathbb{E}[W^{2}_{N_{t}}]=mathbb{E}[(W_{N_{t}}-W_{s}+W_{s})^{2}]$ where $0leq s<N_{t}$.
Using the Brownian motion properties a few more times, this reduces to $mathbb{E}[W^{2}_{N_{t}}]=N_{t}$, which looks like the desired answer but obviously is not. I thought of doing the same trick but with $N_{s}$ instead of $s$, but again no luck. This is a trick I learned during lectures but I'm starting to doubt if it works here aswell.
b) Intuitively, this makes obvious sense. Eventually we will cross hit $a$ and thus the stopping is finite. However, I struggle to make this a solid proof. Perhaps I could use the Law of Large Numbers but I'm not sure if it is applicable here.
Any help is appreciated.
probability-theory brownian-motion stopping-times poisson-process expected-value
edited Jan 1 at 8:49
saz
78.6k858123
asked Dec 31 '18 at 17:23
S. CrimS. Crim
15912
$endgroup$
add a comment |
$begingroup$
Situation:
We have that ${W_{t},tgeq 0}$ is a Brownian motion and
${N_{t},tgeq 0}$ is a Poisson process such that $N_{t}$ follows a
Poisson distribution with parameter $t$. This process is independent
from our Brownian motion.
Question: a) Show that $mathbb{E}[W^{2}_{N_{t}}]=t$
b) For $ageq 0$ we define the stopping time: $tau=inf{tgeq 0:|W_{t}|>a}$. Show that this stopping time is finite.
My attempt: a) Keeping in mind that Brownian motion has independent normally distributed increments, I rewrite the problem into:
$mathbb{E}[W^{2}_{N_{t}}]=mathbb{E}[(W_{N_{t}}-W_{s}+W_{s})^{2}]$ where $0leq s<N_{t}$.
Using the Brownian motion properties a few more times, this reduces to $mathbb{E}[W^{2}_{N_{t}}]=N_{t}$, which looks like the desired answer but obviously is not. I thought of doing the same trick but with $N_{s}$ instead of $s$, but again no luck. This is a trick I learned during lectures but I'm starting to doubt if it works here aswell.
b) Intuitively, this makes obvious sense. Eventually we will cross hit $a$ and thus the stopping is finite. However, I struggle to make this a solid proof. Perhaps I could use the Law of Large Numbers but I'm not sure if it is applicable here.
Any help is appreciated.
probability-theory brownian-motion stopping-times poisson-process expected-value
edited Jan 1 at 8:49
saz
78.6k858123
asked Dec 31 '18 at 17:23
S. CrimS. Crim
15912
$endgroup$
Situation:
We have that ${W_{t},tgeq 0}$ is a Brownian motion and
${N_{t},tgeq 0}$ is a Poisson process such that $N_{t}$ follows a
Poisson distribution with parameter $t$. This process is independent
from our Brownian motion.
Question: a) Show that $mathbb{E}[W^{2}_{N_{t}}]=t$
b) For $ageq 0$ we define the stopping time: $tau=inf{tgeq 0:|W_{t}|>a}$. Show that this stopping time is finite.
My attempt: a) Keeping in mind that Brownian motion has independent normally distributed increments, I rewrite the problem into:
$mathbb{E}[W^{2}_{N_{t}}]=mathbb{E}[(W_{N_{t}}-W_{s}+W_{s})^{2}]$ where $0leq s<N_{t}$.
Using the Brownian motion properties a few more times, this reduces to $mathbb{E}[W^{2}_{N_{t}}]=N_{t}$, which looks like the desired answer but obviously is not. I thought of doing the same trick but with $N_{s}$ instead of $s$, but again no luck. This is a trick I learned during lectures but I'm starting to doubt if it works here aswell.
b) Intuitively, this makes obvious sense. Eventually we will cross hit $a$ and thus the stopping is finite. However, I struggle to make this a solid proof. Perhaps I could use the Law of Large Numbers but I'm not sure if it is applicable here.
