Definition of predictable process
$begingroup$
I am trying to understand the notion of predictable process. Let $(Ω,F_t,P)$ be a filtered measure space, satisfying the usual condition. Things starts with the predictable $sigma$-algebra ${mathcal P}$, which is generated by sets of the form $Atimes (a,b]$ with $Ain{mathcal F}_a$ and $Atimes {0}$ with $Ain{mathcal F}_0$.
My question: is it true that $Sin {mathcal P}$ if and only if $S$ is progressive and ${omega|(omega,t)in S}in{mathcal F}_{t−}$ for all $t$? In another word, is it true that $X$ is predictable if and only if $X$ is progressive and $X$ is adapted to the filtration ${mathcal F}_{t−}$?
The only if part is easy but I am not sure about the if part. I feel that $X$ being ${mathcal F}_{t−}$-measurable seems to be a more "reasonable" definition of "predictable", but maybe I am wrong.
measure-theory stochastic-processes stochastic-calculus
edited Dec 9 '18 at 7:13
saz
79.7k860124
asked Oct 29 '18 at 16:47
Yu DingYu Ding
3365
$endgroup$
add a comment |
$begingroup$
I am trying to understand the notion of predictable process. Let $(Ω,F_t,P)$ be a filtered measure space, satisfying the usual condition. Things starts with the predictable $sigma$-algebra ${mathcal P}$, which is generated by sets of the form $Atimes (a,b]$ with $Ain{mathcal F}_a$ and $Atimes {0}$ with $Ain{mathcal F}_0$.
My question: is it true that $Sin {mathcal P}$ if and only if $S$ is progressive and ${omega|(omega,t)in S}in{mathcal F}_{t−}$ for all $t$? In another word, is it true that $X$ is predictable if and only if $X$ is progressive and $X$ is adapted to the filtration ${mathcal F}_{t−}$?
The only if part is easy but I am not sure about the if part. I feel that $X$ being ${mathcal F}_{t−}$-measurable seems to be a more "reasonable" definition of "predictable", but maybe I am wrong.
measure-theory stochastic-processes stochastic-calculus
edited Dec 9 '18 at 7:13
saz
79.7k860124
asked Oct 29 '18 at 16:47
Yu DingYu Ding
3365
$endgroup$
add a comment |
$begingroup$
I am trying to understand the notion of predictable process. Let $(Ω,F_t,P)$ be a filtered measure space, satisfying the usual condition. Things starts with the predictable $sigma$-algebra ${mathcal P}$, which is generated by sets of the form $Atimes (a,b]$ with $Ain{mathcal F}_a$ and $Atimes {0}$ with $Ain{mathcal F}_0$.
My question: is it true that $Sin {mathcal P}$ if and only if $S$ is progressive and ${omega|(omega,t)in S}in{mathcal F}_{t−}$ for all $t$? In another word, is it true that $X$ is predictable if and only if $X$ is progressive and $X$ is adapted to the filtration ${mathcal F}_{t−}$?
The only if part is easy but I am not sure about the if part. I feel that $X$ being ${mathcal F}_{t−}$-measurable seems to be a more "reasonable" definition of "predictable", but maybe I am wrong.
measure-theory stochastic-processes stochastic-calculus
edited Dec 9 '18 at 7:13
saz
79.7k860124
asked Oct 29 '18 at 16:47
Yu DingYu Ding
3365
$endgroup$
I am trying to understand the notion of predictable process. Let $(Ω,F_t,P)$ be a filtered measure space, satisfying the usual condition. Things starts with the predictable $sigma$-algebra ${mathcal P}$, which is generated by sets of the form $Atimes (a,b]$ with $Ain{mathcal F}_a$ and $Atimes {0}$ with $Ain{mathcal F}_0$.
