Limit of martingale is not $Sigma_n$-measurable












1












$begingroup$


Let $U_nsim$ Unif$(-1,1)$ and $Sigma_n=sigma(U_1,...,U_n)$. Define $X_0=0$ and $$X_n=X_{n-1}+(1-|X_{n-1}|)U_n.$$




Given that for the limit $X$ of $X_n$ we have $mathbb{P}(X=-1)=mathbb{P}(X=1)=frac{1}{2}$, how do we show that ${X=1}notinSigma_n$?




To show that we have a.s. and $L_1$ convergence of $X_n$ to $X$ we can show uniform integrability. To do this I thought we can show that $|X_n|leq 1$ for all $n$ a.s. How do I do this?



If we have $L_1$ convergence then
$$mathbb{E}[X mid Sigma_n]=X_n.$$
Further we have $int_{{X=1}}X , dmathbb{P}=mathbb{P}(X=1)=frac{1}{2}$.

If ${X=1}inSigma_n$, then we would have
$$int_{{X=1}}X , dmathbb{P}=int_{{X=1}}mathbb{E}[X mid Sigma_n] , dmathbb{P}=int_{{X=1}}X_n , dmathbb{P}.$$
Am I approaching this in the right way? How do I continue?






probability-theory measure-theory conditional-expectation martingales







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    1












    $begingroup$


    Let $U_nsim$ Unif$(-1,1)$ and $Sigma_n=sigma(U_1,...,U_n)$. Define $X_0=0$ and $$X_n=X_{n-1}+(1-|X_{n-1}|)U_n.$$




    Given that for the limit $X$ of $X_n$ we have $mathbb{P}(X=-1)=mathbb{P}(X=1)=frac{1}{2}$, how do we show that ${X=1}notinSigma_n$?




    To show that we have a.s. and $L_1$ convergence of $X_n$ to $X$ we can show uniform integrability. To do this I thought we can show that $|X_n|leq 1$ for all $n$ a.s. How do I do this?



    If we have $L_1$ convergence then
    $$mathbb{E}[X mid Sigma_n]=X_n.$$
    Further we have $int_{{X=1}}X , dmathbb{P}=mathbb{P}(X=1)=frac{1}{2}$.

    If ${X=1}inSigma_n$, then we would have
    $$int_{{X=1}}X , dmathbb{P}=int_{{X=1}}mathbb{E}[X mid Sigma_n] , dmathbb{P}=int_{{X=1}}X_n , dmathbb{P}.$$
    Am I approaching this in the right way? How do I continue?






    probability-theory measure-theory conditional-expectation martingales







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $U_nsim$ Unif$(-1,1)$ and $Sigma_n=sigma(U_1,...,U_n)$. Define $X_0=0$ and $$X_n=X_{n-1}+(1-|X_{n-1}|)U_n.$$




      Given that for the limit $X$ of $X_n$ we have $mathbb{P}(X=-1)=mathbb{P}(X=1)=frac{1}{2}$, how do we show that ${X=1}notinSigma_n$?




      To show that we have a.s. and $L_1$ convergence of $X_n$ to $X$ we can show uniform integrability. To do this I thought we can show that $|X_n|leq 1$ for all $n$ a.s. How do I do this?



      If we have $L_1$ convergence then
      $$mathbb{E}[X mid Sigma_n]=X_n.$$
      Further we have $int_{{X=1}}X , dmathbb{P}=mathbb{P}(X=1)=frac{1}{2}$.

      If ${X=1}inSigma_n$, then we would have
      $$int_{{X=1}}X , dmathbb{P}=int_{{X=1}}mathbb{E}[X mid Sigma_n] , dmathbb{P}=int_{{X=1}}X_n , dmathbb{P}.$$
      Am I approaching this in the right way? How do I continue?






      probability-theory measure-theory conditional-expectation martingales







      share|cite|improve this question











      $endgroup$




      Let $U_nsim$ Unif$(-1,1)$ and $Sigma_n=sigma(U_1,...,U_n)$. Define $X_0=0$ and $$X_n=X_{n-1}+(1-|X_{n-1}|)U_n.$$




      Given that for the limit $X$ of $X_n$ we have $mathbb{P}(X=-1)=mathbb{P}(X=1)=frac{1}{2}$, how do we show that ${X=1}notinSigma_n$?




