Mixing
Showing that the projection of a measure on path space onto marginals is continuous
$begingroup$
Let $C([0,T], mathbb{R}^d)$ be the metric space of continuous functions from $[0,T]$ to $mathbb{R}^d$, endowed with the supremum metric. For any metric space $E$, let $mathcal{P}(E)$ be the space of probability measures on $E$, endowed with the weak topology.
For fixed $t in [0,T]$, we define the map
$$ Phi: mathcal{P}(C([0,T],mathbb{R}^d)) to mathcal{P}(mathbb{R}^d); quad quad mu mapsto mu_t,$$
where $mu_t$ is the marginal of $mu$ at time $t$.
Is it true that $Phi$ is continuous? It seems obvious, but I can't see that from the definition.
probability-theory measure-theory weak-convergence
asked Jan 20 at 14:01
RichardRichard
1,2211724
$endgroup$
add a comment |
$begingroup$
Let $C([0,T], mathbb{R}^d)$ be the metric space of continuous functions from $[0,T]$ to $mathbb{R}^d$, endowed with the supremum metric. For any metric space $E$, let $mathcal{P}(E)$ be the space of probability measures on $E$, endowed with the weak topology.
For fixed $t in [0,T]$, we define the map
$$ Phi: mathcal{P}(C([0,T],mathbb{R}^d)) to mathcal{P}(mathbb{R}^d); quad quad mu mapsto mu_t,$$
where $mu_t$ is the marginal of $mu$ at time $t$.
Is it true that $Phi$ is continuous? It seems obvious, but I can't see that from the definition.
probability-theory measure-theory weak-convergence
asked Jan 20 at 14:01
RichardRichard
1,2211724
$endgroup$
add a comment |
$begingroup$
Let $C([0,T], mathbb{R}^d)$ be the metric space of continuous functions from $[0,T]$ to $mathbb{R}^d$, endowed with the supremum metric. For any metric space $E$, let $mathcal{P}(E)$ be the space of probability measures on $E$, endowed with the weak topology.
For fixed $t in [0,T]$, we define the map
$$ Phi: mathcal{P}(C([0,T],mathbb{R}^d)) to mathcal{P}(mathbb{R}^d); quad quad mu mapsto mu_t,$$
where $mu_t$ is the marginal of $mu$ at time $t$.
Is it true that $Phi$ is continuous? It seems obvious, but I can't see that from the definition.
probability-theory measure-theory weak-convergence
asked Jan 20 at 14:01
RichardRichard
1,2211724
$endgroup$
Let $C([0,T], mathbb{R}^d)$ be the metric space of continuous functions from $[0,T]$ to $mathbb{R}^d$, endowed with the supremum metric. For any metric space $E$, let $mathcal{P}(E)$ be the space of probability measures on $E$, endowed with the weak topology.
For fixed $t in [0,T]$, we define the map
$$ Phi: mathcal{P}(C([0,T],mathbb{R}^d)) to mathcal{P}(mathbb{R}^d); quad quad mu mapsto mu_t,$$
where $mu_t$ is the marginal of $mu$ at time $t$.
Is it true that $Phi$ is continuous? It seems obvious, but I can't see that from the definition.
probability-theory measure-theory weak-convergence
probability-theory measure-theory weak-convergence
asked Jan 20 at 14:01
RichardRichard
1,2211724
asked Jan 20 at 14:01
RichardRichard
1,2211724
asked Jan 20 at 14:01
RichardRichard
1,2211724
asked Jan 20 at 14:01
RichardRichard
1,2211724
asked Jan 20 at 14:01
RichardRichard
1,2211724
1,2211724
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
A formal proof proceeds as the following. We denote by $omega$ an element of the sample space $C([0,T])$ and by $omega(t)$ its $t$-coordinate. By definition, we have
$$
mu_t(A)=muleft(left{omegain C([0,T]):omega(t)in Aright}right)=int 1_{omega(t)in A}mathrm dmu
$$ for any fixed $tin [0,T]$ and $Ainmathcal{B}(Bbb R^d)$. This implies
$$
int_{Bbb R^d} 1_Amathrm{d}mu_t = int_{C([0,T])} 1_A(omega(t))mathrm{d}mu(omega)
$$ for all $Ainmathcal{B}(Bbb R^d)$, and thus
$$
int_{Bbb R^d} f;mathrm{d}mu_t=int_{C([0,T])} f(omega(t));mathrm{d}mu(omega)
$$ holds for all bounded continuous $f:Bbb R^dto Bbb R$.
