Mixing
Conditional expectation of Brownian motion given stopping-time sigma algebra
$begingroup$
Let $W$ be a Brownian motion with filtration $(F_t)$. Let $tau$ be a stopping time.
It is well-known by the strong Markov property that the law of $W_{tau+t}-W_tau$ given $F_tau$ is normal with mean zero, variance $t$.
I am interested in a very minor extension of this result and I would like to prove that the law of $W_{tauvee t}-W_tau$ given $F_tau$ is normal with mean zero and variance $(tauvee t)-tau$. How can I do it?
probability-theory stochastic-processes brownian-motion conditional-expectation stopping-times
edited Jan 24 at 6:29
saz
81.7k861128
asked Jan 23 at 15:11
ZoltánZoltán
132
$endgroup$
add a comment |
$begingroup$
Let $W$ be a Brownian motion with filtration $(F_t)$. Let $tau$ be a stopping time.
It is well-known by the strong Markov property that the law of $W_{tau+t}-W_tau$ given $F_tau$ is normal with mean zero, variance $t$.
I am interested in a very minor extension of this result and I would like to prove that the law of $W_{tauvee t}-W_tau$ given $F_tau$ is normal with mean zero and variance $(tauvee t)-tau$. How can I do it?
probability-theory stochastic-processes brownian-motion conditional-expectation stopping-times
edited Jan 24 at 6:29
saz
81.7k861128
asked Jan 23 at 15:11
ZoltánZoltán
132
$endgroup$
$begingroup$
Welcome to MSE. What were your attempts in proving it? And why are you interested in this question?
$endgroup$
– James
Jan 23 at 15:19
$begingroup$
@James I tried to consider a very simple case, when $tau$ takes only two values, but couldn't prove it. The question is natural since I discovered that in general $W_{tau}-W_{sigma}$ can have any distribution if $tau$ and $sigma$ are stopping times (it does not have to be normal given $F_sigma$). I wonder what can we say in my case.
$endgroup$
– Zoltán
Jan 23 at 15:31
add a comment |
$begingroup$
Let $W$ be a Brownian motion with filtration $(F_t)$. Let $tau$ be a stopping time.
It is well-known by the strong Markov property that the law of $W_{tau+t}-W_tau$ given $F_tau$ is normal with mean zero, variance $t$.
I am interested in a very minor extension of this result and I would like to prove that the law of $W_{tauvee t}-W_tau$ given $F_tau$ is normal with mean zero and variance $(tauvee t)-tau$. How can I do it?
probability-theory stochastic-processes brownian-motion conditional-expectation stopping-times
edited Jan 24 at 6:29
saz
81.7k861128
asked Jan 23 at 15:11
ZoltánZoltán
132
$endgroup$
Let $W$ be a Brownian motion with filtration $(F_t)$. Let $tau$ be a stopping time.
It is well-known by the strong Markov property that the law of $W_{tau+t}-W_tau$ given $F_tau$ is normal with mean zero, variance $t$.
I am interested in a very minor extension of this result and I would like to prove that the law of $W_{tauvee t}-W_tau$ given $F_tau$ is normal with mean zero and variance $(tauvee t)-tau$. How can I do it?
probability-theory stochastic-processes brownian-motion conditional-expectation stopping-times
probability-theory stochastic-processes brownian-motion conditional-expectation stopping-times
edited Jan 24 at 6:29
saz
81.7k861128
asked Jan 23 at 15:11
ZoltánZoltán
132
edited Jan 24 at 6:29
saz
81.7k861128
asked Jan 23 at 15:11
ZoltánZoltán
132
edited Jan 24 at 6:29
saz
81.7k861128
edited Jan 24 at 6:29
saz
81.7k861128
edited Jan 24 at 6:29
saz
81.7k861128
81.7k861128
asked Jan 23 at 15:11
ZoltánZoltán
132
asked Jan 23 at 15:11
ZoltánZoltán
132
asked Jan 23 at 15:11
ZoltánZoltán
132
132
$begingroup$
Welcome to MSE. What were your attempts in proving it? And why are you interested in this question?
$endgroup$
– James
Jan 23 at 15:19
$begingroup$
@James I tried to consider a very simple case, when $tau$ takes only two values, but couldn't prove it. The question is natural since I discovered that in general $W_{tau}-W_{sigma}$ can have any distribution if $tau$ and $sigma$ are stopping times (it does not have to be normal given $F_sigma$). I wonder what can we say in my case.