      To show that we have a.s. and $L_1$ convergence of $X_n$ to $X$ we can show uniform integrability. To do this I thought we can show that $|X_n|leq 1$ for all $n$ a.s. How do I do this?



      If we have $L_1$ convergence then
      $$mathbb{E}[X mid Sigma_n]=X_n.$$
      Further we have $int_{{X=1}}X , dmathbb{P}=mathbb{P}(X=1)=frac{1}{2}$.

      If ${X=1}inSigma_n$, then we would have
      $$int_{{X=1}}X , dmathbb{P}=int_{{X=1}}mathbb{E}[X mid Sigma_n] , dmathbb{P}=int_{{X=1}}X_n , dmathbb{P}.$$
      Am I approaching this in the right way? How do I continue?






      probability-theory measure-theory conditional-expectation martingales




      probability-theory measure-theory conditional-expectation martingales






      share|cite|improve this question















      share|cite|improve this question













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      edited Jan 28 at 10:02









      saz

      81.9k862131




      81.9k862131










      asked Jan 27 at 19:26









      Joachim DoyleJoachim Doyle

      818




      818






















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          $begingroup$

          Suppose that ${X=1} in Sigma_n$ for some $n in mathbb{N}$. Since $Sigma_n$ is a $sigma$-algebra this would imply $${X=-1} = {X=1}^c in Sigma_n.$$ Using



          $$X = 1_{{X=1}} - 1_{{X=-1}}$$



          it follows that



          $$mathbb{E}(X mid Sigma_n) = X.$$



          On the other hand, the martingale property also gives



          $$mathbb{E}(X mid Sigma_n) =X_n,$$



          and so $X_n = X$ almost surely. In particular, $$mathbb{P}(X_n=1) = mathbb{P}(X_n = -1) = frac{1}{2}. tag{1}$$ Since the mapping $x mapsto f(x) := x+(1-|x|)u$ satisfies



          $$f(x)=1 iff x=1 quad text{and} quad f(x)=-1 iff x=-1$$



          for any fixed $u in [-1,1]$, we have



          $$X_n(omega) = 1 iff X_{n-1}(omega)=1 quad text{and} quad X_{n}(omega)=1 iff X_{n-1}(omega)=-1$$ Combining this with $(1)$ we get $$mathbb{P}(X_{n-1}=1) =mathbb{P}(X_{n-1} = -1) = frac{1}{2}.$$ Iterating the procedure we find that $$mathbb{P}(X_0=1) = mathbb{P}(X_0=-1) = frac{1}{2}$$ in contradiction to the assumption that $X_0 =0$.






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            $begingroup$

            Suppose that ${X=1} in Sigma_n$ for some $n in mathbb{N}$. Since $Sigma_n$ is a $sigma$-algebra this would imply $${X=-1} = {X=1}^c in Sigma_n.$$ Using



            $$X = 1_{{X=1}} - 1_{{X=-1}}$$



            it follows that



            $$mathbb{E}(X mid Sigma_n) = X.$$



            On the other hand, the martingale property also gives



            $$mathbb{E}(X mid Sigma_n) =X_n,$$



            and so $X_n = X$ almost surely. In particular, $$mathbb{P}(X_n=1) = mathbb{P}(X_n = -1) = frac{1}{2}. tag{1}$$ Since the mapping $x mapsto f(x) := x+(1-|x|)u$ satisfies



            $$f(x)=1 iff x=1 quad text{and} quad f(x)=-1 iff x=-1$$



            for any fixed $u in [-1,1]$, we have



            $$X_n(omega) = 1 iff X_{n-1}(omega)=1 quad text{and} quad X_{n}(omega)=1 iff X_{n-1}(omega)=-1$$ Combining this with $(1)$ we get $$mathbb{P}(X_{n-1}=1) =mathbb{P}(X_{n-1} = -1) = frac{1}{2}.$$ Iterating the procedure we find that $$mathbb{P}(X_0=1) = mathbb{P}(X_0=-1) = frac{1}{2}$$ in contradiction to the assumption that $X_0 =0$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Suppose that ${X=1} in Sigma_n$ for some $n in mathbb{N}$. Since $Sigma_n$ is a $sigma$-algebra this would imply $${X=-1} = {X=1}^c in Sigma_n.$$ Using