Any help is appreciated.
probability-theory brownian-motion stopping-times poisson-process expected-value
probability-theory brownian-motion stopping-times poisson-process expected-value
edited Jan 1 at 8:49
saz
78.6k858123
asked Dec 31 '18 at 17:23
S. CrimS. Crim
15912
edited Jan 1 at 8:49
saz
78.6k858123
asked Dec 31 '18 at 17:23
S. CrimS. Crim
15912
edited Jan 1 at 8:49
saz
78.6k858123
edited Jan 1 at 8:49
saz
78.6k858123
edited Jan 1 at 8:49
saz
78.6k858123
78.6k858123
asked Dec 31 '18 at 17:23
S. CrimS. Crim
15912
asked Dec 31 '18 at 17:23
S. CrimS. Crim
15912
asked Dec 31 '18 at 17:23
S. CrimS. Crim
15912
15912
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that $N_t$ is a random variable and therefore the statement "where $0 leq s < N_t$" doesn't make sense.
$N_t$ takes values in $mathbb{N}_0$ and because of the independence from $(W_t)_{t geq 0}$ we have
$$mathbb{E}(W_{N_t}^2) = mathbb{E} left( sum_{k geq 0} W_k^2 1_{{N_t = k}} right) = sum_{k geq 0} underbrace{mathbb{E}(W_k^2)}_{=k} underbrace{mathbb{P}(N_t=k)}_{=e^{-t} t^k/k!} = t e^{-t} sum_{k geq 1} frac{t^{k-1}}{(k-1)!} =t.$$
Alternatively, we can use the tower property of the conditional expectation to show that
$$mathbb{E}(W_{N_t}^2) = mathbb{E} big[ mathbb{E}(W_{N_t}^2 mid N_t) big] = mathbb{E} big[ underbrace{mathbb{E}(W_k^2)}_{=k} big|_{k=N_t} big] = mathbb{E}(N_t)=t.$$
To prove that the stopping time $tau$ is almost surely finite we can use the optional stopping theorem. Since $M_t := W_t^2-t$ is a martingale, it follows from the optional stopping theorem that $$mathbb{E}(W_{t wedge tau}^2) = mathbb{E}(t wedge tau).$$ Since $|W_{t wedge tau}| leq a$ this gives $$mathbb{E}(t wedge tau) leq a^2,$$ and now the monotone convergence theorem yields $$mathbb{E}(tau) leq a^2$$ which shows, in particular, that $tau<infty$ almost surely.
answered Dec 31 '18 at 17:45
sazsaz
78.6k858123
$endgroup$
$begingroup$
Thanks alot, once again. Your solution for a) using the tower property makes perfect sense. I have a question about your solution for b), though. How do you conclude from $|W_{t wedge tau}| leq a$ that $mathbb{E}(t wedge tau) leq a$? I'm struggling to relate the fact that $|W_{t wedge tau}| leq a$ to $mathbb{E}(W_{t wedge tau}^2)$.
$endgroup$
– S. Crim
Dec 31 '18 at 18:14
1
$begingroup$
@S.Crim If $X$ is a random variable such that $0 leq X leq b$, then $mathbb{E}(X) leq b$, right? That's what I used to estimate $mathbb{E}(W_{t wedge tau}^2)$. (Sorry, there was a square missing in my estimate; should be fine now... perhaps this was the reason for your confusion.)
$endgroup$
– saz
Dec 31 '18 at 18:17
$begingroup$
Shouldn't it be that $mathbb{E}(W_{t wedge tau}^2) leq a^{2}$ then? Ah, you editted it haha! I get it now, thanks alot.
$endgroup$
– S. Crim
Dec 31 '18 at 18:19
1
$begingroup$
@S.Crim Yeah, sorry for the typo.
$endgroup$
– saz
Dec 31 '18 at 18:21
$begingroup$
I just came back to this question after attempting to solve it again and I ran into a problem. How exactly can you say that $|W_{t wedge tau}| leq a$? Isn't it, because of the way our stopping time is defined, that in the case that $t wedge tau = tau$ that we have that $|W_{tau}|>a$? Which would mean that in that case the inequality you gave does not hold?
$endgroup$
– S. Crim
Jan 1 at 16:24
|
show 8 more comments
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1 Answer
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1 Answer
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active
oldest
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votes
active
oldest
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$begingroup$
Note that $N_t$ is a random variable and therefore the statement "where $0 leq s < N_t$" doesn't make sense.
