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Terminology Filtered probability space

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0

$begingroup$

What is the difference between a "filtered probability space" and a "Stochastic base"? I see both terms used but it is not clear to be me if there is a difference. Maybe one is complete but the other need not be?

probability probability-theory stochastic-processes soft-question terminology

share|cite|improve this question

edited Mar 12 at 11:01

Bernard

123k741116

asked Mar 12 at 10:06

AIM_BLBAIM_BLB

2,5172819

$endgroup$

add a comment | 

0

$begingroup$

What is the difference between a "filtered probability space" and a "Stochastic base"? I see both terms used but it is not clear to be me if there is a difference. Maybe one is complete but the other need not be?

probability probability-theory stochastic-processes soft-question terminology

share|cite|improve this question

edited Mar 12 at 11:01

Bernard

123k741116

asked Mar 12 at 10:06

AIM_BLBAIM_BLB

2,5172819

$endgroup$

add a comment | 

$begingroup$

What is the difference between a "filtered probability space" and a "Stochastic base"? I see both terms used but it is not clear to be me if there is a difference. Maybe one is complete but the other need not be?

probability probability-theory stochastic-processes soft-question terminology

share|cite|improve this question

edited Mar 12 at 11:01

Bernard

123k741116

asked Mar 12 at 10:06

AIM_BLBAIM_BLB

2,5172819

$endgroup$

share|cite|improve this question

edited Mar 12 at 11:01

Bernard

123k741116

asked Mar 12 at 10:06

AIM_BLBAIM_BLB

2,5172819

share|cite|improve this question

edited Mar 12 at 11:01

Bernard

123k741116

asked Mar 12 at 10:06

AIM_BLBAIM_BLB

2,5172819

edited Mar 12 at 11:01

Bernard

123k741116

edited Mar 12 at 11:01

Bernard

123k741116

asked Mar 12 at 10:06

AIM_BLBAIM_BLB

2,5172819

asked Mar 12 at 10:06

AIM_BLBAIM_BLB

2,5172819

1 Answer
1

active

oldest

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2

$begingroup$

A filtered probability space is a probability space $(Omega,mathcal F,mathbb P)$ equipped with a filtration $(mathcal F_t)_tinmathbb R_+$.

Some authors define a stochastic basis as a filtered probability space whose filtration is right-continuous, that is
$$
forall tinmathbb R_+,quad F_t=cap_s>tmathcal F_s.
$$
(see for instance Jacod-Shiryaev:Limit theorems for stochastic processes, for a definition)

Other authors might define a stochastic basis as a filtered probability space whose filtration is right-continuous and complete, that is for all $tinmathbb R_+$, $F_t=cap_s>tmathcal F_s$ and $mathcal F_t$ contains all the $mathbb P$-null sets of $mathcal F$.

As far as I am concerned, I never use the term "stochastic basis". I rather say "probability space equipped with a right-continuous filtration" for instance. It's longer, yes, but at least it is unequivocal.

share|cite|improve this answer

edited Mar 12 at 13:05

answered Mar 12 at 11:25

WillWill

1663

New contributor

Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Awesome! I always wondered what it meant. Thanks :).
  $endgroup$
  – AIM_BLB
  Mar 12 at 11:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure that there is such a strong consensus about the use of the term stochastic basis? You're right that Jacod and Shiryaev define it like that - but one should note that in their terminology a filtration is right-continuous by definition. I've also seen people use the term stochastic basis for filtered probability spaces with a right-continuous and complete filtration. My advice would be to always check the individual terminology of the pertaining author. Besides, I whole-heartedly agree with your suggestion not to use the term at all and instead be explicit and precise.
  $endgroup$
  – Mars Plastic
  Mar 12 at 12:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You're right. People I discuss with mostly use the term "stochastic basis" for right-continuous filtration and I have already heard "complete stochastic basis" when they want the filtration to be complete. But indeed it does not seem to be a strong consensus. I'll edit my answer to highlight this ambiguity.
  $endgroup$
  – Will
  Mar 12 at 13:01
  

  
  
  
  

add a comment | 

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2

$begingroup$

A filtered probability space is a probability space $(Omega,mathcal F,mathbb P)$ equipped with a filtration $(mathcal F_t)_tinmathbb R_+$.

Some authors define a stochastic basis as a filtered probability space whose filtration is right-continuous, that is
$$
forall tinmathbb R_+,quad F_t=cap_s>tmathcal F_s.
$$
(see for instance Jacod-Shiryaev:Limit theorems for stochastic processes, for a definition)

Other authors might define a stochastic basis as a filtered probability space whose filtration is right-continuous and complete, that is for all $tinmathbb R_+$, $F_t=cap_s>tmathcal F_s$ and $mathcal F_t$ contains all the $mathbb P$-null sets of $mathcal F$.

