For a Brownian motion $(B_t)_{t geq 0}$, do we have $E[(B_tau - B_sigma)^2]=E[B_tau^2 - B_sigma^2]$ for...
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Let $(Omega,mathcal{F},(mathcal{F}_t)_{t geq 0}, P)$ be a filtered probability space and let $(B_t)_{t geq 0}$ be a Brownian motion on that space. The question is if the following is true:
For two bounded stopping times $sigma leq tau$, we have
$$E[(B_tau-B_sigma)^2] = E[B_tau^2 - B_sigma^2]$$
If this is true, how can I see that it holds?
Thanks a lot in advance!
probability-theory stochastic-processes brownian-motion stopping-times expected-value
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Let $(Omega,mathcal{F},(mathcal{F}_t)_{t geq 0}, P)$ be a filtered probability space and let $(B_t)_{t geq 0}$ be a Brownian motion on that space. The question is if the following is true:
For two bounded stopping times $sigma leq tau$, we have
$$E[(B_tau-B_sigma)^2] = E[B_tau^2 - B_sigma^2]$$
If this is true, how can I see that it holds?
Thanks a lot in advance!
probability-theory stochastic-processes brownian-motion stopping-times expected-value
It would be reasonable to assume that $mathbb{E}(tau)<infty$; otherwise it's not even clear that the expectations are finite.
– saz
Nov 15 at 15:39
Do you have any hypothesis on $sigma$ and $tau$ (boundedness...)? Otherwise, you can just take $sigma = T_1$ and $tau = T_2$, where $T_a$ is the first hitting time of $a$.
– D. Thomine
Nov 15 at 15:40
@D.Thomine ,Yes I forgot to put it here. I assume boundedness of the stopping times
– vaoy
Nov 15 at 15:41
Here's an example to show that some sort of "smallness" hypothesis is needed for (at least) $tau$: Suppose $sigma$ is a finite (even bounded) stopping time with $Bbb E[B_sigma^2]>0$ and take $tau:=inf{t>sigma: B_t=0}$. Then $Bbb P[tau<infty, B_tau=0]=1$. The left side of the equality you request is $Bbb E[B_sigma^2]$ while the right is $-Bbb E[B_sigma^2]$.
– John Dawkins
Nov 15 at 22:33
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(Omega,mathcal{F},(mathcal{F}_t)_{t geq 0}, P)$ be a filtered probability space and let $(B_t)_{t geq 0}$ be a Brownian motion on that space. The question is if the following is true:
For two bounded stopping times $sigma leq tau$, we have
$$E[(B_tau-B_sigma)^2] = E[B_tau^2 - B_sigma^2]$$
If this is true, how can I see that it holds?
Thanks a lot in advance!
probability-theory stochastic-processes brownian-motion stopping-times expected-value
Let $(Omega,mathcal{F},(mathcal{F}_t)_{t geq 0}, P)$ be a filtered probability space and let $(B_t)_{t geq 0}$ be a Brownian motion on that space. The question is if the following is true:
For two bounded stopping times $sigma leq tau$, we have
$$E[(B_tau-B_sigma)^2] = E[B_tau^2 - B_sigma^2]$$
If this is true, how can I see that it holds?
Thanks a lot in advance!
probability-theory stochastic-processes brownian-motion stopping-times expected-value
probability-theory stochastic-processes brownian-motion stopping-times expected-value
edited Nov 15 at 17:20
saz
76.6k755117
76.6k755117
asked Nov 15 at 15:33
vaoy
51729
51729
It would be reasonable to assume that $mathbb{E}(tau)<infty$; otherwise it's not even clear that the expectations are finite.
– saz
Nov 15 at 15:39
Do you have any hypothesis on $sigma$ and $tau$ (boundedness...)? Otherwise, you can just take $sigma = T_1$ and $tau = T_2$, where $T_a$ is the first hitting time of $a$.
– D. Thomine
Nov 15 at 15:40
@D.Thomine ,Yes I forgot to put it here. I assume boundedness of the stopping times
– vaoy
Nov 15 at 15:41
Here's an example to show that some sort of "smallness" hypothesis is needed for (at least) $tau$: Suppose $sigma$ is a finite (even bounded) stopping time with $Bbb E[B_sigma^2]>0$ and take $tau:=inf{t>sigma: B_t=0}$. Then $Bbb P[tau<infty, B_tau=0]=1$. The left side of the equality you request is $Bbb E[B_sigma^2]$ while the right is $-Bbb E[B_sigma^2]$.
