When ${X_t=Y_t, text{for all $t$}}$ is measurable or not? [closed]
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Let $(X_t)$ and $(Y_t)$ two stochastic process on $(Omega ,mathcal F,mathbb P)$. I heard that ${X_t=Y_ttext{ for all }t}$ or $sup_{tin mathbb R}X_t$ are not always measurable but if $X_t$ and $Y_t$ are both a.s. continuous, then there are measurable.
1) Why is it the case ?
2) Could someone give an example where there are not measurable ?
probability measure-theory
asked Dec 3 at 12:51
NewMath
936
closed as off-topic by Did, user302797, José Carlos Santos, Cesareo, Lee David Chung Lin Dec 4 at 12:57
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- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, user302797, José Carlos Santos, Cesareo, Lee David Chung Lin
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up vote
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Let $(X_t)$ and $(Y_t)$ two stochastic process on $(Omega ,mathcal F,mathbb P)$. I heard that ${X_t=Y_ttext{ for all }t}$ or $sup_{tin mathbb R}X_t$ are not always measurable but if $X_t$ and $Y_t$ are both a.s. continuous, then there are measurable.
1) Why is it the case ?
2) Could someone give an example where there are not measurable ?
probability measure-theory
asked Dec 3 at 12:51
NewMath
936
closed as off-topic by Did, user302797, José Carlos Santos, Cesareo, Lee David Chung Lin Dec 4 at 12:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, user302797, José Carlos Santos, Cesareo, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.
Are you sure that ${X_t=Y_ttext{ for all }t}$ measurable ? For $sup_{tinmathbb R}X_t$ it come from the fact that since $(X_t)$ is continuous, $sup_{tinmathbb R}X_t=sup_{qin mathbb Q}X_q$.
– Surb
Dec 3 at 12:58
@Surb: yes I am.
– NewMath
Dec 3 at 13:06
Yes I see, because by continuity ${X_t=Y_ttext{ for all $t$}}=bigcap_{qinmathbb Q}{X_q=Y_q}$.
– Surb
Dec 3 at 13:08
Could you explain why ? @Surb
– NewMath
Dec 3 at 13:08
Because $X_q=Y_q$ for all $qinmathbb Qiff X_t=Y_t$ for all $tin mathbb R$.
– Surb
Dec 3 at 13:09
|
show 2 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $(X_t)$ and $(Y_t)$ two stochastic process on $(Omega ,mathcal F,mathbb P)$. I heard that ${X_t=Y_ttext{ for all }t}$ or $sup_{tin mathbb R}X_t$ are not always measurable but if $X_t$ and $Y_t$ are both a.s. continuous, then there are measurable.
1) Why is it the case ?
2) Could someone give an example where there are not measurable ?
probability measure-theory
asked Dec 3 at 12:51
NewMath
936
Let $(X_t)$ and $(Y_t)$ two stochastic process on $(Omega ,mathcal F,mathbb P)$. I heard that ${X_t=Y_ttext{ for all }t}$ or $sup_{tin mathbb R}X_t$ are not always measurable but if $X_t$ and $Y_t$ are both a.s. continuous, then there are measurable.
1) Why is it the case ?
2) Could someone give an example where there are not measurable ?
probability measure-theory
probability measure-theory
asked Dec 3 at 12:51
NewMath
936
asked Dec 3 at 12:51
NewMath
936
asked Dec 3 at 12:51
NewMath
936
asked Dec 3 at 12:51
NewMath
936
asked Dec 3 at 12:51
NewMath
936
936
closed as off-topic by Did, user302797, José Carlos Santos, Cesareo, Lee David Chung Lin Dec 4 at 12:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, user302797, José Carlos Santos, Cesareo, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, user302797, José Carlos Santos, Cesareo, Lee David Chung Lin Dec 4 at 12:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, user302797, José Carlos Santos, Cesareo, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.
Are you sure that ${X_t=Y_ttext{ for all }t}$ measurable ? For $sup_{tinmathbb R}X_t$ it come from the fact that since $(X_t)$ is continuous, $sup_{tinmathbb R}X_t=sup_{qin mathbb Q}X_q$.
