Two supermartingales and a stopping time












4












$begingroup$


I have the following task to do:



Let $(X_n)$ and $(Y_n)$ be two supermartingales on the probability space $(Omega, mathcal{A}, P)$ and $T$ be a stopping time regarding a filtration $(mathcal{F}_n)$ and $X_T leq Y_T$ on ${T< infty}$.



Define $Z_n = Y_n$ on ${n<T}$ and $Z_n = X_n$ on ${T leq n}$.



Proof that $(Z_n)$ is a supermartingale.



I tried to use $Z_n = Y_nI_{{n<T}} + X_nI_{{Tleq n}}$ and plug that into $mathbb{E}[Z_n|mathcal{F_{n-1}}]$ and use some properties of the conditional expectation, but I don't seem to get to $leq Z_{n-1}$.



Any help is appreciated.






probability probability-theory conditional-expectation martingales stopping-times







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$endgroup$












  • $begingroup$
    On behalf of @hosam ahmad: this is a theorem 8.1 page 498 in Probability: A Graduate Course Authors: Gut, Allan springer.com/la/book/9781441919854
    $endgroup$
    – dantopa
    Dec 20 '18 at 19:40
















4












$begingroup$


I have the following task to do:



Let $(X_n)$ and $(Y_n)$ be two supermartingales on the probability space $(Omega, mathcal{A}, P)$ and $T$ be a stopping time regarding a filtration $(mathcal{F}_n)$ and $X_T leq Y_T$ on ${T< infty}$.



Define $Z_n = Y_n$ on ${n<T}$ and $Z_n = X_n$ on ${T leq n}$.



Proof that $(Z_n)$ is a supermartingale.



I tried to use $Z_n = Y_nI_{{n<T}} + X_nI_{{Tleq n}}$ and plug that into $mathbb{E}[Z_n|mathcal{F_{n-1}}]$ and use some properties of the conditional expectation, but I don't seem to get to $leq Z_{n-1}$.



Any help is appreciated.






probability probability-theory conditional-expectation martingales stopping-times







share|cite|improve this question











$endgroup$












  • $begingroup$
    On behalf of @hosam ahmad: this is a theorem 8.1 page 498 in Probability: A Graduate Course Authors: Gut, Allan springer.com/la/book/9781441919854
    $endgroup$
    – dantopa
    Dec 20 '18 at 19:40














4












4








4





$begingroup$


I have the following task to do:



Let $(X_n)$ and $(Y_n)$ be two supermartingales on the probability space $(Omega, mathcal{A}, P)$ and $T$ be a stopping time regarding a filtration $(mathcal{F}_n)$ and $X_T leq Y_T$ on ${T< infty}$.



Define $Z_n = Y_n$ on ${n<T}$ and $Z_n = X_n$ on ${T leq n}$.



Proof that $(Z_n)$ is a supermartingale.



I tried to use $Z_n = Y_nI_{{n<T}} + X_nI_{{Tleq n}}$ and plug that into $mathbb{E}[Z_n|mathcal{F_{n-1}}]$ and use some properties of the conditional expectation, but I don't seem to get to $leq Z_{n-1}$.



Any help is appreciated.






probability probability-theory conditional-expectation martingales stopping-times







share|cite|improve this question











$endgroup$




I have the following task to do:



Let $(X_n)$ and $(Y_n)$ be two supermartingales on the probability space $(Omega, mathcal{A}, P)$ and $T$ be a stopping time regarding a filtration $(mathcal{F}_n)$ and $X_T leq Y_T$ on ${T< infty}$.



Define $Z_n = Y_n$ on ${n<T}$ and $Z_n = X_n$ on ${T leq n}$.



Proof that $(Z_n)$ is a supermartingale.



I tried to use $Z_n = Y_nI_{{n<T}} + X_nI_{{Tleq n}}$ and plug that into $mathbb{E}[Z_n|mathcal{F_{n-1}}]$ and use some properties of the conditional expectation, but I don't seem to get to $leq Z_{n-1}$.