To prove weak continuity of the map $Phi_t:mumapstomu_t$, we need to show $Phi_t(mu_n)to Phi_t(mu)$ weakly whenever $mu_nto mu$ weakly. Assume $mu_n$ converges to $mu$ weakly, i.e. for every bounded continuous functional $F:C([0,T])toBbb R$, it holds
$$
int_{C([0,T])}F(omega)mathrm{d}mu_n(omega)to int_{C([0,T])}F(omega)mathrm{d}mu(omega).tag{*}
$$ Observe that $omega mapsto f(omega(t))$ is a bounded continuous functional on $C([0,T])$ for any bounded continuous function $f:Bbb R^d to Bbb R$. This is obvious since if $omega_n(cdot) to omega(cdot)$ uniformly on $[0,T]$, then $omega_n(t)toomega(t)$ and $f(omega_n(t))to f(omega(t))$ for each $tin[0,T]$. Thus, $(*)$ implies
$$
int_{Bbb R^d}fmathrm{d}Phi_t(mu_{n})=int_{C([0,T])}f(omega(t))mathrm{d}mu_n(omega)to int_{C([0,T])}f(omega(t))mathrm{d}mu(omega)=int_{Bbb R^d}fmathrm{d}Phi_t(mu)
$$ for all bounded continuous $f$. This proves the claim.
answered Jan 20 at 14:24
SongSong
16.5k1741
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
A formal proof proceeds as the following. We denote by $omega$ an element of the sample space $C([0,T])$ and by $omega(t)$ its $t$-coordinate. By definition, we have
$$
mu_t(A)=muleft(left{omegain C([0,T]):omega(t)in Aright}right)=int 1_{omega(t)in A}mathrm dmu
$$ for any fixed $tin [0,T]$ and $Ainmathcal{B}(Bbb R^d)$. This implies
$$
int_{Bbb R^d} 1_Amathrm{d}mu_t = int_{C([0,T])} 1_A(omega(t))mathrm{d}mu(omega)
$$ for all $Ainmathcal{B}(Bbb R^d)$, and thus
$$
int_{Bbb R^d} f;mathrm{d}mu_t=int_{C([0,T])} f(omega(t));mathrm{d}mu(omega)
$$ holds for all bounded continuous $f:Bbb R^dto Bbb R$.
To prove weak continuity of the map $Phi_t:mumapstomu_t$, we need to show $Phi_t(mu_n)to Phi_t(mu)$ weakly whenever $mu_nto mu$ weakly. Assume $mu_n$ converges to $mu$ weakly, i.e. for every bounded continuous functional $F:C([0,T])toBbb R$, it holds
$$
int_{C([0,T])}F(omega)mathrm{d}mu_n(omega)to int_{C([0,T])}F(omega)mathrm{d}mu(omega).tag{*}
$$ Observe that $omega mapsto f(omega(t))$ is a bounded continuous functional on $C([0,T])$ for any bounded continuous function $f:Bbb R^d to Bbb R$. This is obvious since if $omega_n(cdot) to omega(cdot)$ uniformly on $[0,T]$, then $omega_n(t)toomega(t)$ and $f(omega_n(t))to f(omega(t))$ for each $tin[0,T]$. Thus, $(*)$ implies
$$
int_{Bbb R^d}fmathrm{d}Phi_t(mu_{n})=int_{C([0,T])}f(omega(t))mathrm{d}mu_n(omega)to int_{C([0,T])}f(omega(t))mathrm{d}mu(omega)=int_{Bbb R^d}fmathrm{d}Phi_t(mu)
$$ for all bounded continuous $f$. This proves the claim.
answered Jan 20 at 14:24
SongSong
16.5k1741
$endgroup$
add a comment |
$begingroup$
A formal proof proceeds as the following. We denote by $omega$ an element of the sample space $C([0,T])$ and by $omega(t)$ its $t$-coordinate. By definition, we have
$$
mu_t(A)=muleft(left{omegain C([0,T]):omega(t)in Aright}right)=int 1_{omega(t)in A}mathrm dmu
$$ for any fixed $tin [0,T]$ and $Ainmathcal{B}(Bbb R^d)$. This implies
$$
int_{Bbb R^d} 1_Amathrm{d}mu_t = int_{C([0,T])} 1_A(omega(t))mathrm{d}mu(omega)
$$ for all $Ainmathcal{B}(Bbb R^d)$, and thus
$$
int_{Bbb R^d} f;mathrm{d}mu_t=int_{C([0,T])} f(omega(t));mathrm{d}mu(omega)
$$ holds for all bounded continuous $f:Bbb R^dto Bbb R$.
To prove weak continuity of the map $Phi_t:mumapstomu_t$, we need to show $Phi_t(mu_n)to Phi_t(mu)$ weakly whenever $mu_nto mu$ weakly. Assume $mu_n$ converges to $mu$ weakly, i.e. for every bounded continuous functional $F:C([0,T])toBbb R$, it holds
$$
int_{C([0,T])}F(omega)mathrm{d}mu_n(omega)to int_{C([0,T])}F(omega)mathrm{d}mu(omega).tag{*}
$$ Observe that $omega mapsto f(omega(t))$ is a bounded continuous functional on $C([0,T])$ for any bounded continuous function $f:Bbb R^d to Bbb R$. This is obvious since if $omega_n(cdot) to omega(cdot)$ uniformly on $[0,T]$, then $omega_n(t)toomega(t)$ and $f(omega_n(t))to f(omega(t))$ for each $tin[0,T]$. Thus, $(*)$ implies
$$
int_{Bbb R^d}fmathrm{d}Phi_t(mu_{n})=int_{C([0,T])}f(omega(t))mathrm{d}mu_n(omega)to int_{C([0,T])}f(omega(t))mathrm{d}mu(omega)=int_{Bbb R^d}fmathrm{d}Phi_t(mu)
$$ for all bounded continuous $f$. This proves the claim.
