Filtration in Markov Chains and stopping times The Next CEO of Stack OverflowStopping times of Markov chainsMarkov chains and natural filtrationTime homogeneous Markov chain with random timesA question about the “stopping time with respect to a set” of a Markov chainContinuous time Markov Chain's Natural FiltrationGiven a list L of N elements uniformly sampled from a set A, what is the probability that L contains every element of A?Interchanging limit and expectation for irreducible Markov chainsIrreducible Markov Chain has finite stopping-time to a finite setDetermining whether the following are Markov ChainsStrong Markov property and time-homogeneity

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Filtration in Markov Chains and stopping times



The Next CEO of Stack OverflowStopping times of Markov chainsMarkov chains and natural filtrationTime homogeneous Markov chain with random timesA question about the “stopping time with respect to a set” of a Markov chainContinuous time Markov Chain's Natural FiltrationGiven a list L of N elements uniformly sampled from a set A, what is the probability that L contains every element of A?Interchanging limit and expectation for irreducible Markov chainsIrreducible Markov Chain has finite stopping-time to a finite setDetermining whether the following are Markov ChainsStrong Markov property and time-homogeneity










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$begingroup$


I am trying to get a better understand of filtrations, and I can't seem to find any simple, concrete examples, so I will try to make one here.



Consider a discrete-time Markov chain with states $A,B$ that we run for 3 time steps. Is the following a correct representation of the corresponding filtration?



$$mathcalF_0 = emptyset, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB$$
$$mathcalF_1 = emptyset, AAA,AAB,ABA,ABB, BAA,BAB,BBA,BBB, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB$$
$$mathcalF_2 = emptyset, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB,AAA,AAB,ABA,ABB, BAA,BAB,BBA,BBB, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB$$
$$mathcalF_3 = emptyset, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB,
AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB,AAA,AAB,ABA,ABB, BAA,BAB,BBA,BBB, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB$$



If this is correct, what are some "stopping times" in this example? Suppose $tau_A$ is the first time at which the chain is in state $A$. Then for $tau_A$ to be a stopping time we must have $tau_A=t in mathcalF_t$ for $t=0,1,2,3$? What is the event $tau_A=t$ in the above filtration?










share|cite|improve this question









$endgroup$
















    0












    $begingroup$


    I am trying to get a better understand of filtrations, and I can't seem to find any simple, concrete examples, so I will try to make one here.



    Consider a discrete-time Markov chain with states $A,B$ that we run for 3 time steps. Is the following a correct representation of the corresponding filtration?



    $$mathcalF_0 = emptyset, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB$$
    $$mathcalF_1 = emptyset, AAA,AAB,ABA,ABB, BAA,BAB,BBA,BBB, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB$$
    $$mathcalF_2 = emptyset, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB,AAA,AAB,ABA,ABB, BAA,BAB,BBA,BBB, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB$$
    $$mathcalF_3 = emptyset, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB,
    AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB,AAA,AAB,ABA,ABB, BAA,BAB,BBA,BBB, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB$$



    If this is correct, what are some "stopping times" in this example? Suppose $tau_A$ is the first time at which the chain is in state $A$. Then for $tau_A$ to be a stopping time we must have $tau_A=t in mathcalF_t$ for $t=0,1,2,3$? What is the event $tau_A=t$ in the above filtration?










    share|cite|improve this question









    $endgroup$














      0












      0








      0





      $begingroup$


      I am trying to get a better understand of filtrations, and I can't seem to find any simple, concrete examples, so I will try to make one here.



