Why $P^x left[X_{t+s} in Gamma mid mathcal{F}_sright]=P^x left[X_{t+s} in Gamma mid X_sright]$?
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If we can show that $P^x left[X_{t+s} in Gamma mid mathcal{F}_sright]$ has a $sigma(X_s)$-msb version then why is it true that $P^x left[X_{t+s} in Gamma mid mathcal{F}_sright]=P^x left[X_{t+s} in Gamma mid X_sright], text{ }P^xtext{-a.s}$?
Why does does establishing the measurability imply that $P^x left[X_{t+s} in Gamma mid mathcal{F}_sright]$ has to be of the form $P^xleft[X_{t+s} in Gamma mid X_sright]$.
I was reading proposition 2.5.13 in Karatzas and Shreve where they state this fact
probability-theory markov-process
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If we can show that $P^x left[X_{t+s} in Gamma mid mathcal{F}_sright]$ has a $sigma(X_s)$-msb version then why is it true that $P^x left[X_{t+s} in Gamma mid mathcal{F}_sright]=P^x left[X_{t+s} in Gamma mid X_sright], text{ }P^xtext{-a.s}$?
Why does does establishing the measurability imply that $P^x left[X_{t+s} in Gamma mid mathcal{F}_sright]$ has to be of the form $P^xleft[X_{t+s} in Gamma mid X_sright]$.
I was reading proposition 2.5.13 in Karatzas and Shreve where they state this fact
probability-theory markov-process
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If we can show that $P^x left[X_{t+s} in Gamma mid mathcal{F}_sright]$ has a $sigma(X_s)$-msb version then why is it true that $P^x left[X_{t+s} in Gamma mid mathcal{F}_sright]=P^x left[X_{t+s} in Gamma mid X_sright], text{ }P^xtext{-a.s}$?
Why does does establishing the measurability imply that $P^x left[X_{t+s} in Gamma mid mathcal{F}_sright]$ has to be of the form $P^xleft[X_{t+s} in Gamma mid X_sright]$.
I was reading proposition 2.5.13 in Karatzas and Shreve where they state this fact
probability-theory markov-process
If we can show that $P^x left[X_{t+s} in Gamma mid mathcal{F}_sright]$ has a $sigma(X_s)$-msb version then why is it true that $P^x left[X_{t+s} in Gamma mid mathcal{F}_sright]=P^x left[X_{t+s} in Gamma mid X_sright], text{ }P^xtext{-a.s}$?
Why does does establishing the measurability imply that $P^x left[X_{t+s} in Gamma mid mathcal{F}_sright]$ has to be of the form $P^xleft[X_{t+s} in Gamma mid X_sright]$.
I was reading proposition 2.5.13 in Karatzas and Shreve where they state this fact
probability-theory markov-process
probability-theory markov-process
edited Nov 23 at 15:53
asked Sep 24 at 10:12
user3503589
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1,1731721
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You have to assume that $X_t$ is $mathcal F_t$ measurable for each $t$. Just apply definition of conditional probability. Let $Y$ be a version of $P[X_{t+s} in Gamma |mathcal F_s]$ which is measurable w.r.t. $sigma (X_s)$. To show that $Y=P[X_{t+s} in Gamma | X_s]$ all that you have to show is $P[X_{t+s} in Gamma, A]=int_A Y , dP$ for all $A in sigma (X_s)$ which is immediate because $A in mathcal F_s$.
Thank you for the quick answer .
– user3503589
Sep 24 at 12:59
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have to assume that $X_t$ is $mathcal F_t$ measurable for each $t$. Just apply definition of conditional probability. Let $Y$ be a version of $P[X_{t+s} in Gamma |mathcal F_s]$ which is measurable w.r.t. $sigma (X_s)$. To show that $Y=P[X_{t+s} in Gamma | X_s]$ all that you have to show is $P[X_{t+s} in Gamma, A]=int_A Y , dP$ for all $A in sigma (X_s)$ which is immediate because $A in mathcal F_s$.
Thank you for the quick answer .
– user3503589
Sep 24 at 12:59
add a comment |
up vote
1
down vote
accepted
You have to assume that $X_t$ is $mathcal F_t$ measurable for each $t$. Just apply definition of conditional probability. Let $Y$ be a version of $P[X_{t+s} in Gamma |mathcal F_s]$ which is measurable w.r.t. $sigma (X_s)$. To show that $Y=P[X_{t+s} in Gamma | X_s]$ all that you have to show is $P[X_{t+s} in Gamma, A]=int_A Y , dP$ for all $A in sigma (X_s)$ which is immediate because $A in mathcal F_s$.
Thank you for the quick answer .
– user3503589
Sep 24 at 12:59
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have to assume that $X_t$ is $mathcal F_t$ measurable for each $t$. Just apply definition of conditional probability. Let $Y$ be a version of $P[X_{t+s} in Gamma |mathcal F_s]$ which is measurable w.r.t. $sigma (X_s)$. To show that $Y=P[X_{t+s} in Gamma | X_s]$ all that you have to show is $P[X_{t+s} in Gamma, A]=int_A Y , dP$ for all $A in sigma (X_s)$ which is immediate because $A in mathcal F_s$.
You have to assume that $X_t$ is $mathcal F_t$ measurable for each $t$. Just apply definition of conditional probability. Let $Y$ be a version of $P[X_{t+s} in Gamma |mathcal F_s]$ which is measurable w.r.t. $sigma (X_s)$. To show that $Y=P[X_{t+s} in Gamma | X_s]$ all that you have to show is $P[X_{t+s} in Gamma, A]=int_A Y , dP$ for all $A in sigma (X_s)$ which is immediate because $A in mathcal F_s$.
answered Sep 24 at 10:19
Kavi Rama Murthy
43.3k31751
43.3k31751
Thank you for the quick answer .
– user3503589
Sep 24 at 12:59
add a comment |
Thank you for the quick answer .
– user3503589
Sep 24 at 12:59
Thank you for the quick answer .
– user3503589
Sep 24 at 12:59
Thank you for the quick answer .
– user3503589
Sep 24 at 12:59
add a comment |
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