What is the intuition behind right-continuous filtration?
I cannot understand the concept of it.
So a filtration is right continuous if for every $t$ it holds that:
$mathcal{F_t}=bigcaplimits_{varepsilon>0}mathcal{F_{t+varepsilon}}$
But if for every $t$, then it also holds for $t=0$. And if I choose a large $epsilon$, then it means that at time zero I know every information about the process?
probability stochastic-processes filtrations
asked Jul 14 '16 at 14:50
FelB
213
add a comment |
I cannot understand the concept of it.
So a filtration is right continuous if for every $t$ it holds that:
$mathcal{F_t}=bigcaplimits_{varepsilon>0}mathcal{F_{t+varepsilon}}$
But if for every $t$, then it also holds for $t=0$. And if I choose a large $epsilon$, then it means that at time zero I know every information about the process?
probability stochastic-processes filtrations
asked Jul 14 '16 at 14:50
FelB
213
3
No, because you're taking the intersection over $varepsilon>0$. (To avoid the problem of uncountable intersections we can assume WLOG that $varepsilon$ is rational.)
– Math1000
Jul 14 '16 at 14:56
@Math1000 +1 for the comment, but I don't see any "problem" with uncountable intersections.
– user940
Jul 14 '16 at 15:55
@Math1000 Okay but what is the intuition behind it? why is it useful?
– FelB
Jul 14 '16 at 17:20
@ByronSchmuland Good point, as we are taking the intersection of $sigma$-algebras as opposed to the intersections of elements of $sigma$-algebras...
– Math1000
Jul 14 '16 at 20:45
add a comment |
I cannot understand the concept of it.
So a filtration is right continuous if for every $t$ it holds that:
$mathcal{F_t}=bigcaplimits_{varepsilon>0}mathcal{F_{t+varepsilon}}$
But if for every $t$, then it also holds for $t=0$. And if I choose a large $epsilon$, then it means that at time zero I know every information about the process?
probability stochastic-processes filtrations
asked Jul 14 '16 at 14:50
FelB
213
I cannot understand the concept of it.
So a filtration is right continuous if for every $t$ it holds that:
$mathcal{F_t}=bigcaplimits_{varepsilon>0}mathcal{F_{t+varepsilon}}$
But if for every $t$, then it also holds for $t=0$. And if I choose a large $epsilon$, then it means that at time zero I know every information about the process?
probability stochastic-processes filtrations
probability stochastic-processes filtrations
asked Jul 14 '16 at 14:50
FelB
213
asked Jul 14 '16 at 14:50
FelB
213
edited Jul 14 '16 at 14:54
user940
asked Jul 14 '16 at 14:50
FelB
213
asked Jul 14 '16 at 14:50
FelB
213
asked Jul 14 '16 at 14:50
FelB
213
213
3
No, because you're taking the intersection over $varepsilon>0$. (To avoid the problem of uncountable intersections we can assume WLOG that $varepsilon$ is rational.)
– Math1000
Jul 14 '16 at 14:56
@Math1000 +1 for the comment, but I don't see any "problem" with uncountable intersections.
– user940
Jul 14 '16 at 15:55
@Math1000 Okay but what is the intuition behind it? why is it useful?
– FelB
Jul 14 '16 at 17:20
@ByronSchmuland Good point, as we are taking the intersection of $sigma$-algebras as opposed to the intersections of elements of $sigma$-algebras...
– Math1000
Jul 14 '16 at 20:45
add a comment |
3
No, because you're taking the intersection over $varepsilon>0$. (To avoid the problem of uncountable intersections we can assume WLOG that $varepsilon$ is rational.)
– Math1000
Jul 14 '16 at 14:56
@Math1000 +1 for the comment, but I don't see any "problem" with uncountable intersections.
– user940
Jul 14 '16 at 15:55
@Math1000 Okay but what is the intuition behind it? why is it useful?