$N_t$ takes values in $mathbb{N}_0$ and because of the independence from $(W_t)_{t geq 0}$ we have
$$mathbb{E}(W_{N_t}^2) = mathbb{E} left( sum_{k geq 0} W_k^2 1_{{N_t = k}} right) = sum_{k geq 0} underbrace{mathbb{E}(W_k^2)}_{=k} underbrace{mathbb{P}(N_t=k)}_{=e^{-t} t^k/k!} = t e^{-t} sum_{k geq 1} frac{t^{k-1}}{(k-1)!} =t.$$
Alternatively, we can use the tower property of the conditional expectation to show that
$$mathbb{E}(W_{N_t}^2) = mathbb{E} big[ mathbb{E}(W_{N_t}^2 mid N_t) big] = mathbb{E} big[ underbrace{mathbb{E}(W_k^2)}_{=k} big|_{k=N_t} big] = mathbb{E}(N_t)=t.$$
To prove that the stopping time $tau$ is almost surely finite we can use the optional stopping theorem. Since $M_t := W_t^2-t$ is a martingale, it follows from the optional stopping theorem that $$mathbb{E}(W_{t wedge tau}^2) = mathbb{E}(t wedge tau).$$ Since $|W_{t wedge tau}| leq a$ this gives $$mathbb{E}(t wedge tau) leq a^2,$$ and now the monotone convergence theorem yields $$mathbb{E}(tau) leq a^2$$ which shows, in particular, that $tau<infty$ almost surely.
answered Dec 31 '18 at 17:45
sazsaz
78.6k858123
$endgroup$
$begingroup$
Thanks alot, once again. Your solution for a) using the tower property makes perfect sense. I have a question about your solution for b), though. How do you conclude from $|W_{t wedge tau}| leq a$ that $mathbb{E}(t wedge tau) leq a$? I'm struggling to relate the fact that $|W_{t wedge tau}| leq a$ to $mathbb{E}(W_{t wedge tau}^2)$.
$endgroup$
– S. Crim
Dec 31 '18 at 18:14
1
$begingroup$
@S.Crim If $X$ is a random variable such that $0 leq X leq b$, then $mathbb{E}(X) leq b$, right? That's what I used to estimate $mathbb{E}(W_{t wedge tau}^2)$. (Sorry, there was a square missing in my estimate; should be fine now... perhaps this was the reason for your confusion.)
$endgroup$
– saz
Dec 31 '18 at 18:17
$begingroup$
Shouldn't it be that $mathbb{E}(W_{t wedge tau}^2) leq a^{2}$ then? Ah, you editted it haha! I get it now, thanks alot.
$endgroup$
– S. Crim
Dec 31 '18 at 18:19
1
$begingroup$
@S.Crim Yeah, sorry for the typo.
$endgroup$
– saz
Dec 31 '18 at 18:21
$begingroup$
I just came back to this question after attempting to solve it again and I ran into a problem. How exactly can you say that $|W_{t wedge tau}| leq a$? Isn't it, because of the way our stopping time is defined, that in the case that $t wedge tau = tau$ that we have that $|W_{tau}|>a$? Which would mean that in that case the inequality you gave does not hold?
$endgroup$
– S. Crim
Jan 1 at 16:24
|
show 8 more comments
$begingroup$
Note that $N_t$ is a random variable and therefore the statement "where $0 leq s < N_t$" doesn't make sense.
$N_t$ takes values in $mathbb{N}_0$ and because of the independence from $(W_t)_{t geq 0}$ we have
$$mathbb{E}(W_{N_t}^2) = mathbb{E} left( sum_{k geq 0} W_k^2 1_{{N_t = k}} right) = sum_{k geq 0} underbrace{mathbb{E}(W_k^2)}_{=k} underbrace{mathbb{P}(N_t=k)}_{=e^{-t} t^k/k!} = t e^{-t} sum_{k geq 1} frac{t^{k-1}}{(k-1)!} =t.$$
Alternatively, we can use the tower property of the conditional expectation to show that
$$mathbb{E}(W_{N_t}^2) = mathbb{E} big[ mathbb{E}(W_{N_t}^2 mid N_t) big] = mathbb{E} big[ underbrace{mathbb{E}(W_k^2)}_{=k} big|_{k=N_t} big] = mathbb{E}(N_t)=t.$$
To prove that the stopping time $tau$ is almost surely finite we can use the optional stopping theorem. Since $M_t := W_t^2-t$ is a martingale, it follows from the optional stopping theorem that $$mathbb{E}(W_{t wedge tau}^2) = mathbb{E}(t wedge tau).$$ Since $|W_{t wedge tau}| leq a$ this gives $$mathbb{E}(t wedge tau) leq a^2,$$ and now the monotone convergence theorem yields $$mathbb{E}(tau) leq a^2$$ which shows, in particular, that $tau<infty$ almost surely.