As far as I am concerned, I never use the term "stochastic basis". I rather say "probability space equipped with a right-continuous filtration" for instance. It's longer, yes, but at least it is unequivocal.

share|cite|improve this answer

edited Mar 12 at 13:05

answered Mar 12 at 11:25

WillWill

1663

New contributor

Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Awesome! I always wondered what it meant. Thanks :).
  $endgroup$
  – AIM_BLB
  Mar 12 at 11:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure that there is such a strong consensus about the use of the term stochastic basis? You're right that Jacod and Shiryaev define it like that - but one should note that in their terminology a filtration is right-continuous by definition. I've also seen people use the term stochastic basis for filtered probability spaces with a right-continuous and complete filtration. My advice would be to always check the individual terminology of the pertaining author. Besides, I whole-heartedly agree with your suggestion not to use the term at all and instead be explicit and precise.
  $endgroup$
  – Mars Plastic
  Mar 12 at 12:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You're right. People I discuss with mostly use the term "stochastic basis" for right-continuous filtration and I have already heard "complete stochastic basis" when they want the filtration to be complete. But indeed it does not seem to be a strong consensus. I'll edit my answer to highlight this ambiguity.
  $endgroup$
  – Will
  Mar 12 at 13:01
  

  
  
  
  

add a comment | 

2

$begingroup$

A filtered probability space is a probability space $(Omega,mathcal F,mathbb P)$ equipped with a filtration $(mathcal F_t)_tinmathbb R_+$.

Some authors define a stochastic basis as a filtered probability space whose filtration is right-continuous, that is
$$
forall tinmathbb R_+,quad F_t=cap_s>tmathcal F_s.
$$
(see for instance Jacod-Shiryaev:Limit theorems for stochastic processes, for a definition)

Other authors might define a stochastic basis as a filtered probability space whose filtration is right-continuous and complete, that is for all $tinmathbb R_+$, $F_t=cap_s>tmathcal F_s$ and $mathcal F_t$ contains all the $mathbb P$-null sets of $mathcal F$.

As far as I am concerned, I never use the term "stochastic basis". I rather say "probability space equipped with a right-continuous filtration" for instance. It's longer, yes, but at least it is unequivocal.

share|cite|improve this answer

edited Mar 12 at 13:05

answered Mar 12 at 11:25

WillWill

1663

New contributor

Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Awesome! I always wondered what it meant. Thanks :).
  $endgroup$
  – AIM_BLB
  Mar 12 at 11:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure that there is such a strong consensus about the use of the term stochastic basis? You're right that Jacod and Shiryaev define it like that - but one should note that in their terminology a filtration is right-continuous by definition. I've also seen people use the term stochastic basis for filtered probability spaces with a right-continuous and complete filtration. My advice would be to always check the individual terminology of the pertaining author. Besides, I whole-heartedly agree with your suggestion not to use the term at all and instead be explicit and precise.
  $endgroup$
  – Mars Plastic
  Mar 12 at 12:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You're right. People I discuss with mostly use the term "stochastic basis" for right-continuous filtration and I have already heard "complete stochastic basis" when they want the filtration to be complete. But indeed it does not seem to be a strong consensus. I'll edit my answer to highlight this ambiguity.
  $endgroup$
  – Will
  Mar 12 at 13:01
  

  
  
  
  

add a comment | 

$begingroup$

A filtered probability space is a probability space $(Omega,mathcal F,mathbb P)$ equipped with a filtration $(mathcal F_t)_tinmathbb R_+$.

Some authors define a stochastic basis as a filtered probability space whose filtration is right-continuous, that is
$$
forall tinmathbb R_+,quad F_t=cap_s>tmathcal F_s.
$$
(see for instance Jacod-Shiryaev:Limit theorems for stochastic processes, for a definition)

Other authors might define a stochastic basis as a filtered probability space whose filtration is right-continuous and complete, that is for all $tinmathbb R_+$, $F_t=cap_s>tmathcal F_s$ and $mathcal F_t$ contains all the $mathbb P$-null sets of $mathcal F$.

As far as I am concerned, I never use the term "stochastic basis". I rather say "probability space equipped with a right-continuous filtration" for instance. It's longer, yes, but at least it is unequivocal.

share|cite|improve this answer

edited Mar 12 at 13:05

answered Mar 12 at 11:25

WillWill

1663

New contributor

Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

edited Mar 12 at 13:05

answered Mar 12 at 11:25

WillWill

1663

New contributor

Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 12 at 11:25

WillWill

1663

New contributor

Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 12 at 11:25

WillWill

1663