      Consider a discrete-time Markov chain with states $A,B$ that we run for 3 time steps. Is the following a correct representation of the corresponding filtration?



      $$mathcalF_0 = emptyset, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB$$
      $$mathcalF_1 = emptyset, AAA,AAB,ABA,ABB, BAA,BAB,BBA,BBB, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB$$
      $$mathcalF_2 = emptyset, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB,AAA,AAB,ABA,ABB, BAA,BAB,BBA,BBB, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB$$
      $$mathcalF_3 = emptyset, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB,
      AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB,AAA,AAB,ABA,ABB, BAA,BAB,BBA,BBB, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB$$



      If this is correct, what are some "stopping times" in this example? Suppose $tau_A$ is the first time at which the chain is in state $A$. Then for $tau_A$ to be a stopping time we must have $tau_A=t in mathcalF_t$ for $t=0,1,2,3$? What is the event $tau_A=t$ in the above filtration?










      share|cite|improve this question









      $endgroup$




      I am trying to get a better understand of filtrations, and I can't seem to find any simple, concrete examples, so I will try to make one here.



      Consider a discrete-time Markov chain with states $A,B$ that we run for 3 time steps. Is the following a correct representation of the corresponding filtration?



      $$mathcalF_0 = emptyset, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB$$
      $$mathcalF_1 = emptyset, AAA,AAB,ABA,ABB, BAA,BAB,BBA,BBB, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB$$
      $$mathcalF_2 = emptyset, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB,AAA,AAB,ABA,ABB, BAA,BAB,BBA,BBB, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB$$
      $$mathcalF_3 = emptyset, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB,
      AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB,AAA,AAB,ABA,ABB, BAA,BAB,BBA,BBB, AAA,AAB,ABA,ABB,BAA,BAB,BBA,BBB$$



      If this is correct, what are some "stopping times" in this example? Suppose $tau_A$ is the first time at which the chain is in state $A$. Then for $tau_A$ to be a stopping time we must have $tau_A=t in mathcalF_t$ for $t=0,1,2,3$? What is the event $tau_A=t$ in the above filtration?







      probability markov-chains filtrations






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      asked 2 days ago









      theQmantheQman

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      45038




















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          $begingroup$

          This is correct.



          The event $tau_A=1$ is exactly $AAA,AAB,ABA,ABBinmathcal F_1$. Indeed, this event means that the first step leads to $A$.



          The event $tau_A=2$ is exactly $BAA,BABinmathcal F_2$. This event occures when the chain is at $B$ after the first step, and then at $A$ after the second step.



          Try to find the event $tau_A=3$ in $mathcal F_3$.






          share|cite|improve this answer









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            $begingroup$

            This is correct.



            The event $tau_A=1$ is exactly $AAA,AAB,ABA,ABBinmathcal F_1$. Indeed, this event means that the first step leads to $A$.



            The event $tau_A=2$ is exactly $BAA,BABinmathcal F_2$. This event occures when the chain is at $B$ after the first step, and then at $A$ after the second step.



            Try to find the event $tau_A=3$ in $mathcal F_3$.






            share|cite|improve this answer









            $endgroup$

















              0












              $begingroup$

              This is correct.



              The event $tau_A=1$ is exactly $AAA,AAB,ABA,ABBinmathcal F_1$. Indeed, this event means that the first step leads to $A$.



              The event $tau_A=2$ is exactly $BAA,BABinmathcal F_2$. This event occures when the chain is at $B$ after the first step, and then at $A$ after the second step.



              Try to find the event $tau_A=3$ in $mathcal F_3$.






              share|cite|improve this answer









              $endgroup$















                0












                0








                0





                $begingroup$

                This is correct.



                The event $tau_A=1$ is exactly $AAA,AAB,ABA,ABBinmathcal F_1$. Indeed, this event means that the first step leads to $A$.



                The event $tau_A=2$ is exactly $BAA,BABinmathcal F_2$. This event occures when the chain is at $B$ after the first step, and then at $A$ after the second step.



                Try to find the event $tau_A=3$ in $mathcal F_3$.






                share|cite|improve this answer









                $endgroup$



                This is correct.



                The event $tau_A=1$ is exactly $AAA,AAB,ABA,ABBinmathcal F_1$. Indeed, this event means that the first step leads to $A$.



                The event $tau_A=2$ is exactly $BAA,BABinmathcal F_2$. This event occures when the chain is at $B$ after the first step, and then at $A$ after the second step.



                Try to find the event $tau_A=3$ in $mathcal F_3$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                NChNCh

                6,9453825




                6,9453825



























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