Convergence of $sigma-$algebra for converging stopping time
$begingroup$
Given a filtration, ${mathcal{F}_t},tin[0,infty).$ Let $T_n$ be a sequence of stopping time that converges to $T$ and $T_nle T_{n+1}.$ We have correpsonding $sigma-$algebra, ${mathcal{F}_{T_n}}$ and $mathcal{F}_T.$
Now, denote $mathcal{F}'=sigma(mathcal{F}_{T_n}:n=1,2,cdots),$ i.e., the $sigma-$algebra generated by all $mathcal{F}_{T_n}$.
Q: Will $mathcal{F}'=mathcal{F}_T$ holds?
I believe the condition that the filtration is left continuous is needed, since one can take $T_n,T$ to be constant. Let's assume that.
My try:
- That $mathcal{F}'subsetmathcal{F}_T$ is trivial, since all $mathcal{F}_{T_n}subsetmathcal{F}_T$ by $T_nle T$. We left to show $mathcal{F}_Tsubsetmathcal{F}'.$
- By definition, $Ainmathcal{F}_T$ is equivalent to $Acap{Tle t}inmathcal{F}_t$ for any $t$. How can one deduce from here that $Ain mathcal{F}'.$ Got stuck here.
- In Approximation of a unbounded stopping time and convergence of respective $sigma$-algebras, saz gives an approach for discrete time. For $Ainmathcal{F}_T, A$ can be decomposed as,
$$A = cup_{n=1}^{infty} (Acap{Tle n})cup(Acap{T=infty}) = cup_{n=1}^{infty} A_ncup A_infty.$$
Then show $A_ninmathcal{F}_{T_n}.$ We have $A_ninmathcal{F}_n, $ we require to show $A_ncap{T_nle t}inmathcal{F_t},$ this is so if $tge n$ since ${T_nle t}inmathcal{F}_t$. But how about $t<n?$ I was stuck here.
Update:
One can define, $mathcal{F}_{S-}=$ the $sigma-$algebra generated by $mathcal{F}_{0+}=cap_{s>0}mathcal{F}_s$ and the sets ${S>t}capmathcal{F_t}.$ then when $S=s$ is a constant, $mathcal{F}_{S-}=sigma(mathcal{F}_u:u<s)=mathcal{F}_{s-}.$ So this is simply the generalization of left limit. Then one can prove that,
$$mathcal{F}_{T-}=mathcal{F}'.$$
So now, the question may become to show,
$$ mathcal{F}_{T-}=mathcal{F}_{T} .$$
Any hint is appreciated!
probability-theory measure-theory stopping-times
asked Jan 20 at 9:15
user103567user103567
1136
$endgroup$
add a comment |
$begingroup$
Given a filtration, ${mathcal{F}_t},tin[0,infty).$ Let $T_n$ be a sequence of stopping time that converges to $T$ and $T_nle T_{n+1}.$ We have correpsonding $sigma-$algebra, ${mathcal{F}_{T_n}}$ and $mathcal{F}_T.$
Now, denote $mathcal{F}'=sigma(mathcal{F}_{T_n}:n=1,2,cdots),$ i.e., the $sigma-$algebra generated by all $mathcal{F}_{T_n}$.
Q: Will $mathcal{F}'=mathcal{F}_T$ holds?
I believe the condition that the filtration is left continuous is needed, since one can take $T_n,T$ to be constant. Let's assume that.
My try:
- That $mathcal{F}'subsetmathcal{F}_T$ is trivial, since all $mathcal{F}_{T_n}subsetmathcal{F}_T$ by $T_nle T$. We left to show $mathcal{F}_Tsubsetmathcal{F}'.$
- By definition, $Ainmathcal{F}_T$ is equivalent to $Acap{Tle t}inmathcal{F}_t$ for any $t$. How can one deduce from here that $Ain mathcal{F}'.$ Got stuck here.
- In Approximation of a unbounded stopping time and convergence of respective $sigma$-algebras, saz gives an approach for discrete time. For $Ainmathcal{F}_T, A$ can be decomposed as,
$$A = cup_{n=1}^{infty} (Acap{Tle n})cup(Acap{T=infty}) = cup_{n=1}^{infty} A_ncup A_infty.$$
Then show $A_ninmathcal{F}_{T_n}.$ We have $A_ninmathcal{F}_n, $ we require to show $A_ncap{T_nle t}inmathcal{F_t},$ this is so if $tge n$ since ${T_nle t}inmathcal{F}_t$. But how about $t<n?$ I was stuck here.
