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What is $mathbbE[W(tau)^4]$ for a Wiener process $W$ and a stopping time $tau$?
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I know that $mathbbE[W(t)^2] = t$ and $mathbbE[W(t)^4] = 3t^2$ for a standard Wiener process $W(t)$ when $t$ is a (non-stochastic) number. I also know that $mathbbE[W(tau)^2] = mathbbE[tau]$ for a stopping time $tau$. The question is: can $mathbbE[W(tau)^4]$ be expressed in terms of $mathbbE[tau^n]$, too?
I would also appreciate if an accessible reference is introduced (at the level of Evans's "Introduction to Stochastic Differential Equations"). Also if there is a general result for higher moments of $W(tau)$ that would be great too.
stochastic-processes stochastic-calculus brownian-motion stopping-times
asked Jul 25 at 18:54
Mahdiyar
405
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up vote
1
down vote
favorite
I know that $mathbbE[W(t)^2] = t$ and $mathbbE[W(t)^4] = 3t^2$ for a standard Wiener process $W(t)$ when $t$ is a (non-stochastic) number. I also know that $mathbbE[W(tau)^2] = mathbbE[tau]$ for a stopping time $tau$. The question is: can $mathbbE[W(tau)^4]$ be expressed in terms of $mathbbE[tau^n]$, too?
I would also appreciate if an accessible reference is introduced (at the level of Evans's "Introduction to Stochastic Differential Equations"). Also if there is a general result for higher moments of $W(tau)$ that would be great too.
stochastic-processes stochastic-calculus brownian-motion stopping-times
asked Jul 25 at 18:54
Mahdiyar
405
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I know that $mathbbE[W(t)^2] = t$ and $mathbbE[W(t)^4] = 3t^2$ for a standard Wiener process $W(t)$ when $t$ is a (non-stochastic) number. I also know that $mathbbE[W(tau)^2] = mathbbE[tau]$ for a stopping time $tau$. The question is: can $mathbbE[W(tau)^4]$ be expressed in terms of $mathbbE[tau^n]$, too?
I would also appreciate if an accessible reference is introduced (at the level of Evans's "Introduction to Stochastic Differential Equations"). Also if there is a general result for higher moments of $W(tau)$ that would be great too.
stochastic-processes stochastic-calculus brownian-motion stopping-times
asked Jul 25 at 18:54
Mahdiyar
405
I know that $mathbbE[W(t)^2] = t$ and $mathbbE[W(t)^4] = 3t^2$ for a standard Wiener process $W(t)$ when $t$ is a (non-stochastic) number. I also know that $mathbbE[W(tau)^2] = mathbbE[tau]$ for a stopping time $tau$. The question is: can $mathbbE[W(tau)^4]$ be expressed in terms of $mathbbE[tau^n]$, too?
I would also appreciate if an accessible reference is introduced (at the level of Evans's "Introduction to Stochastic Differential Equations"). Also if there is a general result for higher moments of $W(tau)$ that would be great too.
stochastic-processes stochastic-calculus brownian-motion stopping-times
asked Jul 25 at 18:54
Mahdiyar
405
edited Jul 25 at 18:59
asked Jul 25 at 18:54
Mahdiyar
405
asked Jul 25 at 18:54
Mahdiyar
405
asked Jul 25 at 18:54
Mahdiyar
405
405
add a comment |Â
add a comment |Â
1 Answer
1
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1
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The process $W(t)^4-6tW(t)^2+3t^2$ is a martingale. So if $tau$ is a reasonable stopping time (e.g., bounded) then $Bbb E[W(tau)^4] =6Bbb E[tau W(tau)^2]-3Bbb E[tau^2]$. To go further you need to know something about the joint distribution of $tau$ and $W(tau)^2$.
answered Jul 25 at 19:29
John Dawkins
12.5k1917
Thanks! Honestly, I am not familiar with martingales. But I could find the same result by applying $mathbbE[(int_0^tau FdW)^2] = mathbbE[int_0^tau F^2 dt]$ to $F=W$.