              $$X = 1_{{X=1}} - 1_{{X=-1}}$$



              it follows that



              $$mathbb{E}(X mid Sigma_n) = X.$$



              On the other hand, the martingale property also gives



              $$mathbb{E}(X mid Sigma_n) =X_n,$$



              and so $X_n = X$ almost surely. In particular, $$mathbb{P}(X_n=1) = mathbb{P}(X_n = -1) = frac{1}{2}. tag{1}$$ Since the mapping $x mapsto f(x) := x+(1-|x|)u$ satisfies



              $$f(x)=1 iff x=1 quad text{and} quad f(x)=-1 iff x=-1$$



              for any fixed $u in [-1,1]$, we have



              $$X_n(omega) = 1 iff X_{n-1}(omega)=1 quad text{and} quad X_{n}(omega)=1 iff X_{n-1}(omega)=-1$$ Combining this with $(1)$ we get $$mathbb{P}(X_{n-1}=1) =mathbb{P}(X_{n-1} = -1) = frac{1}{2}.$$ Iterating the procedure we find that $$mathbb{P}(X_0=1) = mathbb{P}(X_0=-1) = frac{1}{2}$$ in contradiction to the assumption that $X_0 =0$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Suppose that ${X=1} in Sigma_n$ for some $n in mathbb{N}$. Since $Sigma_n$ is a $sigma$-algebra this would imply $${X=-1} = {X=1}^c in Sigma_n.$$ Using



                $$X = 1_{{X=1}} - 1_{{X=-1}}$$



                it follows that



                $$mathbb{E}(X mid Sigma_n) = X.$$



                On the other hand, the martingale property also gives



                $$mathbb{E}(X mid Sigma_n) =X_n,$$



                and so $X_n = X$ almost surely. In particular, $$mathbb{P}(X_n=1) = mathbb{P}(X_n = -1) = frac{1}{2}. tag{1}$$ Since the mapping $x mapsto f(x) := x+(1-|x|)u$ satisfies



                $$f(x)=1 iff x=1 quad text{and} quad f(x)=-1 iff x=-1$$



                for any fixed $u in [-1,1]$, we have



                $$X_n(omega) = 1 iff X_{n-1}(omega)=1 quad text{and} quad X_{n}(omega)=1 iff X_{n-1}(omega)=-1$$ Combining this with $(1)$ we get $$mathbb{P}(X_{n-1}=1) =mathbb{P}(X_{n-1} = -1) = frac{1}{2}.$$ Iterating the procedure we find that $$mathbb{P}(X_0=1) = mathbb{P}(X_0=-1) = frac{1}{2}$$ in contradiction to the assumption that $X_0 =0$.






                share|cite|improve this answer









                $endgroup$



                Suppose that ${X=1} in Sigma_n$ for some $n in mathbb{N}$. Since $Sigma_n$ is a $sigma$-algebra this would imply $${X=-1} = {X=1}^c in Sigma_n.$$ Using



                $$X = 1_{{X=1}} - 1_{{X=-1}}$$



                it follows that



                $$mathbb{E}(X mid Sigma_n) = X.$$



                On the other hand, the martingale property also gives



                $$mathbb{E}(X mid Sigma_n) =X_n,$$



                and so $X_n = X$ almost surely. In particular, $$mathbb{P}(X_n=1) = mathbb{P}(X_n = -1) = frac{1}{2}. tag{1}$$ Since the mapping $x mapsto f(x) := x+(1-|x|)u$ satisfies



                $$f(x)=1 iff x=1 quad text{and} quad f(x)=-1 iff x=-1$$



                for any fixed $u in [-1,1]$, we have



                $$X_n(omega) = 1 iff X_{n-1}(omega)=1 quad text{and} quad X_{n}(omega)=1 iff X_{n-1}(omega)=-1$$ Combining this with $(1)$ we get $$mathbb{P}(X_{n-1}=1) =mathbb{P}(X_{n-1} = -1) = frac{1}{2}.$$ Iterating the procedure we find that $$mathbb{P}(X_0=1) = mathbb{P}(X_0=-1) = frac{1}{2}$$ in contradiction to the assumption that $X_0 =0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 27 at 19:51









                sazsaz

                81.9k862131




                81.9k862131






























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