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










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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










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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










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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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      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 $A in mathcal F_s$.






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          • Thank you for the quick answer .
            – user3503589
            Sep 24 at 12:59











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          1 Answer
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          active

          oldest

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          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 $A in mathcal F_s$.






          share|cite|improve this answer





















          • Thank you for the quick answer .
            – user3503589
            Sep 24 at 12:59















          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 $A in mathcal F_s$.






          share|cite|improve this answer





















          • Thank you for the quick answer .
            – user3503589
            Sep 24 at 12:59













          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 $A in mathcal F_s$.






          share|cite|improve this answer












          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 $A in mathcal F_s$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • 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


















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