– Mahdiyar
Jul 25 at 20:03
I have $dX = a e^-3t dt + b dW$ for positive $a$ and $b$. And $tau$ is the first time $X$ reaches a particular $x_f$, given that it starts from $X=0$ at $t=0$. Any idea how that helps finding $mathbbE[tau W^2]$?
– Mahdiyar
Jul 25 at 20:04
I figured out how to compute $mathbbE[tau W^2]$ in my case of interest. It's actually trivial, but just for the reference here it goes: Since $x_f = a(1-e^-tau)/3 + bW(tau)$, I can just solve for $W(tau)$ and insert in $mathbbE[tau W^2]$ so that everything is in terms of $tau$.
– Mahdiyar
Jul 26 at 5:01
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The process $W(t)^4-6tW(t)^2+3t^2$ is a martingale. So if $tau$ is a reasonable stopping time (e.g., bounded) then $Bbb E[W(tau)^4] =6Bbb E[tau W(tau)^2]-3Bbb E[tau^2]$. To go further you need to know something about the joint distribution of $tau$ and $W(tau)^2$.
answered Jul 25 at 19:29
John Dawkins
12.5k1917
Thanks! Honestly, I am not familiar with martingales. But I could find the same result by applying $mathbbE[(int_0^tau FdW)^2] = mathbbE[int_0^tau F^2 dt]$ to $F=W$.
– Mahdiyar
Jul 25 at 20:03
I have $dX = a e^-3t dt + b dW$ for positive $a$ and $b$. And $tau$ is the first time $X$ reaches a particular $x_f$, given that it starts from $X=0$ at $t=0$. Any idea how that helps finding $mathbbE[tau W^2]$?
– Mahdiyar
Jul 25 at 20:04
I figured out how to compute $mathbbE[tau W^2]$ in my case of interest. It's actually trivial, but just for the reference here it goes: Since $x_f = a(1-e^-tau)/3 + bW(tau)$, I can just solve for $W(tau)$ and insert in $mathbbE[tau W^2]$ so that everything is in terms of $tau$.
– Mahdiyar
Jul 26 at 5:01
add a comment |Â
up vote
1
down vote
accepted
The process $W(t)^4-6tW(t)^2+3t^2$ is a martingale. So if $tau$ is a reasonable stopping time (e.g., bounded) then $Bbb E[W(tau)^4] =6Bbb E[tau W(tau)^2]-3Bbb E[tau^2]$. To go further you need to know something about the joint distribution of $tau$ and $W(tau)^2$.
answered Jul 25 at 19:29
John Dawkins
12.5k1917
Thanks! Honestly, I am not familiar with martingales. But I could find the same result by applying $mathbbE[(int_0^tau FdW)^2] = mathbbE[int_0^tau F^2 dt]$ to $F=W$.
– Mahdiyar
Jul 25 at 20:03
I have $dX = a e^-3t dt + b dW$ for positive $a$ and $b$. And $tau$ is the first time $X$ reaches a particular $x_f$, given that it starts from $X=0$ at $t=0$. Any idea how that helps finding $mathbbE[tau W^2]$?
– Mahdiyar
Jul 25 at 20:04
I figured out how to compute $mathbbE[tau W^2]$ in my case of interest. It's actually trivial, but just for the reference here it goes: Since $x_f = a(1-e^-tau)/3 + bW(tau)$, I can just solve for $W(tau)$ and insert in $mathbbE[tau W^2]$ so that everything is in terms of $tau$.
– Mahdiyar
Jul 26 at 5:01
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The process $W(t)^4-6tW(t)^2+3t^2$ is a martingale. So if $tau$ is a reasonable stopping time (e.g., bounded) then $Bbb E[W(tau)^4] =6Bbb E[tau W(tau)^2]-3Bbb E[tau^2]$. To go further you need to know something about the joint distribution of $tau$ and $W(tau)^2$.
answered Jul 25 at 19:29
John Dawkins
12.5k1917
The process $W(t)^4-6tW(t)^2+3t^2$ is a martingale. So if $tau$ is a reasonable stopping time (e.g., bounded) then $Bbb E[W(tau)^4] =6Bbb E[tau W(tau)^2]-3Bbb E[tau^2]$. To go further you need to know something about the joint distribution of $tau$ and $W(tau)^2$.
answered Jul 25 at 19:29
John Dawkins
12.5k1917
answered Jul 25 at 19:29
John Dawkins
12.5k1917
answered Jul 25 at 19:29
John Dawkins
12.5k1917
answered Jul 25 at 19:29
John Dawkins
12.5k1917
12.5k1917
Thanks! Honestly, I am not familiar with martingales. But I could find the same result by applying $mathbbE[(int_0^tau FdW)^2] = mathbbE[int_0^tau F^2 dt]$ to $F=W$.
