Measurable Version of a Banach Space Valued Process
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I'm trying to convince myself on the existence of a measurable version of a process, $X:[0,T]times Omegato E$ which is assumed to be stochastically continuous with $E$ a separable Banach space. Recall that measurability in this context means that it is measurable jointly in $(t,omega)$.
The text I'm following suggests first identifying a sequence of partitions of $[0,T]$ and then using these to construct a sequence of piecewise constant (in $t$) left continuous approximations, $X_m(t,omega)$. This makes sense and these are obviously measurable.
The next step is to look at the set on which these are convergent,
$$
A = {(t,omega): {X_m(t,omega)};text{is Cauchy}},
$$
and define
$$
Y(t,omega) = 1_{A}(t,omega)(lim_{mto infty} X_m(t,omega)).
$$
By Borel-Cantelli, you can then argue that $Y$ is a version of $X$.
However, there is one point that sticks out for me, that I do not know how to address, and the text I'm looking at is very vague about - the measurability of $A$, which is needed to conclude $Y$ is measurable. I'm hoping someone can either clarify why $A$ is obviously measurable, or point me to a reference that clarifies this issue.
measure-theory stochastic-processes
asked Dec 5 '18 at 14:08
user2379888user2379888
24019
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add a comment |
$begingroup$
I'm trying to convince myself on the existence of a measurable version of a process, $X:[0,T]times Omegato E$ which is assumed to be stochastically continuous with $E$ a separable Banach space. Recall that measurability in this context means that it is measurable jointly in $(t,omega)$.
The text I'm following suggests first identifying a sequence of partitions of $[0,T]$ and then using these to construct a sequence of piecewise constant (in $t$) left continuous approximations, $X_m(t,omega)$. This makes sense and these are obviously measurable.
The next step is to look at the set on which these are convergent,
$$
A = {(t,omega): {X_m(t,omega)};text{is Cauchy}},
$$
and define
$$
Y(t,omega) = 1_{A}(t,omega)(lim_{mto infty} X_m(t,omega)).
$$
By Borel-Cantelli, you can then argue that $Y$ is a version of $X$.
However, there is one point that sticks out for me, that I do not know how to address, and the text I'm looking at is very vague about - the measurability of $A$, which is needed to conclude $Y$ is measurable. I'm hoping someone can either clarify why $A$ is obviously measurable, or point me to a reference that clarifies this issue.
measure-theory stochastic-processes
asked Dec 5 '18 at 14:08
user2379888user2379888
24019
$endgroup$
add a comment |
$begingroup$
I'm trying to convince myself on the existence of a measurable version of a process, $X:[0,T]times Omegato E$ which is assumed to be stochastically continuous with $E$ a separable Banach space. Recall that measurability in this context means that it is measurable jointly in $(t,omega)$.
The text I'm following suggests first identifying a sequence of partitions of $[0,T]$ and then using these to construct a sequence of piecewise constant (in $t$) left continuous approximations, $X_m(t,omega)$. This makes sense and these are obviously measurable.
The next step is to look at the set on which these are convergent,
$$
A = {(t,omega): {X_m(t,omega)};text{is Cauchy}},
$$
and define
$$
Y(t,omega) = 1_{A}(t,omega)(lim_{mto infty} X_m(t,omega)).
$$
By Borel-Cantelli, you can then argue that $Y$ is a version of $X$.
However, there is one point that sticks out for me, that I do not know how to address, and the text I'm looking at is very vague about - the measurability of $A$, which is needed to conclude $Y$ is measurable. I'm hoping someone can either clarify why $A$ is obviously measurable, or point me to a reference that clarifies this issue.
measure-theory stochastic-processes
asked Dec 5 '18 at 14:08
user2379888user2379888
24019
$endgroup$
I'm trying to convince myself on the existence of a measurable version of a process, $X:[0,T]times Omegato E$ which is assumed to be stochastically continuous with $E$ a separable Banach space. Recall that measurability in this context means that it is measurable jointly in $(t,omega)$.