$endgroup$
– Zoltán
Jan 23 at 15:31
add a comment |
$begingroup$
Welcome to MSE. What were your attempts in proving it? And why are you interested in this question?
$endgroup$
– James
Jan 23 at 15:19
$begingroup$
@James I tried to consider a very simple case, when $tau$ takes only two values, but couldn't prove it. The question is natural since I discovered that in general $W_{tau}-W_{sigma}$ can have any distribution if $tau$ and $sigma$ are stopping times (it does not have to be normal given $F_sigma$). I wonder what can we say in my case.
$endgroup$
– Zoltán
Jan 23 at 15:31
$begingroup$
Welcome to MSE. What were your attempts in proving it? And why are you interested in this question?
$endgroup$
– James
Jan 23 at 15:19
$begingroup$
Welcome to MSE. What were your attempts in proving it? And why are you interested in this question?
$endgroup$
– James
Jan 23 at 15:19
$begingroup$
@James I tried to consider a very simple case, when $tau$ takes only two values, but couldn't prove it. The question is natural since I discovered that in general $W_{tau}-W_{sigma}$ can have any distribution if $tau$ and $sigma$ are stopping times (it does not have to be normal given $F_sigma$). I wonder what can we say in my case.
$endgroup$
– Zoltán
Jan 23 at 15:31
$begingroup$
@James I tried to consider a very simple case, when $tau$ takes only two values, but couldn't prove it. The question is natural since I discovered that in general $W_{tau}-W_{sigma}$ can have any distribution if $tau$ and $sigma$ are stopping times (it does not have to be normal given $F_sigma$). I wonder what can we say in my case.
$endgroup$
– Zoltán
Jan 23 at 15:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $tau$ is a (finite) stopping time, then the process
$$B_s := B_{s+tau}-B_{tau}, qquad s geq 0$$
is a Brownian motion which is independent of $mathcal{F}_{tau}$. As
$$W_{max{t,tau}}-W_{tau} = B_{max{0,t-tau}},$$
we have
$$mathbb{E}big(f(W_{max{t,tau}}-W_{tau}) mid mathcal{F}_{tau} big) = mathbb{E} big( f(B_{max{0,t-tau}}) mid mathcal{F}_{tau} big)$$
for any bounded measurable function $f$. Since $max{0,t-tau}$ is $mathcal{F}_{tau}$-measurable and $(B_s)_{s geq 0}$ is independent from $mathcal{F}_{tau}$, it follows (see below for details) that
$$mathbb{E}big(f(W_{max{t,tau}}-W_{tau}) mid mathcal{F}_{tau} big) = g(max{0,t-tau}) tag{1}$$ where $$g(s) := mathbb{E}f(B_s).$$
Using this identity for $f(x) := exp(ix xi)$ with $xi in mathbb{R}$ fixed, we get
begin{align*} mathbb{E} left( exp left[i xi (W_{max{t,tau}}-W_{tau}) right] mid mathcal{F}_{tau} right) &= exp left(- frac{max{0,t-tau}}{2} xi^2 right) \ &= exp left(- frac{max{t,tau}-tau}{2} xi^2 right) end{align*}
which proves the assertion.
To prove $(1)$ rigorously, we can use the following property of conditional expectation (which you can find, in this particular formulation, in the book on Brownian motion by Schilling & Partzsch, Lemma A.3).
Proposition: Let $X: (Omega,mathcal{A}) to (D,mathcal{D})$ be a random variable. Assume that $mathcal{X}$, $mathcal{Y}$ are independent $sigma$-algebras such that $X$ is $mathcal{X}/mathcal{D}$-measurable. If $Psi: D times Omega to mathbb{R}$ is bounded and $mathcal{D} otimes mathcal{Y}/mathcal{B}(mathbb{R})$-measurable, then $$mathbb{E}(Psi(X(cdot),cdot) mid mathcal{X}) = g(X)$$ for $$g(x) := mathbb{E}(Psi(x,cdot)).$$
To prove $(1)$ we choose the objects as follows:
$D:= [0,infty)$ endowed with the Borel-$sigma$-algebra (restricted to $[0,infty)$)- $mathcal{X} := mathcal{F}_{tau}$
$mathcal{Y} := sigma(B_s, s geq 0)$,- $X := max{0,t-tau}$
$Psi(x,omega) = f(B_x(omega))$ for $x in D=[0,infty)$
Let's check the assumptions of the proposition: As already mentioned earlier on, the Brownian motion $(B_s)_{s geq 0}$ is independent from $mathcal{F}_{tau}$, i.e. $mathcal{X}$ and $mathcal{Y}$ are independent. Moreover, $tau$ is $mathcal{F}_{tau}$-measurable (i.e. $mathcal{X}$-measurable) and therefore $X$ is $mathcal{X}$-measurable. Moreover, the progressive measurability of $(B_s)_{s geq 0}$ implies that $Psi$ is $mathcal{D} otimes mathcal{Y}/mathcal{B}(mathbb{R})$-measurable.