Any help is appreciated.






probability probability-theory conditional-expectation martingales stopping-times




probability probability-theory conditional-expectation martingales stopping-times






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 9 '18 at 3:37









saz

81.8k862130




81.8k862130










asked Jun 8 '18 at 18:34









thatha

1185




1185












  • $begingroup$
    On behalf of @hosam ahmad: this is a theorem 8.1 page 498 in Probability: A Graduate Course Authors: Gut, Allan springer.com/la/book/9781441919854
    $endgroup$
    – dantopa
    Dec 20 '18 at 19:40


















  • $begingroup$
    On behalf of @hosam ahmad: this is a theorem 8.1 page 498 in Probability: A Graduate Course Authors: Gut, Allan springer.com/la/book/9781441919854
    $endgroup$
    – dantopa
    Dec 20 '18 at 19:40
















$begingroup$
On behalf of @hosam ahmad: this is a theorem 8.1 page 498 in Probability: A Graduate Course Authors: Gut, Allan springer.com/la/book/9781441919854
$endgroup$
– dantopa
Dec 20 '18 at 19:40




$begingroup$
On behalf of @hosam ahmad: this is a theorem 8.1 page 498 in Probability: A Graduate Course Authors: Gut, Allan springer.com/la/book/9781441919854
$endgroup$
– dantopa
Dec 20 '18 at 19:40










2 Answers
2






active

oldest

votes


















2












$begingroup$

Note that $X_T leq Y_T$ implies that



$$begin{align*} X_n 1_{{T leq n}} = X_n 1_{{T leq n-1}} + X_T 1_{{T=n}} &leq X_n 1_{{T leq n-1}}+ Y_T 1_{{T =n}} \ &= X_n 1_{{T leq n-1}}+ Y_n 1_{{T =n}}. end{align*}$$



Thus,



$$Z_n leq Y_{n} 1_{{T leq n-1}^c} + X_n 1_{{T leq n-1}}.$$



Now use that $(Y_{n})_{n in mathbb{N}}$ and $(X_n)_{n in mathbb{N}}$ are supermartingales and the fact that ${T leq n-1} in mathcal{F}_{n-1}$ to conclude that



$$mathbb{E}(Z_n mid mathcal{F}_{n-1}) leq Z_{n-1}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your answer, but I have to ask how do you conclude $Z_n leq Y_n 1_{{Tleq n-1}^c} + X_n1_{{Tleq n-1}}$? I have $Z_n leq Y_n 1_{{nleq T}} + X_n1_{{Tleq n-1}}$.
    $endgroup$
    – tha
    Jun 8 '18 at 21:41








  • 1




    $begingroup$
    @tha That's exactly the same thing, isn't it? Note that ${n leq T} = {T geq n} = {T leq n-1}^c$.
    $endgroup$
    – saz
    Jun 9 '18 at 3:37










  • $begingroup$
    Oh yes, that was my fault. Everything is good, thank you.
    $endgroup$
    – tha
    Jun 9 '18 at 8:39



















0












$begingroup$

this is a theorem 8.1 page 498 in Probability: A Graduate Course
Authors: Gut, Allan



https://www.springer.com/la/book/9781441919854






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    2












    $begingroup$

    Note that $X_T leq Y_T$ implies that



    $$begin{align*} X_n 1_{{T leq n}} = X_n 1_{{T leq n-1}} + X_T 1_{{T=n}} &leq X_n 1_{{T leq n-1}}+ Y_T 1_{{T =n}} \ &= X_n 1_{{T leq n-1}}+ Y_n 1_{{T =n}}. end{align*}$$



    Thus,



    $$Z_n leq Y_{n} 1_{{T leq n-1}^c} + X_n 1_{{T leq n-1}}.$$



    Now use that $(Y_{n})_{n in mathbb{N}}$ and $(X_n)_{n in mathbb{N}}$ are supermartingales and the fact that ${T leq n-1} in mathcal{F}_{n-1}$ to conclude that



    $$mathbb{E}(Z_n mid mathcal{F}_{n-1}) leq Z_{n-1}.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for your answer, but I have to ask how do you conclude $Z_n leq Y_n 1_{{Tleq n-1}^c} + X_n1_{{Tleq n-1}}$? I have $Z_n leq Y_n 1_{{nleq T}} + X_n1_{{Tleq n-1}}$.
      $endgroup$
      – tha
      Jun 8 '18 at 21:41








