Stopping time with finite expectation












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Given a filtered space $left(S, mathcal{Sigma}, {mathcal{Sigma_n: nge 0}}, mathbb{P}right)$, and a stopping time $tau: S to {0, 1,2, ldots}$ with respect to ${mathcal{Sigma_n: nge 0}}$, suppose that there exist $minmathbb{N}$ and $epsilon$ such that, for every $nge 0$:
$$mathbb{E}left[mathbb{I}_{tau le n+m} | mathcal{Sigma_n}right] > epsilon quad text{a.s.}.$$
Prove that $mathbb{E}[tau]<infty$.




I think one should use $E[tau] < infty Longleftrightarrow sumlimits_{n=0}^infty mathbb{P}(tau ge n)< infty.$
However, I cannot find a tight enough bound for $mathbb{P}(tau ge n)$. The best I can do is
$$mathbb{P}(tau > n+m) = 1- mathbb{P}(tau le n+m) = 1 - mathbb{E}left[mathbb{I}_{tau le n+m}right] = 1-mathbb{E}mathbb{E}left[mathbb{I}_{tau le n+m} | mathcal{Sigma_n}right] le 1- mathbb{E}[epsilon] =1 - epsilon.$$
A hint is provided:




For every $k = 1, 2, 3, ldots,$ $$mathbb{P}(tau > (k+1)m) = mathbb{P}(tau > (k+1)m quad text{and} quad tau > km).$$
Use induction to show that $mathbb{P}(tau > km) le (1-epsilon)^k.$







probability-theory measure-theory stopping-times







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    $begingroup$



    Given a filtered space $left(S, mathcal{Sigma}, {mathcal{Sigma_n: nge 0}}, mathbb{P}right)$, and a stopping time $tau: S to {0, 1,2, ldots}$ with respect to ${mathcal{Sigma_n: nge 0}}$, suppose that there exist $minmathbb{N}$ and $epsilon$ such that, for every $nge 0$:
    $$mathbb{E}left[mathbb{I}_{tau le n+m} | mathcal{Sigma_n}right] > epsilon quad text{a.s.}.$$
    Prove that $mathbb{E}[tau]<infty$.




    I think one should use $E[tau] < infty Longleftrightarrow sumlimits_{n=0}^infty mathbb{P}(tau ge n)< infty.$
    However, I cannot find a tight enough bound for $mathbb{P}(tau ge n)$. The best I can do is
    $$mathbb{P}(tau > n+m) = 1- mathbb{P}(tau le n+m) = 1 - mathbb{E}left[mathbb{I}_{tau le n+m}right] = 1-mathbb{E}mathbb{E}left[mathbb{I}_{tau le n+m} | mathcal{Sigma_n}right] le 1- mathbb{E}[epsilon] =1 - epsilon.$$
    A hint is provided:




    For every $k = 1, 2, 3, ldots,$ $$mathbb{P}(tau > (k+1)m) = mathbb{P}(tau > (k+1)m quad text{and} quad tau > km).$$
    Use induction to show that $mathbb{P}(tau > km) le (1-epsilon)^k.$







    probability-theory measure-theory stopping-times







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      $begingroup$



      Given a filtered space $left(S, mathcal{Sigma}, {mathcal{Sigma_n: nge 0}}, mathbb{P}right)$, and a stopping time $tau: S to {0, 1,2, ldots}$ with respect to ${mathcal{Sigma_n: nge 0}}$, suppose that there exist $minmathbb{N}$ and $epsilon$ such that, for every $nge 0$:
      $$mathbb{E}left[mathbb{I}_{tau le n+m} | mathcal{Sigma_n}right] > epsilon quad text{a.s.}.$$
      Prove that $mathbb{E}[tau]<infty$.