Since we have verified all assumptions, we may apply the above proposition and this gives exactly $(1)$.
Remark: The above reasoning shows, more generally, that $W_{sigma}-W_{tau}$ given $mathcal{F}_{tau}$ is Gaussian with mean $0$ and variance $sigma-tau$ for any stopping time $sigma$ which is $mathcal{F}_{tau}$-measurable and satisfies $sigma geq tau$.
answered Jan 23 at 16:50
sazsaz
81.7k861128
$endgroup$
$begingroup$
Wow! Thank you so much for this very detailed proof, I will now use it as an example for my future studies! Amazing! By myself I managed to get equation (1), but didn't know how to proceed and prove it formally. Once again, thank you very very much!
$endgroup$
– Zoltán
Jan 23 at 17:09
$begingroup$
Also do you maybe know what is going on if we don't have independence? Is it true that $E(psi(X(⋅),⋅)∣mathcal{X})=g(X)$, where $g(x)=E(psi(x,⋅)∣mathcal{X})$? In which book can I find these type of statements?
$endgroup$
– Zoltán
Jan 23 at 17:40
$begingroup$
@Zoltán You are welcome. And no, without independence we cannot expect that a statement of this form holds.
$endgroup$
– saz
Jan 23 at 18:15
$begingroup$
Do you maybe now a simple counterexample for this? Let's say I would like $E(f(X,Y)|Z)$ to be different from $g(X)$, where $g(x):=E (f(x,Y)|Z)$, and $X$ is $Z$-measurable (but no independence assumption on $Y$ and $Z$). Is there any simple situation where this equality breaks?
$endgroup$
– Zoltán
Jan 24 at 12:10
$begingroup$
@Zoltán Take a look at this example
$endgroup$
– saz
Jan 24 at 14:40
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
If $tau$ is a (finite) stopping time, then the process
$$B_s := B_{s+tau}-B_{tau}, qquad s geq 0$$
is a Brownian motion which is independent of $mathcal{F}_{tau}$. As
$$W_{max{t,tau}}-W_{tau} = B_{max{0,t-tau}},$$
we have
$$mathbb{E}big(f(W_{max{t,tau}}-W_{tau}) mid mathcal{F}_{tau} big) = mathbb{E} big( f(B_{max{0,t-tau}}) mid mathcal{F}_{tau} big)$$
for any bounded measurable function $f$. Since $max{0,t-tau}$ is $mathcal{F}_{tau}$-measurable and $(B_s)_{s geq 0}$ is independent from $mathcal{F}_{tau}$, it follows (see below for details) that
$$mathbb{E}big(f(W_{max{t,tau}}-W_{tau}) mid mathcal{F}_{tau} big) = g(max{0,t-tau}) tag{1}$$ where $$g(s) := mathbb{E}f(B_s).$$
Using this identity for $f(x) := exp(ix xi)$ with $xi in mathbb{R}$ fixed, we get
begin{align*} mathbb{E} left( exp left[i xi (W_{max{t,tau}}-W_{tau}) right] mid mathcal{F}_{tau} right) &= exp left(- frac{max{0,t-tau}}{2} xi^2 right) \ &= exp left(- frac{max{t,tau}-tau}{2} xi^2 right) end{align*}
which proves the assertion.
To prove $(1)$ rigorously, we can use the following property of conditional expectation (which you can find, in this particular formulation, in the book on Brownian motion by Schilling & Partzsch, Lemma A.3).