answered Dec 31 '18 at 17:45
sazsaz
78.6k858123
$endgroup$
$begingroup$
Thanks alot, once again. Your solution for a) using the tower property makes perfect sense. I have a question about your solution for b), though. How do you conclude from $|W_{t wedge tau}| leq a$ that $mathbb{E}(t wedge tau) leq a$? I'm struggling to relate the fact that $|W_{t wedge tau}| leq a$ to $mathbb{E}(W_{t wedge tau}^2)$.
$endgroup$
– S. Crim
Dec 31 '18 at 18:14
1
$begingroup$
@S.Crim If $X$ is a random variable such that $0 leq X leq b$, then $mathbb{E}(X) leq b$, right? That's what I used to estimate $mathbb{E}(W_{t wedge tau}^2)$. (Sorry, there was a square missing in my estimate; should be fine now... perhaps this was the reason for your confusion.)
$endgroup$
– saz
Dec 31 '18 at 18:17
$begingroup$
Shouldn't it be that $mathbb{E}(W_{t wedge tau}^2) leq a^{2}$ then? Ah, you editted it haha! I get it now, thanks alot.
$endgroup$
– S. Crim
Dec 31 '18 at 18:19
1
$begingroup$
@S.Crim Yeah, sorry for the typo.
$endgroup$
– saz
Dec 31 '18 at 18:21
$begingroup$
I just came back to this question after attempting to solve it again and I ran into a problem. How exactly can you say that $|W_{t wedge tau}| leq a$? Isn't it, because of the way our stopping time is defined, that in the case that $t wedge tau = tau$ that we have that $|W_{tau}|>a$? Which would mean that in that case the inequality you gave does not hold?
$endgroup$
– S. Crim
Jan 1 at 16:24
|
show 8 more comments
$begingroup$
Note that $N_t$ is a random variable and therefore the statement "where $0 leq s < N_t$" doesn't make sense.
$N_t$ takes values in $mathbb{N}_0$ and because of the independence from $(W_t)_{t geq 0}$ we have
$$mathbb{E}(W_{N_t}^2) = mathbb{E} left( sum_{k geq 0} W_k^2 1_{{N_t = k}} right) = sum_{k geq 0} underbrace{mathbb{E}(W_k^2)}_{=k} underbrace{mathbb{P}(N_t=k)}_{=e^{-t} t^k/k!} = t e^{-t} sum_{k geq 1} frac{t^{k-1}}{(k-1)!} =t.$$
Alternatively, we can use the tower property of the conditional expectation to show that
$$mathbb{E}(W_{N_t}^2) = mathbb{E} big[ mathbb{E}(W_{N_t}^2 mid N_t) big] = mathbb{E} big[ underbrace{mathbb{E}(W_k^2)}_{=k} big|_{k=N_t} big] = mathbb{E}(N_t)=t.$$
To prove that the stopping time $tau$ is almost surely finite we can use the optional stopping theorem. Since $M_t := W_t^2-t$ is a martingale, it follows from the optional stopping theorem that $$mathbb{E}(W_{t wedge tau}^2) = mathbb{E}(t wedge tau).$$ Since $|W_{t wedge tau}| leq a$ this gives $$mathbb{E}(t wedge tau) leq a^2,$$ and now the monotone convergence theorem yields $$mathbb{E}(tau) leq a^2$$ which shows, in particular, that $tau<infty$ almost surely.
answered Dec 31 '18 at 17:45
sazsaz
78.6k858123
$endgroup$
Note that $N_t$ is a random variable and therefore the statement "where $0 leq s < N_t$" doesn't make sense.