answered Jan 20 at 14:24
SongSong
16.5k1741
$endgroup$
add a comment |
$begingroup$
A formal proof proceeds as the following. We denote by $omega$ an element of the sample space $C([0,T])$ and by $omega(t)$ its $t$-coordinate. By definition, we have
$$
mu_t(A)=muleft(left{omegain C([0,T]):omega(t)in Aright}right)=int 1_{omega(t)in A}mathrm dmu
$$ for any fixed $tin [0,T]$ and $Ainmathcal{B}(Bbb R^d)$. This implies
$$
int_{Bbb R^d} 1_Amathrm{d}mu_t = int_{C([0,T])} 1_A(omega(t))mathrm{d}mu(omega)
$$ for all $Ainmathcal{B}(Bbb R^d)$, and thus
$$
int_{Bbb R^d} f;mathrm{d}mu_t=int_{C([0,T])} f(omega(t));mathrm{d}mu(omega)
$$ holds for all bounded continuous $f:Bbb R^dto Bbb R$.
To prove weak continuity of the map $Phi_t:mumapstomu_t$, we need to show $Phi_t(mu_n)to Phi_t(mu)$ weakly whenever $mu_nto mu$ weakly. Assume $mu_n$ converges to $mu$ weakly, i.e. for every bounded continuous functional $F:C([0,T])toBbb R$, it holds
$$
int_{C([0,T])}F(omega)mathrm{d}mu_n(omega)to int_{C([0,T])}F(omega)mathrm{d}mu(omega).tag{*}
$$ Observe that $omega mapsto f(omega(t))$ is a bounded continuous functional on $C([0,T])$ for any bounded continuous function $f:Bbb R^d to Bbb R$. This is obvious since if $omega_n(cdot) to omega(cdot)$ uniformly on $[0,T]$, then $omega_n(t)toomega(t)$ and $f(omega_n(t))to f(omega(t))$ for each $tin[0,T]$. Thus, $(*)$ implies
$$
int_{Bbb R^d}fmathrm{d}Phi_t(mu_{n})=int_{C([0,T])}f(omega(t))mathrm{d}mu_n(omega)to int_{C([0,T])}f(omega(t))mathrm{d}mu(omega)=int_{Bbb R^d}fmathrm{d}Phi_t(mu)
$$ for all bounded continuous $f$. This proves the claim.
answered Jan 20 at 14:24
SongSong
16.5k1741
$endgroup$
A formal proof proceeds as the following. We denote by $omega$ an element of the sample space $C([0,T])$ and by $omega(t)$ its $t$-coordinate. By definition, we have
$$
mu_t(A)=muleft(left{omegain C([0,T]):omega(t)in Aright}right)=int 1_{omega(t)in A}mathrm dmu
$$ for any fixed $tin [0,T]$ and $Ainmathcal{B}(Bbb R^d)$. This implies
$$
int_{Bbb R^d} 1_Amathrm{d}mu_t = int_{C([0,T])} 1_A(omega(t))mathrm{d}mu(omega)
$$ for all $Ainmathcal{B}(Bbb R^d)$, and thus
$$
int_{Bbb R^d} f;mathrm{d}mu_t=int_{C([0,T])} f(omega(t));mathrm{d}mu(omega)
$$ holds for all bounded continuous $f:Bbb R^dto Bbb R$.
To prove weak continuity of the map $Phi_t:mumapstomu_t$, we need to show $Phi_t(mu_n)to Phi_t(mu)$ weakly whenever $mu_nto mu$ weakly. Assume $mu_n$ converges to $mu$ weakly, i.e. for every bounded continuous functional $F:C([0,T])toBbb R$, it holds
$$
int_{C([0,T])}F(omega)mathrm{d}mu_n(omega)to int_{C([0,T])}F(omega)mathrm{d}mu(omega).tag{*}
$$ Observe that $omega mapsto f(omega(t))$ is a bounded continuous functional on $C([0,T])$ for any bounded continuous function $f:Bbb R^d to Bbb R$. This is obvious since if $omega_n(cdot) to omega(cdot)$ uniformly on $[0,T]$, then $omega_n(t)toomega(t)$ and $f(omega_n(t))to f(omega(t))$ for each $tin[0,T]$. Thus, $(*)$ implies
$$
int_{Bbb R^d}fmathrm{d}Phi_t(mu_{n})=int_{C([0,T])}f(omega(t))mathrm{d}mu_n(omega)to int_{C([0,T])}f(omega(t))mathrm{d}mu(omega)=int_{Bbb R^d}fmathrm{d}Phi_t(mu)
$$ for all bounded continuous $f$. This proves the claim.
answered Jan 20 at 14:24
SongSong
16.5k1741
edited Jan 20 at 14:39
answered Jan 20 at 14:24
SongSong
16.5k1741
answered Jan 20 at 14:24
SongSong
16.5k1741
answered Jan 20 at 14:24
SongSong
16.5k1741
16.5k1741
add a comment |
add a comment |
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