Proposition: Let $X: (Omega,mathcal{A}) to (D,mathcal{D})$ be a random variable. Assume that $mathcal{X}$, $mathcal{Y}$ are independent $sigma$-algebras such that $X$ is $mathcal{X}/mathcal{D}$-measurable. If $Psi: D times Omega to mathbb{R}$ is bounded and $mathcal{D} otimes mathcal{Y}/mathcal{B}(mathbb{R})$-measurable, then $$mathbb{E}(Psi(X(cdot),cdot) mid mathcal{X}) = g(X)$$ for $$g(x) := mathbb{E}(Psi(x,cdot)).$$
To prove $(1)$ we choose the objects as follows:
$D:= [0,infty)$ endowed with the Borel-$sigma$-algebra (restricted to $[0,infty)$)- $mathcal{X} := mathcal{F}_{tau}$
$mathcal{Y} := sigma(B_s, s geq 0)$,- $X := max{0,t-tau}$
$Psi(x,omega) = f(B_x(omega))$ for $x in D=[0,infty)$
Let's check the assumptions of the proposition: As already mentioned earlier on, the Brownian motion $(B_s)_{s geq 0}$ is independent from $mathcal{F}_{tau}$, i.e. $mathcal{X}$ and $mathcal{Y}$ are independent. Moreover, $tau$ is $mathcal{F}_{tau}$-measurable (i.e. $mathcal{X}$-measurable) and therefore $X$ is $mathcal{X}$-measurable. Moreover, the progressive measurability of $(B_s)_{s geq 0}$ implies that $Psi$ is $mathcal{D} otimes mathcal{Y}/mathcal{B}(mathbb{R})$-measurable.
Since we have verified all assumptions, we may apply the above proposition and this gives exactly $(1)$.
Remark: The above reasoning shows, more generally, that $W_{sigma}-W_{tau}$ given $mathcal{F}_{tau}$ is Gaussian with mean $0$ and variance $sigma-tau$ for any stopping time $sigma$ which is $mathcal{F}_{tau}$-measurable and satisfies $sigma geq tau$.
answered Jan 23 at 16:50
sazsaz
81.7k861128
$endgroup$
$begingroup$
Wow! Thank you so much for this very detailed proof, I will now use it as an example for my future studies! Amazing! By myself I managed to get equation (1), but didn't know how to proceed and prove it formally. Once again, thank you very very much!
$endgroup$
– Zoltán
Jan 23 at 17:09
$begingroup$
Also do you maybe know what is going on if we don't have independence? Is it true that $E(psi(X(⋅),⋅)∣mathcal{X})=g(X)$, where $g(x)=E(psi(x,⋅)∣mathcal{X})$? In which book can I find these type of statements?
$endgroup$
– Zoltán
Jan 23 at 17:40
$begingroup$
@Zoltán You are welcome. And no, without independence we cannot expect that a statement of this form holds.
$endgroup$
– saz
Jan 23 at 18:15
$begingroup$
Do you maybe now a simple counterexample for this? Let's say I would like $E(f(X,Y)|Z)$ to be different from $g(X)$, where $g(x):=E (f(x,Y)|Z)$, and $X$ is $Z$-measurable (but no independence assumption on $Y$ and $Z$). Is there any simple situation where this equality breaks?
$endgroup$
– Zoltán
Jan 24 at 12:10
$begingroup$
@Zoltán Take a look at this example
$endgroup$
– saz
Jan 24 at 14:40
add a comment |
$begingroup$
If $tau$ is a (finite) stopping time, then the process
$$B_s := B_{s+tau}-B_{tau}, qquad s geq 0$$
is a Brownian motion which is independent of $mathcal{F}_{tau}$. As
$$W_{max{t,tau}}-W_{tau} = B_{max{0,t-tau}},$$
we have
$$mathbb{E}big(f(W_{max{t,tau}}-W_{tau}) mid mathcal{F}_{tau} big) = mathbb{E} big( f(B_{max{0,t-tau}}) mid mathcal{F}_{tau} big)$$
for any bounded measurable function $f$. Since $max{0,t-tau}$ is $mathcal{F}_{tau}$-measurable and $(B_s)_{s geq 0}$ is independent from $mathcal{F}_{tau}$, it follows (see below for details) that
$$mathbb{E}big(f(W_{max{t,tau}}-W_{tau}) mid mathcal{F}_{tau} big) = g(max{0,t-tau}) tag{1}$$ where $$g(s) := mathbb{E}f(B_s).$$
Using this identity for $f(x) := exp(ix xi)$ with $xi in mathbb{R}$ fixed, we get
begin{align*} mathbb{E} left( exp left[i xi (W_{max{t,tau}}-W_{tau}) right] mid mathcal{F}_{tau} right) &= exp left(- frac{max{0,t-tau}}{2} xi^2 right) \ &= exp left(- frac{max{t,tau}-tau}{2} xi^2 right) end{align*}
which proves the assertion.