– Surb
Dec 3 at 12:58
@Surb: yes I am.
– NewMath
Dec 3 at 13:06
Yes I see, because by continuity ${X_t=Y_ttext{ for all $t$}}=bigcap_{qinmathbb Q}{X_q=Y_q}$.
– Surb
Dec 3 at 13:08
Could you explain why ? @Surb
– NewMath
Dec 3 at 13:08
Because $X_q=Y_q$ for all $qinmathbb Qiff X_t=Y_t$ for all $tin mathbb R$.
– Surb
Dec 3 at 13:09
|
show 2 more comments
Are you sure that ${X_t=Y_ttext{ for all }t}$ measurable ? For $sup_{tinmathbb R}X_t$ it come from the fact that since $(X_t)$ is continuous, $sup_{tinmathbb R}X_t=sup_{qin mathbb Q}X_q$.
– Surb
Dec 3 at 12:58
@Surb: yes I am.
– NewMath
Dec 3 at 13:06
Yes I see, because by continuity ${X_t=Y_ttext{ for all $t$}}=bigcap_{qinmathbb Q}{X_q=Y_q}$.
– Surb
Dec 3 at 13:08
Could you explain why ? @Surb
– NewMath
Dec 3 at 13:08
Because $X_q=Y_q$ for all $qinmathbb Qiff X_t=Y_t$ for all $tin mathbb R$.
– Surb
Dec 3 at 13:09
Are you sure that ${X_t=Y_ttext{ for all }t}$ measurable ? For $sup_{tinmathbb R}X_t$ it come from the fact that since $(X_t)$ is continuous, $sup_{tinmathbb R}X_t=sup_{qin mathbb Q}X_q$.
– Surb
Dec 3 at 12:58
Are you sure that ${X_t=Y_ttext{ for all }t}$ measurable ? For $sup_{tinmathbb R}X_t$ it come from the fact that since $(X_t)$ is continuous, $sup_{tinmathbb R}X_t=sup_{qin mathbb Q}X_q$.
– Surb
Dec 3 at 12:58
@Surb: yes I am.
– NewMath
Dec 3 at 13:06
@Surb: yes I am.
– NewMath
Dec 3 at 13:06
Yes I see, because by continuity ${X_t=Y_ttext{ for all $t$}}=bigcap_{qinmathbb Q}{X_q=Y_q}$.
– Surb
Dec 3 at 13:08
Yes I see, because by continuity ${X_t=Y_ttext{ for all $t$}}=bigcap_{qinmathbb Q}{X_q=Y_q}$.
– Surb
Dec 3 at 13:08
Could you explain why ? @Surb
– NewMath
Dec 3 at 13:08
Could you explain why ? @Surb
– NewMath
Dec 3 at 13:08
Because $X_q=Y_q$ for all $qinmathbb Qiff X_t=Y_t$ for all $tin mathbb R$.
– Surb
Dec 3 at 13:09
Because $X_q=Y_q$ for all $qinmathbb Qiff X_t=Y_t$ for all $tin mathbb R$.
– Surb
Dec 3 at 13:09
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Are you sure that ${X_t=Y_ttext{ for all }t}$ measurable ? For $sup_{tinmathbb R}X_t$ it come from the fact that since $(X_t)$ is continuous, $sup_{tinmathbb R}X_t=sup_{qin mathbb Q}X_q$.
– Surb
Dec 3 at 12:58
@Surb: yes I am.
– NewMath
Dec 3 at 13:06
Yes I see, because by continuity ${X_t=Y_ttext{ for all $t$}}=bigcap_{qinmathbb Q}{X_q=Y_q}$.
– Surb
Dec 3 at 13:08
Could you explain why ? @Surb
– NewMath
Dec 3 at 13:08
Because $X_q=Y_q$ for all $qinmathbb Qiff X_t=Y_t$ for all $tin mathbb R$.
– Surb
Dec 3 at 13:09