My question: is it true that $Sin {mathcal P}$ if and only if $S$ is progressive and ${omega|(omega,t)in S}in{mathcal F}_{t−}$ for all $t$? In another word, is it true that $X$ is predictable if and only if $X$ is progressive and $X$ is adapted to the filtration ${mathcal F}_{t−}$?
The only if part is easy but I am not sure about the if part. I feel that $X$ being ${mathcal F}_{t−}$-measurable seems to be a more "reasonable" definition of "predictable", but maybe I am wrong.
measure-theory stochastic-processes stochastic-calculus
measure-theory stochastic-processes stochastic-calculus
edited Dec 9 '18 at 7:13
saz
79.7k860124
asked Oct 29 '18 at 16:47
Yu DingYu Ding
3365
edited Dec 9 '18 at 7:13
saz
79.7k860124
asked Oct 29 '18 at 16:47
Yu DingYu Ding
3365
edited Dec 9 '18 at 7:13
saz
79.7k860124
edited Dec 9 '18 at 7:13
saz
79.7k860124
edited Dec 9 '18 at 7:13
saz
79.7k860124
79.7k860124
asked Oct 29 '18 at 16:47
Yu DingYu Ding
3365
asked Oct 29 '18 at 16:47
Yu DingYu Ding
3365
asked Oct 29 '18 at 16:47
Yu DingYu Ding
3365
3365
add a comment |
add a comment |
1 Answer
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$begingroup$
No, it's not true.
First of all, recall that any adapted càdlàg process $(X_t)_{t geq 0}$ is progressively measurable. This means that for any such process $(X_t)_{t geq 0}$ your assertion reads
$(X_t)_{t geq 0}$ is predictable $iff$ $X_{t}$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$.
Now consider for instance a Poisson process $(X_t)_{t geq 0}$, and let $(mathcal{F}_t)_{t geq 0}$ be its completed canonical filtration. By the very definition of $mathcal{F}_{t-}$, we know that $X_{t-} = lim_{s uparrow t} X_s$ is $mathcal{F}_{t-}$-measurable. Since $X_t = X_{t-}$ almost surely we find that $X_t$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$. However, $(X_t)_{t geq 0}$ is not predictable. Indeed: If $(X_t)_{t geq 0}$ was predictable, then
$$M_t := X_t -t mathbb{E}(X_1), qquad t geq 0,$$
would be a predictable martingale which would imply that $(M_t)_{t geq 0}$ has continuous sample paths (see e.g. here) which is clearly not true; hence $(X_t)_{t geq 0}$ is not predictable.
answered Dec 8 '18 at 19:44
sazsaz
79.7k860124
$endgroup$
add a comment |
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1 Answer
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$begingroup$
No, it's not true.
First of all, recall that any adapted càdlàg process $(X_t)_{t geq 0}$ is progressively measurable. This means that for any such process $(X_t)_{t geq 0}$ your assertion reads
$(X_t)_{t geq 0}$ is predictable $iff$ $X_{t}$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$.
Now consider for instance a Poisson process $(X_t)_{t geq 0}$, and let $(mathcal{F}_t)_{t geq 0}$ be its completed canonical filtration. By the very definition of $mathcal{F}_{t-}$, we know that $X_{t-} = lim_{s uparrow t} X_s$ is $mathcal{F}_{t-}$-measurable. Since $X_t = X_{t-}$ almost surely we find that $X_t$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$. However, $(X_t)_{t geq 0}$ is not predictable. Indeed: If $(X_t)_{t geq 0}$ was predictable, then
$$M_t := X_t -t mathbb{E}(X_1), qquad t geq 0,$$
would be a predictable martingale which would imply that $(M_t)_{t geq 0}$ has continuous sample paths (see e.g. here) which is clearly not true; hence $(X_t)_{t geq 0}$ is not predictable.
answered Dec 8 '18 at 19:44
sazsaz
79.7k860124
$endgroup$
add a comment |
$begingroup$
No, it's not true.