To prove $(1)$ rigorously, we can use the following property of conditional expectation (which you can find, in this particular formulation, in the book on Brownian motion by Schilling & Partzsch, Lemma A.3).
Proposition: Let $X: (Omega,mathcal{A}) to (D,mathcal{D})$ be a random variable. Assume that $mathcal{X}$, $mathcal{Y}$ are independent $sigma$-algebras such that $X$ is $mathcal{X}/mathcal{D}$-measurable. If $Psi: D times Omega to mathbb{R}$ is bounded and $mathcal{D} otimes mathcal{Y}/mathcal{B}(mathbb{R})$-measurable, then $$mathbb{E}(Psi(X(cdot),cdot) mid mathcal{X}) = g(X)$$ for $$g(x) := mathbb{E}(Psi(x,cdot)).$$
To prove $(1)$ we choose the objects as follows:
$D:= [0,infty)$ endowed with the Borel-$sigma$-algebra (restricted to $[0,infty)$)- $mathcal{X} := mathcal{F}_{tau}$
$mathcal{Y} := sigma(B_s, s geq 0)$,- $X := max{0,t-tau}$
$Psi(x,omega) = f(B_x(omega))$ for $x in D=[0,infty)$
Let's check the assumptions of the proposition: As already mentioned earlier on, the Brownian motion $(B_s)_{s geq 0}$ is independent from $mathcal{F}_{tau}$, i.e. $mathcal{X}$ and $mathcal{Y}$ are independent. Moreover, $tau$ is $mathcal{F}_{tau}$-measurable (i.e. $mathcal{X}$-measurable) and therefore $X$ is $mathcal{X}$-measurable. Moreover, the progressive measurability of $(B_s)_{s geq 0}$ implies that $Psi$ is $mathcal{D} otimes mathcal{Y}/mathcal{B}(mathbb{R})$-measurable.
Since we have verified all assumptions, we may apply the above proposition and this gives exactly $(1)$.
Remark: The above reasoning shows, more generally, that $W_{sigma}-W_{tau}$ given $mathcal{F}_{tau}$ is Gaussian with mean $0$ and variance $sigma-tau$ for any stopping time $sigma$ which is $mathcal{F}_{tau}$-measurable and satisfies $sigma geq tau$.
answered Jan 23 at 16:50
sazsaz
81.7k861128
$endgroup$
$begingroup$
Wow! Thank you so much for this very detailed proof, I will now use it as an example for my future studies! Amazing! By myself I managed to get equation (1), but didn't know how to proceed and prove it formally. Once again, thank you very very much!
$endgroup$
– Zoltán
Jan 23 at 17:09
$begingroup$
Also do you maybe know what is going on if we don't have independence? Is it true that $E(psi(X(⋅),⋅)∣mathcal{X})=g(X)$, where $g(x)=E(psi(x,⋅)∣mathcal{X})$? In which book can I find these type of statements?
$endgroup$
– Zoltán
Jan 23 at 17:40
$begingroup$
@Zoltán You are welcome. And no, without independence we cannot expect that a statement of this form holds.
$endgroup$
– saz
Jan 23 at 18:15
$begingroup$
Do you maybe now a simple counterexample for this? Let's say I would like $E(f(X,Y)|Z)$ to be different from $g(X)$, where $g(x):=E (f(x,Y)|Z)$, and $X$ is $Z$-measurable (but no independence assumption on $Y$ and $Z$). Is there any simple situation where this equality breaks?