    • 1




      $begingroup$
      @tha That's exactly the same thing, isn't it? Note that ${n leq T} = {T geq n} = {T leq n-1}^c$.
      $endgroup$
      – saz
      Jun 9 '18 at 3:37










    • $begingroup$
      Oh yes, that was my fault. Everything is good, thank you.
      $endgroup$
      – tha
      Jun 9 '18 at 8:39
















    2












    $begingroup$

    Note that $X_T leq Y_T$ implies that



    $$begin{align*} X_n 1_{{T leq n}} = X_n 1_{{T leq n-1}} + X_T 1_{{T=n}} &leq X_n 1_{{T leq n-1}}+ Y_T 1_{{T =n}} \ &= X_n 1_{{T leq n-1}}+ Y_n 1_{{T =n}}. end{align*}$$



    Thus,



    $$Z_n leq Y_{n} 1_{{T leq n-1}^c} + X_n 1_{{T leq n-1}}.$$



    Now use that $(Y_{n})_{n in mathbb{N}}$ and $(X_n)_{n in mathbb{N}}$ are supermartingales and the fact that ${T leq n-1} in mathcal{F}_{n-1}$ to conclude that



    $$mathbb{E}(Z_n mid mathcal{F}_{n-1}) leq Z_{n-1}.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for your answer, but I have to ask how do you conclude $Z_n leq Y_n 1_{{Tleq n-1}^c} + X_n1_{{Tleq n-1}}$? I have $Z_n leq Y_n 1_{{nleq T}} + X_n1_{{Tleq n-1}}$.
      $endgroup$
      – tha
      Jun 8 '18 at 21:41








    • 1




      $begingroup$
      @tha That's exactly the same thing, isn't it? Note that ${n leq T} = {T geq n} = {T leq n-1}^c$.
      $endgroup$
      – saz
      Jun 9 '18 at 3:37










    • $begingroup$
      Oh yes, that was my fault. Everything is good, thank you.
      $endgroup$
      – tha
      Jun 9 '18 at 8:39














    2












    2








    2





    $begingroup$

    Note that $X_T leq Y_T$ implies that



    $$begin{align*} X_n 1_{{T leq n}} = X_n 1_{{T leq n-1}} + X_T 1_{{T=n}} &leq X_n 1_{{T leq n-1}}+ Y_T 1_{{T =n}} \ &= X_n 1_{{T leq n-1}}+ Y_n 1_{{T =n}}. end{align*}$$



    Thus,



    $$Z_n leq Y_{n} 1_{{T leq n-1}^c} + X_n 1_{{T leq n-1}}.$$



    Now use that $(Y_{n})_{n in mathbb{N}}$ and $(X_n)_{n in mathbb{N}}$ are supermartingales and the fact that ${T leq n-1} in mathcal{F}_{n-1}$ to conclude that



    $$mathbb{E}(Z_n mid mathcal{F}_{n-1}) leq Z_{n-1}.$$






    share|cite|improve this answer











    $endgroup$



    Note that $X_T leq Y_T$ implies that



    $$begin{align*} X_n 1_{{T leq n}} = X_n 1_{{T leq n-1}} + X_T 1_{{T=n}} &leq X_n 1_{{T leq n-1}}+ Y_T 1_{{T =n}} \ &= X_n 1_{{T leq n-1}}+ Y_n 1_{{T =n}}. end{align*}$$