$N_t$ takes values in $mathbb{N}_0$ and because of the independence from $(W_t)_{t geq 0}$ we have
$$mathbb{E}(W_{N_t}^2) = mathbb{E} left( sum_{k geq 0} W_k^2 1_{{N_t = k}} right) = sum_{k geq 0} underbrace{mathbb{E}(W_k^2)}_{=k} underbrace{mathbb{P}(N_t=k)}_{=e^{-t} t^k/k!} = t e^{-t} sum_{k geq 1} frac{t^{k-1}}{(k-1)!} =t.$$
Alternatively, we can use the tower property of the conditional expectation to show that
$$mathbb{E}(W_{N_t}^2) = mathbb{E} big[ mathbb{E}(W_{N_t}^2 mid N_t) big] = mathbb{E} big[ underbrace{mathbb{E}(W_k^2)}_{=k} big|_{k=N_t} big] = mathbb{E}(N_t)=t.$$
To prove that the stopping time $tau$ is almost surely finite we can use the optional stopping theorem. Since $M_t := W_t^2-t$ is a martingale, it follows from the optional stopping theorem that $$mathbb{E}(W_{t wedge tau}^2) = mathbb{E}(t wedge tau).$$ Since $|W_{t wedge tau}| leq a$ this gives $$mathbb{E}(t wedge tau) leq a^2,$$ and now the monotone convergence theorem yields $$mathbb{E}(tau) leq a^2$$ which shows, in particular, that $tau<infty$ almost surely.
answered Dec 31 '18 at 17:45
sazsaz
78.6k858123
edited Dec 31 '18 at 18:18
answered Dec 31 '18 at 17:45
sazsaz
78.6k858123
answered Dec 31 '18 at 17:45
sazsaz
78.6k858123
answered Dec 31 '18 at 17:45
sazsaz
78.6k858123
78.6k858123
$begingroup$
Thanks alot, once again. Your solution for a) using the tower property makes perfect sense. I have a question about your solution for b), though. How do you conclude from $|W_{t wedge tau}| leq a$ that $mathbb{E}(t wedge tau) leq a$? I'm struggling to relate the fact that $|W_{t wedge tau}| leq a$ to $mathbb{E}(W_{t wedge tau}^2)$.
$endgroup$
– S. Crim
Dec 31 '18 at 18:14
1
$begingroup$
@S.Crim If $X$ is a random variable such that $0 leq X leq b$, then $mathbb{E}(X) leq b$, right? That's what I used to estimate $mathbb{E}(W_{t wedge tau}^2)$. (Sorry, there was a square missing in my estimate; should be fine now... perhaps this was the reason for your confusion.)
$endgroup$
– saz
Dec 31 '18 at 18:17
$begingroup$
Shouldn't it be that $mathbb{E}(W_{t wedge tau}^2) leq a^{2}$ then? Ah, you editted it haha! I get it now, thanks alot.
$endgroup$
– S. Crim
Dec 31 '18 at 18:19
1
$begingroup$
@S.Crim Yeah, sorry for the typo.
$endgroup$
– saz
Dec 31 '18 at 18:21
$begingroup$
I just came back to this question after attempting to solve it again and I ran into a problem. How exactly can you say that $|W_{t wedge tau}| leq a$? Isn't it, because of the way our stopping time is defined, that in the case that $t wedge tau = tau$ that we have that $|W_{tau}|>a$? Which would mean that in that case the inequality you gave does not hold?
$endgroup$
– S. Crim
Jan 1 at 16:24
|
show 8 more comments
$begingroup$
Thanks alot, once again. Your solution for a) using the tower property makes perfect sense. I have a question about your solution for b), though. How do you conclude from $|W_{t wedge tau}| leq a$ that $mathbb{E}(t wedge tau) leq a$? I'm struggling to relate the fact that $|W_{t wedge tau}| leq a$ to $mathbb{E}(W_{t wedge tau}^2)$.
$endgroup$
– S. Crim
Dec 31 '18 at 18:14
1
$begingroup$
@S.Crim If $X$ is a random variable such that $0 leq X leq b$, then $mathbb{E}(X) leq b$, right? That's what I used to estimate $mathbb{E}(W_{t wedge tau}^2)$. (Sorry, there was a square missing in my estimate; should be fine now... perhaps this was the reason for your confusion.)