The text I'm following suggests first identifying a sequence of partitions of $[0,T]$ and then using these to construct a sequence of piecewise constant (in $t$) left continuous approximations, $X_m(t,omega)$. This makes sense and these are obviously measurable.
The next step is to look at the set on which these are convergent,
$$
A = {(t,omega): {X_m(t,omega)};text{is Cauchy}},
$$
and define
$$
Y(t,omega) = 1_{A}(t,omega)(lim_{mto infty} X_m(t,omega)).
$$
By Borel-Cantelli, you can then argue that $Y$ is a version of $X$.
However, there is one point that sticks out for me, that I do not know how to address, and the text I'm looking at is very vague about - the measurability of $A$, which is needed to conclude $Y$ is measurable. I'm hoping someone can either clarify why $A$ is obviously measurable, or point me to a reference that clarifies this issue.
measure-theory stochastic-processes
measure-theory stochastic-processes
asked Dec 5 '18 at 14:08
user2379888user2379888
24019
asked Dec 5 '18 at 14:08
user2379888user2379888
24019
asked Dec 5 '18 at 14:08
user2379888user2379888
24019
asked Dec 5 '18 at 14:08
user2379888user2379888
24019
asked Dec 5 '18 at 14:08
user2379888user2379888
24019
24019
add a comment |
add a comment |
1 Answer
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You can easily check that all $X_m$ are measurable according to the product-$sigma$-algebra $mathbb{B}([0,T]) otimes mathcal{F}.$ The set $A$ can be written as
$$A= bigcap_{k=1}^infty bigcup_{N=1}^infty bigcap_{n>m>N }{ |X_n -X_m| <1/k}$$
and is therefore measurable.
answered Dec 6 '18 at 7:44
p4schp4sch
4,995217
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can easily check that all $X_m$ are measurable according to the product-$sigma$-algebra $mathbb{B}([0,T]) otimes mathcal{F}.$ The set $A$ can be written as
$$A= bigcap_{k=1}^infty bigcup_{N=1}^infty bigcap_{n>m>N }{ |X_n -X_m| <1/k}$$
and is therefore measurable.
answered Dec 6 '18 at 7:44
p4schp4sch
4,995217
$endgroup$
add a comment |
$begingroup$
You can easily check that all $X_m$ are measurable according to the product-$sigma$-algebra $mathbb{B}([0,T]) otimes mathcal{F}.$ The set $A$ can be written as
$$A= bigcap_{k=1}^infty bigcup_{N=1}^infty bigcap_{n>m>N }{ |X_n -X_m| <1/k}$$
and is therefore measurable.
answered Dec 6 '18 at 7:44
p4schp4sch
4,995217
$endgroup$
add a comment |
$begingroup$
You can easily check that all $X_m$ are measurable according to the product-$sigma$-algebra $mathbb{B}([0,T]) otimes mathcal{F}.$ The set $A$ can be written as
$$A= bigcap_{k=1}^infty bigcup_{N=1}^infty bigcap_{n>m>N }{ |X_n -X_m| <1/k}$$
and is therefore measurable.
answered Dec 6 '18 at 7:44
p4schp4sch
4,995217
$endgroup$
You can easily check that all $X_m$ are measurable according to the product-$sigma$-algebra $mathbb{B}([0,T]) otimes mathcal{F}.$ The set $A$ can be written as
$$A= bigcap_{k=1}^infty bigcup_{N=1}^infty bigcap_{n>m>N }{ |X_n -X_m| <1/k}$$
and is therefore measurable.
answered Dec 6 '18 at 7:44
p4schp4sch
4,995217
answered Dec 6 '18 at 7:44
p4schp4sch
4,995217
answered Dec 6 '18 at 7:44
p4schp4sch
4,995217
answered Dec 6 '18 at 7:44
p4schp4sch
4,995217
4,995217
add a comment |
add a comment |
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