– John Dawkins
Nov 15 at 22:33
add a comment |
It would be reasonable to assume that $mathbb{E}(tau)<infty$; otherwise it's not even clear that the expectations are finite.
– saz
Nov 15 at 15:39
Do you have any hypothesis on $sigma$ and $tau$ (boundedness...)? Otherwise, you can just take $sigma = T_1$ and $tau = T_2$, where $T_a$ is the first hitting time of $a$.
– D. Thomine
Nov 15 at 15:40
@D.Thomine ,Yes I forgot to put it here. I assume boundedness of the stopping times
– vaoy
Nov 15 at 15:41
Here's an example to show that some sort of "smallness" hypothesis is needed for (at least) $tau$: Suppose $sigma$ is a finite (even bounded) stopping time with $Bbb E[B_sigma^2]>0$ and take $tau:=inf{t>sigma: B_t=0}$. Then $Bbb P[tau<infty, B_tau=0]=1$. The left side of the equality you request is $Bbb E[B_sigma^2]$ while the right is $-Bbb E[B_sigma^2]$.
– John Dawkins
Nov 15 at 22:33
It would be reasonable to assume that $mathbb{E}(tau)<infty$; otherwise it's not even clear that the expectations are finite.
– saz
Nov 15 at 15:39
It would be reasonable to assume that $mathbb{E}(tau)<infty$; otherwise it's not even clear that the expectations are finite.
– saz
Nov 15 at 15:39
Do you have any hypothesis on $sigma$ and $tau$ (boundedness...)? Otherwise, you can just take $sigma = T_1$ and $tau = T_2$, where $T_a$ is the first hitting time of $a$.
– D. Thomine
Nov 15 at 15:40
Do you have any hypothesis on $sigma$ and $tau$ (boundedness...)? Otherwise, you can just take $sigma = T_1$ and $tau = T_2$, where $T_a$ is the first hitting time of $a$.
– D. Thomine
Nov 15 at 15:40
@D.Thomine ,Yes I forgot to put it here. I assume boundedness of the stopping times
– vaoy
Nov 15 at 15:41
@D.Thomine ,Yes I forgot to put it here. I assume boundedness of the stopping times
– vaoy
Nov 15 at 15:41
Here's an example to show that some sort of "smallness" hypothesis is needed for (at least) $tau$: Suppose $sigma$ is a finite (even bounded) stopping time with $Bbb E[B_sigma^2]>0$ and take $tau:=inf{t>sigma: B_t=0}$. Then $Bbb P[tau<infty, B_tau=0]=1$. The left side of the equality you request is $Bbb E[B_sigma^2]$ while the right is $-Bbb E[B_sigma^2]$.
– John Dawkins
Nov 15 at 22:33
Here's an example to show that some sort of "smallness" hypothesis is needed for (at least) $tau$: Suppose $sigma$ is a finite (even bounded) stopping time with $Bbb E[B_sigma^2]>0$ and take $tau:=inf{t>sigma: B_t=0}$. Then $Bbb P[tau<infty, B_tau=0]=1$. The left side of the equality you request is $Bbb E[B_sigma^2]$ while the right is $-Bbb E[B_sigma^2]$.
– John Dawkins
Nov 15 at 22:33
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
For a Brownian motion $(B_t)_{t geq 0}$ it is well-known that $B_{tau} in L^2(mathbb{P})$ for any bounded stopping time $tau$, this follows e.g. from Wald's equation and ensures that all the expectations which appear in the remaining part of the answer are finite.
Recall the optional stopping theorem for martingales with continuous sample paths:
Let $(M_t,mathcal{F}_t)_{t geq 0}$ be a martingale with continuous sample paths. If $sigma,tau$ are bounded stopping times such that $sigma leq tau$ then $$mathbb{E}(M_{tau} mid mathcal{F}_{sigma}) = M_{sigma}$$ (in particular, $M_{tau} in L^1(mathbb{P})$).