      I think one should use $E[tau] < infty Longleftrightarrow sumlimits_{n=0}^infty mathbb{P}(tau ge n)< infty.$
      However, I cannot find a tight enough bound for $mathbb{P}(tau ge n)$. The best I can do is
      $$mathbb{P}(tau > n+m) = 1- mathbb{P}(tau le n+m) = 1 - mathbb{E}left[mathbb{I}_{tau le n+m}right] = 1-mathbb{E}mathbb{E}left[mathbb{I}_{tau le n+m} | mathcal{Sigma_n}right] le 1- mathbb{E}[epsilon] =1 - epsilon.$$
      A hint is provided:




      For every $k = 1, 2, 3, ldots,$ $$mathbb{P}(tau > (k+1)m) = mathbb{P}(tau > (k+1)m quad text{and} quad tau > km).$$
      Use induction to show that $mathbb{P}(tau > km) le (1-epsilon)^k.$







      probability-theory measure-theory stopping-times







      share|cite|improve this question











      $endgroup$





      Given a filtered space $left(S, mathcal{Sigma}, {mathcal{Sigma_n: nge 0}}, mathbb{P}right)$, and a stopping time $tau: S to {0, 1,2, ldots}$ with respect to ${mathcal{Sigma_n: nge 0}}$, suppose that there exist $minmathbb{N}$ and $epsilon$ such that, for every $nge 0$:
      $$mathbb{E}left[mathbb{I}_{tau le n+m} | mathcal{Sigma_n}right] > epsilon quad text{a.s.}.$$
      Prove that $mathbb{E}[tau]<infty$.




      I think one should use $E[tau] < infty Longleftrightarrow sumlimits_{n=0}^infty mathbb{P}(tau ge n)< infty.$
      However, I cannot find a tight enough bound for $mathbb{P}(tau ge n)$. The best I can do is
      $$mathbb{P}(tau > n+m) = 1- mathbb{P}(tau le n+m) = 1 - mathbb{E}left[mathbb{I}_{tau le n+m}right] = 1-mathbb{E}mathbb{E}left[mathbb{I}_{tau le n+m} | mathcal{Sigma_n}right] le 1- mathbb{E}[epsilon] =1 - epsilon.$$
      A hint is provided:




      For every $k = 1, 2, 3, ldots,$ $$mathbb{P}(tau > (k+1)m) = mathbb{P}(tau > (k+1)m quad text{and} quad tau > km).$$
      Use induction to show that $mathbb{P}(tau > km) le (1-epsilon)^k.$







      probability-theory measure-theory stopping-times




      probability-theory measure-theory stopping-times






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      edited Nov 29 '18 at 15:05







      math_enthuthiast

















      asked Nov 28 '18 at 22:00









      math_enthuthiastmath_enthuthiast

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          $begingroup$

          Since ${tau>km} in mathcal{Sigma_{km}}:$
          $$mathbb{P}(tau le (k+1)m, tau >km) =mathbb{E}[I_{tau le (k+1)m, tau >km}]= mathbb{E}mathbb{E}[I_{tau le (k+1)m, tau >km}|mathcal{Sigma_{km}}] = mathbb{E}mathbb{E}[I_{tau le (k+1)m} I_{tau >km}|mathcal{Sigma_{km}}] = mathbb{E}left[I_{tau >km}mathbb{E}[I_{tau le (k+1)m} |mathcal{Sigma_{km}}]right] > mathbb{E}left[I_{tau >km}epsilonright] = mathbb{P}(tau>km)epsilon.$$
          On the other hand
          $$
          mathbb{P}(tau le (k+1)m, tau >km) = mathbb{P}(km < tau le (k+1)m ) = mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m),
          $$
          hence
          $$
          mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m) > mathbb{P}(tau>km)epsilon,
          $$
          or
          $$
          mathbb{P}(tau>(k+1)m) le (1-epsilon)mathbb{P}(tau>km).
          $$
          Induction gives the desired result.






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            $begingroup$

            Since ${tau>km} in mathcal{Sigma_{km}}:$
            $$mathbb{P}(tau le (k+1)m, tau >km) =mathbb{E}[I_{tau le (k+1)m, tau >km}]= mathbb{E}mathbb{E}[I_{tau le (k+1)m, tau >km}|mathcal{Sigma_{km}}] = mathbb{E}mathbb{E}[I_{tau le (k+1)m} I_{tau >km}|mathcal{Sigma_{km}}] = mathbb{E}left[I_{tau >km}mathbb{E}[I_{tau le (k+1)m} |mathcal{Sigma_{km}}]right] > mathbb{E}left[I_{tau >km}epsilonright] = mathbb{P}(tau>km)epsilon.$$
            On the other hand
            $$
            mathbb{P}(tau le (k+1)m, tau >km) = mathbb{P}(km < tau le (k+1)m ) = mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m),
            $$
            hence
            $$
            mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m) > mathbb{P}(tau>km)epsilon,
            $$
            or
            $$
            mathbb{P}(tau>(k+1)m) le (1-epsilon)mathbb{P}(tau>km).
            $$
            Induction gives the desired result.