Update:
One can define, $mathcal{F}_{S-}=$ the $sigma-$algebra generated by $mathcal{F}_{0+}=cap_{s>0}mathcal{F}_s$ and the sets ${S>t}capmathcal{F_t}.$ then when $S=s$ is a constant, $mathcal{F}_{S-}=sigma(mathcal{F}_u:u<s)=mathcal{F}_{s-}.$ So this is simply the generalization of left limit. Then one can prove that,
$$mathcal{F}_{T-}=mathcal{F}'.$$
So now, the question may become to show,
$$ mathcal{F}_{T-}=mathcal{F}_{T} .$$
Any hint is appreciated!
probability-theory measure-theory stopping-times
asked Jan 20 at 9:15
user103567user103567
1136
$endgroup$
add a comment |
$begingroup$
Given a filtration, ${mathcal{F}_t},tin[0,infty).$ Let $T_n$ be a sequence of stopping time that converges to $T$ and $T_nle T_{n+1}.$ We have correpsonding $sigma-$algebra, ${mathcal{F}_{T_n}}$ and $mathcal{F}_T.$
Now, denote $mathcal{F}'=sigma(mathcal{F}_{T_n}:n=1,2,cdots),$ i.e., the $sigma-$algebra generated by all $mathcal{F}_{T_n}$.
Q: Will $mathcal{F}'=mathcal{F}_T$ holds?
I believe the condition that the filtration is left continuous is needed, since one can take $T_n,T$ to be constant. Let's assume that.
My try:
- That $mathcal{F}'subsetmathcal{F}_T$ is trivial, since all $mathcal{F}_{T_n}subsetmathcal{F}_T$ by $T_nle T$. We left to show $mathcal{F}_Tsubsetmathcal{F}'.$
- By definition, $Ainmathcal{F}_T$ is equivalent to $Acap{Tle t}inmathcal{F}_t$ for any $t$. How can one deduce from here that $Ain mathcal{F}'.$ Got stuck here.
- In Approximation of a unbounded stopping time and convergence of respective $sigma$-algebras, saz gives an approach for discrete time. For $Ainmathcal{F}_T, A$ can be decomposed as,
$$A = cup_{n=1}^{infty} (Acap{Tle n})cup(Acap{T=infty}) = cup_{n=1}^{infty} A_ncup A_infty.$$
Then show $A_ninmathcal{F}_{T_n}.$ We have $A_ninmathcal{F}_n, $ we require to show $A_ncap{T_nle t}inmathcal{F_t},$ this is so if $tge n$ since ${T_nle t}inmathcal{F}_t$. But how about $t<n?$ I was stuck here.
Update:
One can define, $mathcal{F}_{S-}=$ the $sigma-$algebra generated by $mathcal{F}_{0+}=cap_{s>0}mathcal{F}_s$ and the sets ${S>t}capmathcal{F_t}.$ then when $S=s$ is a constant, $mathcal{F}_{S-}=sigma(mathcal{F}_u:u<s)=mathcal{F}_{s-}.$ So this is simply the generalization of left limit. Then one can prove that,
$$mathcal{F}_{T-}=mathcal{F}'.$$
So now, the question may become to show,
$$ mathcal{F}_{T-}=mathcal{F}_{T} .$$
Any hint is appreciated!
probability-theory measure-theory stopping-times
asked Jan 20 at 9:15
user103567user103567
1136
$endgroup$
Given a filtration, ${mathcal{F}_t},tin[0,infty).$ Let $T_n$ be a sequence of stopping time that converges to $T$ and $T_nle T_{n+1}.$ We have correpsonding $sigma-$algebra, ${mathcal{F}_{T_n}}$ and $mathcal{F}_T.$
Now, denote $mathcal{F}'=sigma(mathcal{F}_{T_n}:n=1,2,cdots),$ i.e., the $sigma-$algebra generated by all $mathcal{F}_{T_n}$.
Q: Will $mathcal{F}'=mathcal{F}_T$ holds?
I believe the condition that the filtration is left continuous is needed, since one can take $T_n,T$ to be constant. Let's assume that.
My try:
- That $mathcal{F}'subsetmathcal{F}_T$ is trivial, since all $mathcal{F}_{T_n}subsetmathcal{F}_T$ by $T_nle T$. We left to show $mathcal{F}_Tsubsetmathcal{F}'.$
- By definition, $Ainmathcal{F}_T$ is equivalent to $Acap{Tle t}inmathcal{F}_t$ for any $t$. How can one deduce from here that $Ain mathcal{F}'.$ Got stuck here.
- In Approximation of a unbounded stopping time and convergence of respective $sigma$-algebras, saz gives an approach for discrete time. For $Ainmathcal{F}_T, A$ can be decomposed as,
$$A = cup_{n=1}^{infty} (Acap{Tle n})cup(Acap{T=infty}) = cup_{n=1}^{infty} A_ncup A_infty.$$
Then show $A_ninmathcal{F}_{T_n}.$ We have $A_ninmathcal{F}_n, $ we require to show $A_ncap{T_nle t}inmathcal{F_t},$ this is so if $tge n$ since ${T_nle t}inmathcal{F}_t$. But how about $t<n?$ I was stuck here.
