Progressive measurability implies adaptedness
$begingroup$
I've read that every progressively measurable process is also adapted, but I can't prove it using the definition of measurability.
Can anyone give me a proof of this result ?
stochastic-processes
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migrated from mathoverflow.net Dec 17 '18 at 3:57
This question came from our site for professional mathematicians.
add a comment |
$begingroup$
I've read that every progressively measurable process is also adapted, but I can't prove it using the definition of measurability.
Can anyone give me a proof of this result ?
stochastic-processes
$endgroup$
migrated from mathoverflow.net Dec 17 '18 at 3:57
This question came from our site for professional mathematicians.
2
$begingroup$
In short, for all $T$ fixed, you define $G( omega ) = (T, omega)$, then combine it with the progressive measurability. I am not sure if mathoverflow is the right place for this question.
$endgroup$
– Taro NGUYEN
Nov 10 '18 at 23:59
add a comment |
$begingroup$
I've read that every progressively measurable process is also adapted, but I can't prove it using the definition of measurability.
Can anyone give me a proof of this result ?
stochastic-processes
$endgroup$
I've read that every progressively measurable process is also adapted, but I can't prove it using the definition of measurability.
Can anyone give me a proof of this result ?
stochastic-processes
stochastic-processes
asked Nov 10 '18 at 21:46
Ali Jouahri
migrated from mathoverflow.net Dec 17 '18 at 3:57
This question came from our site for professional mathematicians.
migrated from mathoverflow.net Dec 17 '18 at 3:57
This question came from our site for professional mathematicians.
2
$begingroup$
In short, for all $T$ fixed, you define $G( omega ) = (T, omega)$, then combine it with the progressive measurability. I am not sure if mathoverflow is the right place for this question.
$endgroup$
– Taro NGUYEN
Nov 10 '18 at 23:59
add a comment |
2
$begingroup$
In short, for all $T$ fixed, you define $G( omega ) = (T, omega)$, then combine it with the progressive measurability. I am not sure if mathoverflow is the right place for this question.
$endgroup$
– Taro NGUYEN
Nov 10 '18 at 23:59
2
2
$begingroup$
In short, for all $T$ fixed, you define $G( omega ) = (T, omega)$, then combine it with the progressive measurability. I am not sure if mathoverflow is the right place for this question.
$endgroup$
– Taro NGUYEN
Nov 10 '18 at 23:59
$begingroup$
In short, for all $T$ fixed, you define $G( omega ) = (T, omega)$, then combine it with the progressive measurability. I am not sure if mathoverflow is the right place for this question.
$endgroup$
– Taro NGUYEN
Nov 10 '18 at 23:59
add a comment |
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$begingroup$
In short, for all $T$ fixed, you define $G( omega ) = (T, omega)$, then combine it with the progressive measurability. I am not sure if mathoverflow is the right place for this question.
$endgroup$
– Taro NGUYEN
Nov 10 '18 at 23:59