The Brownian motion is a martingale with continuous sample paths, and therefore we may apply the above result to obtain $$mathbb{E}(B_{tau} mid mathcal{F}_{sigma}) = B_{sigma}$$ for any two bounded stopping times $sigma leq tau$. Combining this with the fact that $B_{sigma}$ is $mathcal{F}_{sigma}$-measurable, we find that
$$begin{align*} mathbb{E}(B_{sigma} B_{tau}) &= mathbb{E} big[ mathbb{E}(B_{sigma} B_{tau} mid mathcal{F}_{sigma}) big] \ &= mathbb{E} big[B_{sigma} mathbb{E} (B_{tau} mid mathcal{F}_{sigma}) big] \ &= mathbb{E}(B_{sigma}^2). end{align*}$$
Hence,
$$mathbb{E}((B_{tau}-B_{sigma})^2) = mathbb{E}(B_{tau}^2)-2 mathbb{E}(B_{tau} B_{sigma}) + mathbb{E}(B_{sigma}^2) = mathbb{E}(B_{tau}^2)-mathbb{E}(B_{sigma}^2).$$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
For a Brownian motion $(B_t)_{t geq 0}$ it is well-known that $B_{tau} in L^2(mathbb{P})$ for any bounded stopping time $tau$, this follows e.g. from Wald's equation and ensures that all the expectations which appear in the remaining part of the answer are finite.
Recall the optional stopping theorem for martingales with continuous sample paths:
Let $(M_t,mathcal{F}_t)_{t geq 0}$ be a martingale with continuous sample paths. If $sigma,tau$ are bounded stopping times such that $sigma leq tau$ then $$mathbb{E}(M_{tau} mid mathcal{F}_{sigma}) = M_{sigma}$$ (in particular, $M_{tau} in L^1(mathbb{P})$).
The Brownian motion is a martingale with continuous sample paths, and therefore we may apply the above result to obtain $$mathbb{E}(B_{tau} mid mathcal{F}_{sigma}) = B_{sigma}$$ for any two bounded stopping times $sigma leq tau$. Combining this with the fact that $B_{sigma}$ is $mathcal{F}_{sigma}$-measurable, we find that
$$begin{align*} mathbb{E}(B_{sigma} B_{tau}) &= mathbb{E} big[ mathbb{E}(B_{sigma} B_{tau} mid mathcal{F}_{sigma}) big] \ &= mathbb{E} big[B_{sigma} mathbb{E} (B_{tau} mid mathcal{F}_{sigma}) big] \ &= mathbb{E}(B_{sigma}^2). end{align*}$$
Hence,
$$mathbb{E}((B_{tau}-B_{sigma})^2) = mathbb{E}(B_{tau}^2)-2 mathbb{E}(B_{tau} B_{sigma}) + mathbb{E}(B_{sigma}^2) = mathbb{E}(B_{tau}^2)-mathbb{E}(B_{sigma}^2).$$
add a comment |
up vote
3
down vote
accepted
For a Brownian motion $(B_t)_{t geq 0}$ it is well-known that $B_{tau} in L^2(mathbb{P})$ for any bounded stopping time $tau$, this follows e.g. from Wald's equation and ensures that all the expectations which appear in the remaining part of the answer are finite.
Recall the optional stopping theorem for martingales with continuous sample paths:
Let $(M_t,mathcal{F}_t)_{t geq 0}$ be a martingale with continuous sample paths. If $sigma,tau$ are bounded stopping times such that $sigma leq tau$ then $$mathbb{E}(M_{tau} mid mathcal{F}_{sigma}) = M_{sigma}$$ (in particular, $M_{tau} in L^1(mathbb{P})$).
The Brownian motion is a martingale with continuous sample paths, and therefore we may apply the above result to obtain $$mathbb{E}(B_{tau} mid mathcal{F}_{sigma}) = B_{sigma}$$ for any two bounded stopping times $sigma leq tau$. Combining this with the fact that $B_{sigma}$ is $mathcal{F}_{sigma}$-measurable, we find that
$$begin{align*} mathbb{E}(B_{sigma} B_{tau}) &= mathbb{E} big[ mathbb{E}(B_{sigma} B_{tau} mid mathcal{F}_{sigma}) big] \ &= mathbb{E} big[B_{sigma} mathbb{E} (B_{tau} mid mathcal{F}_{sigma}) big] \ &= mathbb{E}(B_{sigma}^2). end{align*}$$
Hence,
$$mathbb{E}((B_{tau}-B_{sigma})^2) = mathbb{E}(B_{tau}^2)-2 mathbb{E}(B_{tau} B_{sigma}) + mathbb{E}(B_{sigma}^2) = mathbb{E}(B_{tau}^2)-mathbb{E}(B_{sigma}^2).$$
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
For a Brownian motion $(B_t)_{t geq 0}$ it is well-known that $B_{tau} in L^2(mathbb{P})$ for any bounded stopping time $tau$, this follows e.g. from Wald's equation and ensures that all the expectations which appear in the remaining part of the answer are finite.