Update:
One can define, $mathcal{F}_{S-}=$ the $sigma-$algebra generated by $mathcal{F}_{0+}=cap_{s>0}mathcal{F}_s$ and the sets ${S>t}capmathcal{F_t}.$ then when $S=s$ is a constant, $mathcal{F}_{S-}=sigma(mathcal{F}_u:u<s)=mathcal{F}_{s-}.$ So this is simply the generalization of left limit. Then one can prove that,
$$mathcal{F}_{T-}=mathcal{F}'.$$
So now, the question may become to show,
$$ mathcal{F}_{T-}=mathcal{F}_{T} .$$
Any hint is appreciated!
probability-theory measure-theory stopping-times
probability-theory measure-theory stopping-times
asked Jan 20 at 9:15
user103567user103567
1136
asked Jan 20 at 9:15
user103567user103567
1136
edited Jan 20 at 10:04
user103567
asked Jan 20 at 9:15
user103567user103567
1136
asked Jan 20 at 9:15
user103567user103567
1136
asked Jan 20 at 9:15
user103567user103567
1136
1136
add a comment |
add a comment |
1 Answer
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$begingroup$
The inclusion $mathcal F'subset mathcal F_T$ may be strict.
Example: Let $(Omega,mathcal F,Bbb P)$ be a probability space on which is defined a standard normal random variable $Z$. (So that $Bbb P[Z=0]=0$.) Let $(mathcal F_t)$ be the filtration generated by the process $X$ defined by
$$
X_t=cases{1,&$0le t<1$,cr Z,&$tge 1$.cr}
$$
For $T_n$ take the constant $1-1/n$, and then $T=1$. Yiu have $mathcal F_{T_n}=mathcal F' = {emptyset,Omega}$ for all $n$, but $mathcal F_T=sigma{Z}$.
The key here is that $T$ is predictable, and "something new" happens at time $T$.
answered Jan 20 at 17:28
John DawkinsJohn Dawkins
13.1k11017
$endgroup$
$begingroup$
I agree with you that the inclusion may be strict, but your example doesn't work because the OP is assuming that $mathcal{F}_t = mathcal{F}_{t-}$ for all $t$.
$endgroup$
– saz
Jan 20 at 17:56
$begingroup$
Missed that. To make an example with random $T$, start a Brownian motion at 0 and run it until the time $T$ that it first hits 1. At that time have the process jump to a random position independent of the past, with a standard normal law, and then stay there forever.
$endgroup$
– John Dawkins
Jan 21 at 14:57
add a comment |
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$begingroup$
The inclusion $mathcal F'subset mathcal F_T$ may be strict.
Example: Let $(Omega,mathcal F,Bbb P)$ be a probability space on which is defined a standard normal random variable $Z$. (So that $Bbb P[Z=0]=0$.) Let $(mathcal F_t)$ be the filtration generated by the process $X$ defined by
$$
X_t=cases{1,&$0le t<1$,cr Z,&$tge 1$.cr}
$$
For $T_n$ take the constant $1-1/n$, and then $T=1$. Yiu have $mathcal F_{T_n}=mathcal F' = {emptyset,Omega}$ for all $n$, but $mathcal F_T=sigma{Z}$.
The key here is that $T$ is predictable, and "something new" happens at time $T$.
answered Jan 20 at 17:28
John DawkinsJohn Dawkins
13.1k11017
$endgroup$
$begingroup$
I agree with you that the inclusion may be strict, but your example doesn't work because the OP is assuming that $mathcal{F}_t = mathcal{F}_{t-}$ for all $t$.
$endgroup$
– saz
Jan 20 at 17:56
$begingroup$
Missed that. To make an example with random $T$, start a Brownian motion at 0 and run it until the time $T$ that it first hits 1. At that time have the process jump to a random position independent of the past, with a standard normal law, and then stay there forever.
$endgroup$
– John Dawkins
Jan 21 at 14:57
add a comment |
$begingroup$
The inclusion $mathcal F'subset mathcal F_T$ may be strict.
Example: Let $(Omega,mathcal F,Bbb P)$ be a probability space on which is defined a standard normal random variable $Z$. (So that $Bbb P[Z=0]=0$.) Let $(mathcal F_t)$ be the filtration generated by the process $X$ defined by
$$
X_t=cases{1,&$0le t<1$,cr Z,&$tge 1$.cr}
$$
For $T_n$ take the constant $1-1/n$, and then $T=1$. Yiu have $mathcal F_{T_n}=mathcal F' = {emptyset,Omega}$ for all $n$, but $mathcal F_T=sigma{Z}$.