    Thus,



    $$Z_n leq Y_{n} 1_{{T leq n-1}^c} + X_n 1_{{T leq n-1}}.$$



    Now use that $(Y_{n})_{n in mathbb{N}}$ and $(X_n)_{n in mathbb{N}}$ are supermartingales and the fact that ${T leq n-1} in mathcal{F}_{n-1}$ to conclude that



    $$mathbb{E}(Z_n mid mathcal{F}_{n-1}) leq Z_{n-1}.$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jun 8 '18 at 20:03

























    answered Jun 8 '18 at 19:38









    sazsaz

    81.8k862130




    81.8k862130












    • $begingroup$
      Thank you for your answer, but I have to ask how do you conclude $Z_n leq Y_n 1_{{Tleq n-1}^c} + X_n1_{{Tleq n-1}}$? I have $Z_n leq Y_n 1_{{nleq T}} + X_n1_{{Tleq n-1}}$.
      $endgroup$
      – tha
      Jun 8 '18 at 21:41








    • 1




      $begingroup$
      @tha That's exactly the same thing, isn't it? Note that ${n leq T} = {T geq n} = {T leq n-1}^c$.
      $endgroup$
      – saz
      Jun 9 '18 at 3:37










    • $begingroup$
      Oh yes, that was my fault. Everything is good, thank you.
      $endgroup$
      – tha
      Jun 9 '18 at 8:39


















    • $begingroup$
      Thank you for your answer, but I have to ask how do you conclude $Z_n leq Y_n 1_{{Tleq n-1}^c} + X_n1_{{Tleq n-1}}$? I have $Z_n leq Y_n 1_{{nleq T}} + X_n1_{{Tleq n-1}}$.
      $endgroup$
      – tha
      Jun 8 '18 at 21:41








    • 1




      $begingroup$
      @tha That's exactly the same thing, isn't it? Note that ${n leq T} = {T geq n} = {T leq n-1}^c$.
      $endgroup$
      – saz
      Jun 9 '18 at 3:37










    • $begingroup$
      Oh yes, that was my fault. Everything is good, thank you.
      $endgroup$
      – tha
      Jun 9 '18 at 8:39
















    $begingroup$
    Thank you for your answer, but I have to ask how do you conclude $Z_n leq Y_n 1_{{Tleq n-1}^c} + X_n1_{{Tleq n-1}}$? I have $Z_n leq Y_n 1_{{nleq T}} + X_n1_{{Tleq n-1}}$.
    $endgroup$
    – tha
    Jun 8 '18 at 21:41






    $begingroup$
    Thank you for your answer, but I have to ask how do you conclude $Z_n leq Y_n 1_{{Tleq n-1}^c} + X_n1_{{Tleq n-1}}$? I have $Z_n leq Y_n 1_{{nleq T}} + X_n1_{{Tleq n-1}}$.
    $endgroup$
    – tha
    Jun 8 '18 at 21:41






    1




    1




    $begingroup$
    @tha That's exactly the same thing, isn't it? Note that ${n leq T} = {T geq n} = {T leq n-1}^c$.
    $endgroup$
    – saz
    Jun 9 '18 at 3:37




    $begingroup$
    @tha That's exactly the same thing, isn't it? Note that ${n leq T} = {T geq n} = {T leq n-1}^c$.
    $endgroup$
    – saz
    Jun 9 '18 at 3:37












    $begingroup$
    Oh yes, that was my fault. Everything is good, thank you.
    $endgroup$
    – tha
    Jun 9 '18 at 8:39




    $begingroup$
    Oh yes, that was my fault. Everything is good, thank you.
    $endgroup$
    – tha
    Jun 9 '18 at 8:39











    0












    $begingroup$

    this is a theorem 8.1 page 498 in Probability: A Graduate Course
    Authors: Gut, Allan



    https://www.springer.com/la/book/9781441919854






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      this is a theorem 8.1 page 498 in Probability: A Graduate Course
      Authors: Gut, Allan



      https://www.springer.com/la/book/9781441919854






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        this is a theorem 8.1 page 498 in Probability: A Graduate Course
        Authors: Gut, Allan



        https://www.springer.com/la/book/9781441919854






        share|cite|improve this answer









        $endgroup$



        this is a theorem 8.1 page 498 in Probability: A Graduate Course
        Authors: Gut, Allan



        https://www.springer.com/la/book/9781441919854







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 20 '18 at 18:10









        hosam ahmadhosam ahmad

        84




        84






























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