$endgroup$
– Zoltán
Jan 24 at 12:10
$begingroup$
@Zoltán Take a look at this example
$endgroup$
– saz
Jan 24 at 14:40
add a comment |
$begingroup$
If $tau$ is a (finite) stopping time, then the process
$$B_s := B_{s+tau}-B_{tau}, qquad s geq 0$$
is a Brownian motion which is independent of $mathcal{F}_{tau}$. As
$$W_{max{t,tau}}-W_{tau} = B_{max{0,t-tau}},$$
we have
$$mathbb{E}big(f(W_{max{t,tau}}-W_{tau}) mid mathcal{F}_{tau} big) = mathbb{E} big( f(B_{max{0,t-tau}}) mid mathcal{F}_{tau} big)$$
for any bounded measurable function $f$. Since $max{0,t-tau}$ is $mathcal{F}_{tau}$-measurable and $(B_s)_{s geq 0}$ is independent from $mathcal{F}_{tau}$, it follows (see below for details) that
$$mathbb{E}big(f(W_{max{t,tau}}-W_{tau}) mid mathcal{F}_{tau} big) = g(max{0,t-tau}) tag{1}$$ where $$g(s) := mathbb{E}f(B_s).$$
Using this identity for $f(x) := exp(ix xi)$ with $xi in mathbb{R}$ fixed, we get
begin{align*} mathbb{E} left( exp left[i xi (W_{max{t,tau}}-W_{tau}) right] mid mathcal{F}_{tau} right) &= exp left(- frac{max{0,t-tau}}{2} xi^2 right) \ &= exp left(- frac{max{t,tau}-tau}{2} xi^2 right) end{align*}
which proves the assertion.
To prove $(1)$ rigorously, we can use the following property of conditional expectation (which you can find, in this particular formulation, in the book on Brownian motion by Schilling & Partzsch, Lemma A.3).
Proposition: Let $X: (Omega,mathcal{A}) to (D,mathcal{D})$ be a random variable. Assume that $mathcal{X}$, $mathcal{Y}$ are independent $sigma$-algebras such that $X$ is $mathcal{X}/mathcal{D}$-measurable. If $Psi: D times Omega to mathbb{R}$ is bounded and $mathcal{D} otimes mathcal{Y}/mathcal{B}(mathbb{R})$-measurable, then $$mathbb{E}(Psi(X(cdot),cdot) mid mathcal{X}) = g(X)$$ for $$g(x) := mathbb{E}(Psi(x,cdot)).$$
To prove $(1)$ we choose the objects as follows:
$D:= [0,infty)$ endowed with the Borel-$sigma$-algebra (restricted to $[0,infty)$)- $mathcal{X} := mathcal{F}_{tau}$
$mathcal{Y} := sigma(B_s, s geq 0)$,- $X := max{0,t-tau}$
$Psi(x,omega) = f(B_x(omega))$ for $x in D=[0,infty)$
Let's check the assumptions of the proposition: As already mentioned earlier on, the Brownian motion $(B_s)_{s geq 0}$ is independent from $mathcal{F}_{tau}$, i.e. $mathcal{X}$ and $mathcal{Y}$ are independent. Moreover, $tau$ is $mathcal{F}_{tau}$-measurable (i.e. $mathcal{X}$-measurable) and therefore $X$ is $mathcal{X}$-measurable. Moreover, the progressive measurability of $(B_s)_{s geq 0}$ implies that $Psi$ is $mathcal{D} otimes mathcal{Y}/mathcal{B}(mathbb{R})$-measurable.
Since we have verified all assumptions, we may apply the above proposition and this gives exactly $(1)$.
Remark: The above reasoning shows, more generally, that $W_{sigma}-W_{tau}$ given $mathcal{F}_{tau}$ is Gaussian with mean $0$ and variance $sigma-tau$ for any stopping time $sigma$ which is $mathcal{F}_{tau}$-measurable and satisfies $sigma geq tau$.
answered Jan 23 at 16:50
sazsaz
81.7k861128
$endgroup$
If $tau$ is a (finite) stopping time, then the process
$$B_s := B_{s+tau}-B_{tau}, qquad s geq 0$$
is a Brownian motion which is independent of $mathcal{F}_{tau}$. As
$$W_{max{t,tau}}-W_{tau} = B_{max{0,t-tau}},$$
we have
$$mathbb{E}big(f(W_{max{t,tau}}-W_{tau}) mid mathcal{F}_{tau} big) = mathbb{E} big( f(B_{max{0,t-tau}}) mid mathcal{F}_{tau} big)$$
for any bounded measurable function $f$. Since $max{0,t-tau}$ is $mathcal{F}_{tau}$-measurable and $(B_s)_{s geq 0}$ is independent from $mathcal{F}_{tau}$, it follows (see below for details) that
$$mathbb{E}big(f(W_{max{t,tau}}-W_{tau}) mid mathcal{F}_{tau} big) = g(max{0,t-tau}) tag{1}$$ where $$g(s) := mathbb{E}f(B_s).$$
Using this identity for $f(x) := exp(ix xi)$ with $xi in mathbb{R}$ fixed, we get
begin{align*} mathbb{E} left( exp left[i xi (W_{max{t,tau}}-W_{tau}) right] mid mathcal{F}_{tau} right) &= exp left(- frac{max{0,t-tau}}{2} xi^2 right) \ &= exp left(- frac{max{t,tau}-tau}{2} xi^2 right) end{align*}
which proves the assertion.