– Mahdiyar
Jul 25 at 20:03
I have $dX = a e^-3t dt + b dW$ for positive $a$ and $b$. And $tau$ is the first time $X$ reaches a particular $x_f$, given that it starts from $X=0$ at $t=0$. Any idea how that helps finding $mathbbE[tau W^2]$?
– Mahdiyar
Jul 25 at 20:04
I figured out how to compute $mathbbE[tau W^2]$ in my case of interest. It's actually trivial, but just for the reference here it goes: Since $x_f = a(1-e^-tau)/3 + bW(tau)$, I can just solve for $W(tau)$ and insert in $mathbbE[tau W^2]$ so that everything is in terms of $tau$.
– Mahdiyar
Jul 26 at 5:01
add a comment |Â
Thanks! Honestly, I am not familiar with martingales. But I could find the same result by applying $mathbbE[(int_0^tau FdW)^2] = mathbbE[int_0^tau F^2 dt]$ to $F=W$.
– Mahdiyar
Jul 25 at 20:03
I have $dX = a e^-3t dt + b dW$ for positive $a$ and $b$. And $tau$ is the first time $X$ reaches a particular $x_f$, given that it starts from $X=0$ at $t=0$. Any idea how that helps finding $mathbbE[tau W^2]$?
– Mahdiyar
Jul 25 at 20:04
I figured out how to compute $mathbbE[tau W^2]$ in my case of interest. It's actually trivial, but just for the reference here it goes: Since $x_f = a(1-e^-tau)/3 + bW(tau)$, I can just solve for $W(tau)$ and insert in $mathbbE[tau W^2]$ so that everything is in terms of $tau$.
– Mahdiyar
Jul 26 at 5:01
Thanks! Honestly, I am not familiar with martingales. But I could find the same result by applying $mathbbE[(int_0^tau FdW)^2] = mathbbE[int_0^tau F^2 dt]$ to $F=W$.
– Mahdiyar
Jul 25 at 20:03
Thanks! Honestly, I am not familiar with martingales. But I could find the same result by applying $mathbbE[(int_0^tau FdW)^2] = mathbbE[int_0^tau F^2 dt]$ to $F=W$.
– Mahdiyar
Jul 25 at 20:03
I have $dX = a e^-3t dt + b dW$ for positive $a$ and $b$. And $tau$ is the first time $X$ reaches a particular $x_f$, given that it starts from $X=0$ at $t=0$. Any idea how that helps finding $mathbbE[tau W^2]$?
– Mahdiyar
Jul 25 at 20:04
I have $dX = a e^-3t dt + b dW$ for positive $a$ and $b$. And $tau$ is the first time $X$ reaches a particular $x_f$, given that it starts from $X=0$ at $t=0$. Any idea how that helps finding $mathbbE[tau W^2]$?
– Mahdiyar
Jul 25 at 20:04
I figured out how to compute $mathbbE[tau W^2]$ in my case of interest. It's actually trivial, but just for the reference here it goes: Since $x_f = a(1-e^-tau)/3 + bW(tau)$, I can just solve for $W(tau)$ and insert in $mathbbE[tau W^2]$ so that everything is in terms of $tau$.
– Mahdiyar
Jul 26 at 5:01
I figured out how to compute $mathbbE[tau W^2]$ in my case of interest. It's actually trivial, but just for the reference here it goes: Since $x_f = a(1-e^-tau)/3 + bW(tau)$, I can just solve for $W(tau)$ and insert in $mathbbE[tau W^2]$ so that everything is in terms of $tau$.
– Mahdiyar
Jul 26 at 5:01
add a comment |Â
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