– FelB
Jul 14 '16 at 17:20
@ByronSchmuland Good point, as we are taking the intersection of $sigma$-algebras as opposed to the intersections of elements of $sigma$-algebras...
– Math1000
Jul 14 '16 at 20:45
3
3
No, because you're taking the intersection over $varepsilon>0$. (To avoid the problem of uncountable intersections we can assume WLOG that $varepsilon$ is rational.)
– Math1000
Jul 14 '16 at 14:56
No, because you're taking the intersection over $varepsilon>0$. (To avoid the problem of uncountable intersections we can assume WLOG that $varepsilon$ is rational.)
– Math1000
Jul 14 '16 at 14:56
@Math1000 +1 for the comment, but I don't see any "problem" with uncountable intersections.
– user940
Jul 14 '16 at 15:55
@Math1000 +1 for the comment, but I don't see any "problem" with uncountable intersections.
– user940
Jul 14 '16 at 15:55
@Math1000 Okay but what is the intuition behind it? why is it useful?
– FelB
Jul 14 '16 at 17:20
@Math1000 Okay but what is the intuition behind it? why is it useful?
– FelB
Jul 14 '16 at 17:20
@ByronSchmuland Good point, as we are taking the intersection of $sigma$-algebras as opposed to the intersections of elements of $sigma$-algebras...
– Math1000
Jul 14 '16 at 20:45
@ByronSchmuland Good point, as we are taking the intersection of $sigma$-algebras as opposed to the intersections of elements of $sigma$-algebras...
– Math1000
Jul 14 '16 at 20:45
add a comment |
1 Answer
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The idea is that you gain no additional information by taking an infinitesimal step forward in time.
Remember that an $mathit{intersection}$ means that we are taking only the elements contained in EVERY set in the intersection. So, if we think of each $F_t$ as the information contained in the system up to time $t$, the intersection $cap_{epsilon > 0} mathcal{F}_{t+epsilon}$ contains only the information in EVERY $mathcal{F}_{t+epsilon}$ for every possible value of $epsilon > 0$. That is, only the information contained up until $t+epsilon$ for every $epsilon > 0$, in particular, any arbitrarily small $epsilon$. So, in this intersection, we have added only the information gained by taking an infinitesimally small step forward in time.
Thus, the idea of right continuity, $mathcal{F}_t=cap_{epsilon > 0} mathcal{F}_{t+epsilon}$ is that no information is added in this infinitesimal step. In other words, there are no instantaneous developments of the system, it evolves in a continuous fashion going forward in time.
(Much credit for this answer is due to Huyen Pham, whose book I'm currently using to review some of this material.)
answered Sep 18 '16 at 23:46
Charles Beer
414
add a comment |
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The idea is that you gain no additional information by taking an infinitesimal step forward in time.
Remember that an $mathit{intersection}$ means that we are taking only the elements contained in EVERY set in the intersection. So, if we think of each $F_t$ as the information contained in the system up to time $t$, the intersection $cap_{epsilon > 0} mathcal{F}_{t+epsilon}$ contains only the information in EVERY $mathcal{F}_{t+epsilon}$ for every possible value of $epsilon > 0$. That is, only the information contained up until $t+epsilon$ for every $epsilon > 0$, in particular, any arbitrarily small $epsilon$. So, in this intersection, we have added only the information gained by taking an infinitesimally small step forward in time.
Thus, the idea of right continuity, $mathcal{F}_t=cap_{epsilon > 0} mathcal{F}_{t+epsilon}$ is that no information is added in this infinitesimal step. In other words, there are no instantaneous developments of the system, it evolves in a continuous fashion going forward in time.
(Much credit for this answer is due to Huyen Pham, whose book I'm currently using to review some of this material.)
answered Sep 18 '16 at 23:46
Charles Beer
414
add a comment |
The idea is that you gain no additional information by taking an infinitesimal step forward in time.