To prove $(1)$ rigorously, we can use the following property of conditional expectation (which you can find, in this particular formulation, in the book on Brownian motion by Schilling & Partzsch, Lemma A.3).
Proposition: Let $X: (Omega,mathcal{A}) to (D,mathcal{D})$ be a random variable. Assume that $mathcal{X}$, $mathcal{Y}$ are independent $sigma$-algebras such that $X$ is $mathcal{X}/mathcal{D}$-measurable. If $Psi: D times Omega to mathbb{R}$ is bounded and $mathcal{D} otimes mathcal{Y}/mathcal{B}(mathbb{R})$-measurable, then $$mathbb{E}(Psi(X(cdot),cdot) mid mathcal{X}) = g(X)$$ for $$g(x) := mathbb{E}(Psi(x,cdot)).$$
To prove $(1)$ we choose the objects as follows:
$D:= [0,infty)$ endowed with the Borel-$sigma$-algebra (restricted to $[0,infty)$)- $mathcal{X} := mathcal{F}_{tau}$
$mathcal{Y} := sigma(B_s, s geq 0)$,- $X := max{0,t-tau}$
$Psi(x,omega) = f(B_x(omega))$ for $x in D=[0,infty)$
Let's check the assumptions of the proposition: As already mentioned earlier on, the Brownian motion $(B_s)_{s geq 0}$ is independent from $mathcal{F}_{tau}$, i.e. $mathcal{X}$ and $mathcal{Y}$ are independent. Moreover, $tau$ is $mathcal{F}_{tau}$-measurable (i.e. $mathcal{X}$-measurable) and therefore $X$ is $mathcal{X}$-measurable. Moreover, the progressive measurability of $(B_s)_{s geq 0}$ implies that $Psi$ is $mathcal{D} otimes mathcal{Y}/mathcal{B}(mathbb{R})$-measurable.
Since we have verified all assumptions, we may apply the above proposition and this gives exactly $(1)$.
Remark: The above reasoning shows, more generally, that $W_{sigma}-W_{tau}$ given $mathcal{F}_{tau}$ is Gaussian with mean $0$ and variance $sigma-tau$ for any stopping time $sigma$ which is $mathcal{F}_{tau}$-measurable and satisfies $sigma geq tau$.
answered Jan 23 at 16:50
sazsaz
81.7k861128
edited Jan 25 at 16:41
answered Jan 23 at 16:50
sazsaz
81.7k861128
answered Jan 23 at 16:50
sazsaz
81.7k861128
answered Jan 23 at 16:50
sazsaz
81.7k861128
81.7k861128
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Wow! Thank you so much for this very detailed proof, I will now use it as an example for my future studies! Amazing! By myself I managed to get equation (1), but didn't know how to proceed and prove it formally. Once again, thank you very very much!
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– Zoltán
Jan 23 at 17:09
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Also do you maybe know what is going on if we don't have independence? Is it true that $E(psi(X(⋅),⋅)∣mathcal{X})=g(X)$, where $g(x)=E(psi(x,⋅)∣mathcal{X})$? In which book can I find these type of statements?
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– Zoltán
Jan 23 at 17:40
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@Zoltán You are welcome. And no, without independence we cannot expect that a statement of this form holds.
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– saz
Jan 23 at 18:15
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Do you maybe now a simple counterexample for this? Let's say I would like $E(f(X,Y)|Z)$ to be different from $g(X)$, where $g(x):=E (f(x,Y)|Z)$, and $X$ is $Z$-measurable (but no independence assumption on $Y$ and $Z$). Is there any simple situation where this equality breaks?