Recall the optional stopping theorem for martingales with continuous sample paths:
Let $(M_t,mathcal{F}_t)_{t geq 0}$ be a martingale with continuous sample paths. If $sigma,tau$ are bounded stopping times such that $sigma leq tau$ then $$mathbb{E}(M_{tau} mid mathcal{F}_{sigma}) = M_{sigma}$$ (in particular, $M_{tau} in L^1(mathbb{P})$).
The Brownian motion is a martingale with continuous sample paths, and therefore we may apply the above result to obtain $$mathbb{E}(B_{tau} mid mathcal{F}_{sigma}) = B_{sigma}$$ for any two bounded stopping times $sigma leq tau$. Combining this with the fact that $B_{sigma}$ is $mathcal{F}_{sigma}$-measurable, we find that
$$begin{align*} mathbb{E}(B_{sigma} B_{tau}) &= mathbb{E} big[ mathbb{E}(B_{sigma} B_{tau} mid mathcal{F}_{sigma}) big] \ &= mathbb{E} big[B_{sigma} mathbb{E} (B_{tau} mid mathcal{F}_{sigma}) big] \ &= mathbb{E}(B_{sigma}^2). end{align*}$$
Hence,
$$mathbb{E}((B_{tau}-B_{sigma})^2) = mathbb{E}(B_{tau}^2)-2 mathbb{E}(B_{tau} B_{sigma}) + mathbb{E}(B_{sigma}^2) = mathbb{E}(B_{tau}^2)-mathbb{E}(B_{sigma}^2).$$
For a Brownian motion $(B_t)_{t geq 0}$ it is well-known that $B_{tau} in L^2(mathbb{P})$ for any bounded stopping time $tau$, this follows e.g. from Wald's equation and ensures that all the expectations which appear in the remaining part of the answer are finite.
Recall the optional stopping theorem for martingales with continuous sample paths:
Let $(M_t,mathcal{F}_t)_{t geq 0}$ be a martingale with continuous sample paths. If $sigma,tau$ are bounded stopping times such that $sigma leq tau$ then $$mathbb{E}(M_{tau} mid mathcal{F}_{sigma}) = M_{sigma}$$ (in particular, $M_{tau} in L^1(mathbb{P})$).
The Brownian motion is a martingale with continuous sample paths, and therefore we may apply the above result to obtain $$mathbb{E}(B_{tau} mid mathcal{F}_{sigma}) = B_{sigma}$$ for any two bounded stopping times $sigma leq tau$. Combining this with the fact that $B_{sigma}$ is $mathcal{F}_{sigma}$-measurable, we find that
$$begin{align*} mathbb{E}(B_{sigma} B_{tau}) &= mathbb{E} big[ mathbb{E}(B_{sigma} B_{tau} mid mathcal{F}_{sigma}) big] \ &= mathbb{E} big[B_{sigma} mathbb{E} (B_{tau} mid mathcal{F}_{sigma}) big] \ &= mathbb{E}(B_{sigma}^2). end{align*}$$
Hence,
$$mathbb{E}((B_{tau}-B_{sigma})^2) = mathbb{E}(B_{tau}^2)-2 mathbb{E}(B_{tau} B_{sigma}) + mathbb{E}(B_{sigma}^2) = mathbb{E}(B_{tau}^2)-mathbb{E}(B_{sigma}^2).$$
edited Nov 16 at 13:32
answered Nov 15 at 16:16
saz
76.6k755117
76.6k755117
add a comment |
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It would be reasonable to assume that $mathbb{E}(tau)<infty$; otherwise it's not even clear that the expectations are finite.
– saz
Nov 15 at 15:39
Do you have any hypothesis on $sigma$ and $tau$ (boundedness...)? Otherwise, you can just take $sigma = T_1$ and $tau = T_2$, where $T_a$ is the first hitting time of $a$.
– D. Thomine
Nov 15 at 15:40
@D.Thomine ,Yes I forgot to put it here. I assume boundedness of the stopping times
– vaoy
Nov 15 at 15:41
Here's an example to show that some sort of "smallness" hypothesis is needed for (at least) $tau$: Suppose $sigma$ is a finite (even bounded) stopping time with $Bbb E[B_sigma^2]>0$ and take $tau:=inf{t>sigma: B_t=0}$. Then $Bbb P[tau<infty, B_tau=0]=1$. The left side of the equality you request is $Bbb E[B_sigma^2]$ while the right is $-Bbb E[B_sigma^2]$.
– John Dawkins
Nov 15 at 22:33