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              $begingroup$

              Since ${tau>km} in mathcal{Sigma_{km}}:$
              $$mathbb{P}(tau le (k+1)m, tau >km) =mathbb{E}[I_{tau le (k+1)m, tau >km}]= mathbb{E}mathbb{E}[I_{tau le (k+1)m, tau >km}|mathcal{Sigma_{km}}] = mathbb{E}mathbb{E}[I_{tau le (k+1)m} I_{tau >km}|mathcal{Sigma_{km}}] = mathbb{E}left[I_{tau >km}mathbb{E}[I_{tau le (k+1)m} |mathcal{Sigma_{km}}]right] > mathbb{E}left[I_{tau >km}epsilonright] = mathbb{P}(tau>km)epsilon.$$
              On the other hand
              $$
              mathbb{P}(tau le (k+1)m, tau >km) = mathbb{P}(km < tau le (k+1)m ) = mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m),
              $$
              hence
              $$
              mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m) > mathbb{P}(tau>km)epsilon,
              $$
              or
              $$
              mathbb{P}(tau>(k+1)m) le (1-epsilon)mathbb{P}(tau>km).
              $$
              Induction gives the desired result.






              share|cite|improve this answer











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                $begingroup$

                Since ${tau>km} in mathcal{Sigma_{km}}:$
                $$mathbb{P}(tau le (k+1)m, tau >km) =mathbb{E}[I_{tau le (k+1)m, tau >km}]= mathbb{E}mathbb{E}[I_{tau le (k+1)m, tau >km}|mathcal{Sigma_{km}}] = mathbb{E}mathbb{E}[I_{tau le (k+1)m} I_{tau >km}|mathcal{Sigma_{km}}] = mathbb{E}left[I_{tau >km}mathbb{E}[I_{tau le (k+1)m} |mathcal{Sigma_{km}}]right] > mathbb{E}left[I_{tau >km}epsilonright] = mathbb{P}(tau>km)epsilon.$$
                On the other hand
                $$
                mathbb{P}(tau le (k+1)m, tau >km) = mathbb{P}(km < tau le (k+1)m ) = mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m),
                $$
                hence
                $$
                mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m) > mathbb{P}(tau>km)epsilon,
                $$
                or
                $$
                mathbb{P}(tau>(k+1)m) le (1-epsilon)mathbb{P}(tau>km).
                $$
                Induction gives the desired result.






                share|cite|improve this answer











                $endgroup$



                Since ${tau>km} in mathcal{Sigma_{km}}:$
                $$mathbb{P}(tau le (k+1)m, tau >km) =mathbb{E}[I_{tau le (k+1)m, tau >km}]= mathbb{E}mathbb{E}[I_{tau le (k+1)m, tau >km}|mathcal{Sigma_{km}}] = mathbb{E}mathbb{E}[I_{tau le (k+1)m} I_{tau >km}|mathcal{Sigma_{km}}] = mathbb{E}left[I_{tau >km}mathbb{E}[I_{tau le (k+1)m} |mathcal{Sigma_{km}}]right] > mathbb{E}left[I_{tau >km}epsilonright] = mathbb{P}(tau>km)epsilon.$$
                On the other hand
                $$
                mathbb{P}(tau le (k+1)m, tau >km) = mathbb{P}(km < tau le (k+1)m ) = mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m),
                $$
                hence
                $$
                mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m) > mathbb{P}(tau>km)epsilon,
                $$
                or
                $$
                mathbb{P}(tau>(k+1)m) le (1-epsilon)mathbb{P}(tau>km).
                $$
                Induction gives the desired result.







                share|cite|improve this answer














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                share|cite|improve this answer








                edited Nov 29 '18 at 15:05

























                answered Nov 29 '18 at 0:56









                math_enthuthiastmath_enthuthiast

                16910




                16910






























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