First of all, recall that any adapted càdlàg process $(X_t)_{t geq 0}$ is progressively measurable. This means that for any such process $(X_t)_{t geq 0}$ your assertion reads
$(X_t)_{t geq 0}$ is predictable $iff$ $X_{t}$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$.
Now consider for instance a Poisson process $(X_t)_{t geq 0}$, and let $(mathcal{F}_t)_{t geq 0}$ be its completed canonical filtration. By the very definition of $mathcal{F}_{t-}$, we know that $X_{t-} = lim_{s uparrow t} X_s$ is $mathcal{F}_{t-}$-measurable. Since $X_t = X_{t-}$ almost surely we find that $X_t$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$. However, $(X_t)_{t geq 0}$ is not predictable. Indeed: If $(X_t)_{t geq 0}$ was predictable, then
$$M_t := X_t -t mathbb{E}(X_1), qquad t geq 0,$$
would be a predictable martingale which would imply that $(M_t)_{t geq 0}$ has continuous sample paths (see e.g. here) which is clearly not true; hence $(X_t)_{t geq 0}$ is not predictable.
answered Dec 8 '18 at 19:44
sazsaz
79.7k860124
$endgroup$
add a comment |
$begingroup$
No, it's not true.
First of all, recall that any adapted càdlàg process $(X_t)_{t geq 0}$ is progressively measurable. This means that for any such process $(X_t)_{t geq 0}$ your assertion reads
$(X_t)_{t geq 0}$ is predictable $iff$ $X_{t}$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$.
Now consider for instance a Poisson process $(X_t)_{t geq 0}$, and let $(mathcal{F}_t)_{t geq 0}$ be its completed canonical filtration. By the very definition of $mathcal{F}_{t-}$, we know that $X_{t-} = lim_{s uparrow t} X_s$ is $mathcal{F}_{t-}$-measurable. Since $X_t = X_{t-}$ almost surely we find that $X_t$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$. However, $(X_t)_{t geq 0}$ is not predictable. Indeed: If $(X_t)_{t geq 0}$ was predictable, then
$$M_t := X_t -t mathbb{E}(X_1), qquad t geq 0,$$
would be a predictable martingale which would imply that $(M_t)_{t geq 0}$ has continuous sample paths (see e.g. here) which is clearly not true; hence $(X_t)_{t geq 0}$ is not predictable.
answered Dec 8 '18 at 19:44
sazsaz
79.7k860124
$endgroup$
No, it's not true.
First of all, recall that any adapted càdlàg process $(X_t)_{t geq 0}$ is progressively measurable. This means that for any such process $(X_t)_{t geq 0}$ your assertion reads
$(X_t)_{t geq 0}$ is predictable $iff$ $X_{t}$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$.
Now consider for instance a Poisson process $(X_t)_{t geq 0}$, and let $(mathcal{F}_t)_{t geq 0}$ be its completed canonical filtration. By the very definition of $mathcal{F}_{t-}$, we know that $X_{t-} = lim_{s uparrow t} X_s$ is $mathcal{F}_{t-}$-measurable. Since $X_t = X_{t-}$ almost surely we find that $X_t$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$. However, $(X_t)_{t geq 0}$ is not predictable. Indeed: If $(X_t)_{t geq 0}$ was predictable, then
$$M_t := X_t -t mathbb{E}(X_1), qquad t geq 0,$$
would be a predictable martingale which would imply that $(M_t)_{t geq 0}$ has continuous sample paths (see e.g. here) which is clearly not true; hence $(X_t)_{t geq 0}$ is not predictable.
answered Dec 8 '18 at 19:44
sazsaz
79.7k860124
edited Dec 9 '18 at 18:10
answered Dec 8 '18 at 19:44
sazsaz
79.7k860124
answered Dec 8 '18 at 19:44
sazsaz
79.7k860124
answered Dec 8 '18 at 19:44
sazsaz
79.7k860124
79.7k860124
add a comment |
add a comment |
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