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– Zoltán
Jan 24 at 12:10
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@Zoltán Take a look at this example
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– saz
Jan 24 at 14:40
add a comment |
$begingroup$
Wow! Thank you so much for this very detailed proof, I will now use it as an example for my future studies! Amazing! By myself I managed to get equation (1), but didn't know how to proceed and prove it formally. Once again, thank you very very much!
$endgroup$
– Zoltán
Jan 23 at 17:09
$begingroup$
Also do you maybe know what is going on if we don't have independence? Is it true that $E(psi(X(⋅),⋅)∣mathcal{X})=g(X)$, where $g(x)=E(psi(x,⋅)∣mathcal{X})$? In which book can I find these type of statements?
$endgroup$
– Zoltán
Jan 23 at 17:40
$begingroup$
@Zoltán You are welcome. And no, without independence we cannot expect that a statement of this form holds.
$endgroup$
– saz
Jan 23 at 18:15
$begingroup$
Do you maybe now a simple counterexample for this? Let's say I would like $E(f(X,Y)|Z)$ to be different from $g(X)$, where $g(x):=E (f(x,Y)|Z)$, and $X$ is $Z$-measurable (but no independence assumption on $Y$ and $Z$). Is there any simple situation where this equality breaks?
$endgroup$
– Zoltán
Jan 24 at 12:10
$begingroup$
@Zoltán Take a look at this example
$endgroup$
– saz
Jan 24 at 14:40
$begingroup$
Wow! Thank you so much for this very detailed proof, I will now use it as an example for my future studies! Amazing! By myself I managed to get equation (1), but didn't know how to proceed and prove it formally. Once again, thank you very very much!
$endgroup$
– Zoltán
Jan 23 at 17:09
$begingroup$
Wow! Thank you so much for this very detailed proof, I will now use it as an example for my future studies! Amazing! By myself I managed to get equation (1), but didn't know how to proceed and prove it formally. Once again, thank you very very much!
$endgroup$
– Zoltán
Jan 23 at 17:09
$begingroup$
Also do you maybe know what is going on if we don't have independence? Is it true that $E(psi(X(⋅),⋅)∣mathcal{X})=g(X)$, where $g(x)=E(psi(x,⋅)∣mathcal{X})$? In which book can I find these type of statements?
$endgroup$
– Zoltán
Jan 23 at 17:40
$begingroup$
Also do you maybe know what is going on if we don't have independence? Is it true that $E(psi(X(⋅),⋅)∣mathcal{X})=g(X)$, where $g(x)=E(psi(x,⋅)∣mathcal{X})$? In which book can I find these type of statements?
$endgroup$
– Zoltán
Jan 23 at 17:40
$begingroup$
@Zoltán You are welcome. And no, without independence we cannot expect that a statement of this form holds.
$endgroup$
– saz
Jan 23 at 18:15
$begingroup$
@Zoltán You are welcome. And no, without independence we cannot expect that a statement of this form holds.
$endgroup$
– saz
Jan 23 at 18:15
$begingroup$
Do you maybe now a simple counterexample for this? Let's say I would like $E(f(X,Y)|Z)$ to be different from $g(X)$, where $g(x):=E (f(x,Y)|Z)$, and $X$ is $Z$-measurable (but no independence assumption on $Y$ and $Z$). Is there any simple situation where this equality breaks?
$endgroup$
– Zoltán
Jan 24 at 12:10
$begingroup$
Do you maybe now a simple counterexample for this? Let's say I would like $E(f(X,Y)|Z)$ to be different from $g(X)$, where $g(x):=E (f(x,Y)|Z)$, and $X$ is $Z$-measurable (but no independence assumption on $Y$ and $Z$). Is there any simple situation where this equality breaks?
$endgroup$
– Zoltán
Jan 24 at 12:10
$begingroup$
@Zoltán Take a look at this example
$endgroup$
– saz
Jan 24 at 14:40
$begingroup$
@Zoltán Take a look at this example
$endgroup$
– saz
Jan 24 at 14:40
add a comment |
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Welcome to MSE. What were your attempts in proving it? And why are you interested in this question?
$endgroup$
– James
Jan 23 at 15:19
$begingroup$
@James I tried to consider a very simple case, when $tau$ takes only two values, but couldn't prove it. The question is natural since I discovered that in general $W_{tau}-W_{sigma}$ can have any distribution if $tau$ and $sigma$ are stopping times (it does not have to be normal given $F_sigma$). I wonder what can we say in my case.
$endgroup$
– Zoltán
Jan 23 at 15:31