$endgroup$
– saz
Dec 31 '18 at 18:17
$begingroup$
Shouldn't it be that $mathbb{E}(W_{t wedge tau}^2) leq a^{2}$ then? Ah, you editted it haha! I get it now, thanks alot.
$endgroup$
– S. Crim
Dec 31 '18 at 18:19
1
$begingroup$
@S.Crim Yeah, sorry for the typo.
$endgroup$
– saz
Dec 31 '18 at 18:21
$begingroup$
I just came back to this question after attempting to solve it again and I ran into a problem. How exactly can you say that $|W_{t wedge tau}| leq a$? Isn't it, because of the way our stopping time is defined, that in the case that $t wedge tau = tau$ that we have that $|W_{tau}|>a$? Which would mean that in that case the inequality you gave does not hold?
$endgroup$
– S. Crim
Jan 1 at 16:24
$begingroup$
Thanks alot, once again. Your solution for a) using the tower property makes perfect sense. I have a question about your solution for b), though. How do you conclude from $|W_{t wedge tau}| leq a$ that $mathbb{E}(t wedge tau) leq a$? I'm struggling to relate the fact that $|W_{t wedge tau}| leq a$ to $mathbb{E}(W_{t wedge tau}^2)$.
$endgroup$
– S. Crim
Dec 31 '18 at 18:14
$begingroup$
Thanks alot, once again. Your solution for a) using the tower property makes perfect sense. I have a question about your solution for b), though. How do you conclude from $|W_{t wedge tau}| leq a$ that $mathbb{E}(t wedge tau) leq a$? I'm struggling to relate the fact that $|W_{t wedge tau}| leq a$ to $mathbb{E}(W_{t wedge tau}^2)$.
$endgroup$
– S. Crim
Dec 31 '18 at 18:14
1
1
$begingroup$
@S.Crim If $X$ is a random variable such that $0 leq X leq b$, then $mathbb{E}(X) leq b$, right? That's what I used to estimate $mathbb{E}(W_{t wedge tau}^2)$. (Sorry, there was a square missing in my estimate; should be fine now... perhaps this was the reason for your confusion.)
$endgroup$
– saz
Dec 31 '18 at 18:17
$begingroup$
@S.Crim If $X$ is a random variable such that $0 leq X leq b$, then $mathbb{E}(X) leq b$, right? That's what I used to estimate $mathbb{E}(W_{t wedge tau}^2)$. (Sorry, there was a square missing in my estimate; should be fine now... perhaps this was the reason for your confusion.)
$endgroup$
– saz
Dec 31 '18 at 18:17
$begingroup$
Shouldn't it be that $mathbb{E}(W_{t wedge tau}^2) leq a^{2}$ then? Ah, you editted it haha! I get it now, thanks alot.
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– S. Crim
Dec 31 '18 at 18:19
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Shouldn't it be that $mathbb{E}(W_{t wedge tau}^2) leq a^{2}$ then? Ah, you editted it haha! I get it now, thanks alot.
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– S. Crim
Dec 31 '18 at 18:19
1
1
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@S.Crim Yeah, sorry for the typo.
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– saz
Dec 31 '18 at 18:21
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@S.Crim Yeah, sorry for the typo.
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– saz
Dec 31 '18 at 18:21
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I just came back to this question after attempting to solve it again and I ran into a problem. How exactly can you say that $|W_{t wedge tau}| leq a$? Isn't it, because of the way our stopping time is defined, that in the case that $t wedge tau = tau$ that we have that $|W_{tau}|>a$? Which would mean that in that case the inequality you gave does not hold?
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– S. Crim
Jan 1 at 16:24
$begingroup$
I just came back to this question after attempting to solve it again and I ran into a problem. How exactly can you say that $|W_{t wedge tau}| leq a$? Isn't it, because of the way our stopping time is defined, that in the case that $t wedge tau = tau$ that we have that $|W_{tau}|>a$? Which would mean that in that case the inequality you gave does not hold?
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– S. Crim
Jan 1 at 16:24
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