Remember that an $mathit{intersection}$ means that we are taking only the elements contained in EVERY set in the intersection. So, if we think of each $F_t$ as the information contained in the system up to time $t$, the intersection $cap_{epsilon > 0} mathcal{F}_{t+epsilon}$ contains only the information in EVERY $mathcal{F}_{t+epsilon}$ for every possible value of $epsilon > 0$. That is, only the information contained up until $t+epsilon$ for every $epsilon > 0$, in particular, any arbitrarily small $epsilon$. So, in this intersection, we have added only the information gained by taking an infinitesimally small step forward in time.
Thus, the idea of right continuity, $mathcal{F}_t=cap_{epsilon > 0} mathcal{F}_{t+epsilon}$ is that no information is added in this infinitesimal step. In other words, there are no instantaneous developments of the system, it evolves in a continuous fashion going forward in time.
(Much credit for this answer is due to Huyen Pham, whose book I'm currently using to review some of this material.)
answered Sep 18 '16 at 23:46
Charles Beer
414
add a comment |
The idea is that you gain no additional information by taking an infinitesimal step forward in time.
Remember that an $mathit{intersection}$ means that we are taking only the elements contained in EVERY set in the intersection. So, if we think of each $F_t$ as the information contained in the system up to time $t$, the intersection $cap_{epsilon > 0} mathcal{F}_{t+epsilon}$ contains only the information in EVERY $mathcal{F}_{t+epsilon}$ for every possible value of $epsilon > 0$. That is, only the information contained up until $t+epsilon$ for every $epsilon > 0$, in particular, any arbitrarily small $epsilon$. So, in this intersection, we have added only the information gained by taking an infinitesimally small step forward in time.
Thus, the idea of right continuity, $mathcal{F}_t=cap_{epsilon > 0} mathcal{F}_{t+epsilon}$ is that no information is added in this infinitesimal step. In other words, there are no instantaneous developments of the system, it evolves in a continuous fashion going forward in time.
(Much credit for this answer is due to Huyen Pham, whose book I'm currently using to review some of this material.)
answered Sep 18 '16 at 23:46
Charles Beer
414
The idea is that you gain no additional information by taking an infinitesimal step forward in time.
Remember that an $mathit{intersection}$ means that we are taking only the elements contained in EVERY set in the intersection. So, if we think of each $F_t$ as the information contained in the system up to time $t$, the intersection $cap_{epsilon > 0} mathcal{F}_{t+epsilon}$ contains only the information in EVERY $mathcal{F}_{t+epsilon}$ for every possible value of $epsilon > 0$. That is, only the information contained up until $t+epsilon$ for every $epsilon > 0$, in particular, any arbitrarily small $epsilon$. So, in this intersection, we have added only the information gained by taking an infinitesimally small step forward in time.
Thus, the idea of right continuity, $mathcal{F}_t=cap_{epsilon > 0} mathcal{F}_{t+epsilon}$ is that no information is added in this infinitesimal step. In other words, there are no instantaneous developments of the system, it evolves in a continuous fashion going forward in time.
(Much credit for this answer is due to Huyen Pham, whose book I'm currently using to review some of this material.)
answered Sep 18 '16 at 23:46
Charles Beer
414
answered Sep 18 '16 at 23:46
Charles Beer
414
answered Sep 18 '16 at 23:46
Charles Beer
414
answered Sep 18 '16 at 23:46
Charles Beer
414
414
add a comment |
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No, because you're taking the intersection over $varepsilon>0$. (To avoid the problem of uncountable intersections we can assume WLOG that $varepsilon$ is rational.)
– Math1000
Jul 14 '16 at 14:56
@Math1000 +1 for the comment, but I don't see any "problem" with uncountable intersections.
– user940
Jul 14 '16 at 15:55
@Math1000 Okay but what is the intuition behind it? why is it useful?
– FelB
Jul 14 '16 at 17:20
@ByronSchmuland Good point, as we are taking the intersection of $sigma$-algebras as opposed to the intersections of elements of $sigma$-algebras...
– Math1000
Jul 14 '16 at 20:45