The key here is that $T$ is predictable, and "something new" happens at time $T$.
answered Jan 20 at 17:28
John DawkinsJohn Dawkins
13.1k11017
$endgroup$
$begingroup$
I agree with you that the inclusion may be strict, but your example doesn't work because the OP is assuming that $mathcal{F}_t = mathcal{F}_{t-}$ for all $t$.
$endgroup$
– saz
Jan 20 at 17:56
$begingroup$
Missed that. To make an example with random $T$, start a Brownian motion at 0 and run it until the time $T$ that it first hits 1. At that time have the process jump to a random position independent of the past, with a standard normal law, and then stay there forever.
$endgroup$
– John Dawkins
Jan 21 at 14:57
add a comment |
$begingroup$
The inclusion $mathcal F'subset mathcal F_T$ may be strict.
Example: Let $(Omega,mathcal F,Bbb P)$ be a probability space on which is defined a standard normal random variable $Z$. (So that $Bbb P[Z=0]=0$.) Let $(mathcal F_t)$ be the filtration generated by the process $X$ defined by
$$
X_t=cases{1,&$0le t<1$,cr Z,&$tge 1$.cr}
$$
For $T_n$ take the constant $1-1/n$, and then $T=1$. Yiu have $mathcal F_{T_n}=mathcal F' = {emptyset,Omega}$ for all $n$, but $mathcal F_T=sigma{Z}$.
The key here is that $T$ is predictable, and "something new" happens at time $T$.
answered Jan 20 at 17:28
John DawkinsJohn Dawkins
13.1k11017
$endgroup$
The inclusion $mathcal F'subset mathcal F_T$ may be strict.
Example: Let $(Omega,mathcal F,Bbb P)$ be a probability space on which is defined a standard normal random variable $Z$. (So that $Bbb P[Z=0]=0$.) Let $(mathcal F_t)$ be the filtration generated by the process $X$ defined by
$$
X_t=cases{1,&$0le t<1$,cr Z,&$tge 1$.cr}
$$
For $T_n$ take the constant $1-1/n$, and then $T=1$. Yiu have $mathcal F_{T_n}=mathcal F' = {emptyset,Omega}$ for all $n$, but $mathcal F_T=sigma{Z}$.
The key here is that $T$ is predictable, and "something new" happens at time $T$.
answered Jan 20 at 17:28
John DawkinsJohn Dawkins
13.1k11017
answered Jan 20 at 17:28
John DawkinsJohn Dawkins
13.1k11017
answered Jan 20 at 17:28
John DawkinsJohn Dawkins
13.1k11017
answered Jan 20 at 17:28
John DawkinsJohn Dawkins
13.1k11017
13.1k11017
$begingroup$
I agree with you that the inclusion may be strict, but your example doesn't work because the OP is assuming that $mathcal{F}_t = mathcal{F}_{t-}$ for all $t$.
$endgroup$
– saz
Jan 20 at 17:56
$begingroup$
Missed that. To make an example with random $T$, start a Brownian motion at 0 and run it until the time $T$ that it first hits 1. At that time have the process jump to a random position independent of the past, with a standard normal law, and then stay there forever.
$endgroup$
– John Dawkins
Jan 21 at 14:57
add a comment |
$begingroup$
I agree with you that the inclusion may be strict, but your example doesn't work because the OP is assuming that $mathcal{F}_t = mathcal{F}_{t-}$ for all $t$.
$endgroup$
– saz
Jan 20 at 17:56
$begingroup$
Missed that. To make an example with random $T$, start a Brownian motion at 0 and run it until the time $T$ that it first hits 1. At that time have the process jump to a random position independent of the past, with a standard normal law, and then stay there forever.
$endgroup$
– John Dawkins
Jan 21 at 14:57
$begingroup$
I agree with you that the inclusion may be strict, but your example doesn't work because the OP is assuming that $mathcal{F}_t = mathcal{F}_{t-}$ for all $t$.
$endgroup$
– saz
Jan 20 at 17:56
$begingroup$
I agree with you that the inclusion may be strict, but your example doesn't work because the OP is assuming that $mathcal{F}_t = mathcal{F}_{t-}$ for all $t$.
$endgroup$
– saz
Jan 20 at 17:56
$begingroup$
Missed that. To make an example with random $T$, start a Brownian motion at 0 and run it until the time $T$ that it first hits 1. At that time have the process jump to a random position independent of the past, with a standard normal law, and then stay there forever.
$endgroup$
– John Dawkins
Jan 21 at 14:57
$begingroup$
Missed that. To make an example with random $T$, start a Brownian motion at 0 and run it until the time $T$ that it first hits 1. At that time have the process jump to a random position independent of the past, with a standard normal law, and then stay there forever.
$endgroup$
– John Dawkins
Jan 21 at 14:57
add a comment |
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