Show that i.i.d. process is ergodic












1














Let





  • $(Omega,mathcal A,operatorname P)$ be a probability space


  • $(E,mathcal E)$ be a measurable space


  • $(X_n)_{ninmathbb N_0}$ be an $(E,mathcal E)$-valued i.i.d. process on $(Omega,mathcal A,operatorname P)$


  • $mu$ denote the distribution of $X_0$ under $operatorname P$


Assume that there is a Markov kernel $pi$ with source $(E,mathcal E)$ and target $left(E^{mathbb N_0},mathcal E^{otimesmathbb N_0}right)$ with $$operatorname Pleft[Xin Bmid X_0right]=pi(X_0,B);;;text{almost surely for all }Binmathcal E^{otimesmathbb N_0}tag1.$$ Now, let $$tau:E^{mathbb N_0}to E^{mathbb N_0};,;;;(x_n)_{ninmathbb N_0}mapsto(x_{n+1})_{ninmathbb N_0}$$ and $operatorname P_mu:=mupi$ denote the composition of $mu$ and $pi$, i.e. $$operatorname P_mu[B]=intmu({rm d}x_0)pi(x_0,B);;;text{for all }(x_0,B)in Etimesmathcal E^{otimesmathbb N_0}tag2.$$



Note that $tau$ is $operatorname P_mu$-preserving, i.e. the distribution of $tau$ under $operatorname P$ is $operatorname P$.




How can we show that $operatorname P_mu$ is $tau$-ergodic, i.e. $$mathcal I:=left{Binmathcal E^{otimes I}:tau^{-1}(B)=Bright}$$ is $operatorname P_mu$-trivial (each event has $operatorname P_mu$-measure $0$ or $1$)?




Seems to be an easy task, but I don't see where we can make use of the i.i.d.-property of $X$.






probability-theory stochastic-processes markov-chains markov-process ergodic-theory







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    1














    Let





    • $(Omega,mathcal A,operatorname P)$ be a probability space


    • $(E,mathcal E)$ be a measurable space


    • $(X_n)_{ninmathbb N_0}$ be an $(E,mathcal E)$-valued i.i.d. process on $(Omega,mathcal A,operatorname P)$


    • $mu$ denote the distribution of $X_0$ under $operatorname P$


    Assume that there is a Markov kernel $pi$ with source $(E,mathcal E)$ and target $left(E^{mathbb N_0},mathcal E^{otimesmathbb N_0}right)$ with $$operatorname Pleft[Xin Bmid X_0right]=pi(X_0,B);;;text{almost surely for all }Binmathcal E^{otimesmathbb N_0}tag1.$$ Now, let $$tau:E^{mathbb N_0}to E^{mathbb N_0};,;;;(x_n)_{ninmathbb N_0}mapsto(x_{n+1})_{ninmathbb N_0}$$ and $operatorname P_mu:=mupi$ denote the composition of $mu$ and $pi$, i.e. $$operatorname P_mu[B]=intmu({rm d}x_0)pi(x_0,B);;;text{for all }(x_0,B)in Etimesmathcal E^{otimesmathbb N_0}tag2.$$



    Note that $tau$ is $operatorname P_mu$-preserving, i.e. the distribution of $tau$ under $operatorname P$ is $operatorname P$.




    How can we show that $operatorname P_mu$ is $tau$-ergodic, i.e. $$mathcal I:=left{Binmathcal E^{otimes I}:tau^{-1}(B)=Bright}$$ is $operatorname P_mu$-trivial (each event has $operatorname P_mu$-measure $0$ or $1$)?




    Seems to be an easy task, but I don't see where we can make use of the i.i.d.-property of $X$.






    probability-theory stochastic-processes markov-chains markov-process ergodic-theory







    share|cite|improve this question

























      1












      1








      1







      Let





      • $(Omega,mathcal A,operatorname P)$ be a probability space


      • $(E,mathcal E)$ be a measurable space


      • $(X_n)_{ninmathbb N_0}$ be an $(E,mathcal E)$-valued i.i.d. process on $(Omega,mathcal A,operatorname P)$


      • $mu$ denote the distribution of $X_0$ under $operatorname P$


      Assume that there is a Markov kernel $pi$ with source $(E,mathcal E)$ and target $left(E^{mathbb N_0},mathcal E^{otimesmathbb N_0}right)$ with $$operatorname Pleft[Xin Bmid X_0right]=pi(X_0,B);;;text{almost surely for all }Binmathcal E^{otimesmathbb N_0}tag1.$$ Now, let $$tau:E^{mathbb N_0}to E^{mathbb N_0};,;;;(x_n)_{ninmathbb N_0}mapsto(x_{n+1})_{ninmathbb N_0}$$ and $operatorname P_mu:=mupi$ denote the composition of $mu$ and $pi$, i.e. $$operatorname P_mu[B]=intmu({rm d}x_0)pi(x_0,B);;;text{for all }(x_0,B)in Etimesmathcal E^{otimesmathbb N_0}tag2.$$



      Note that $tau$ is $operatorname P_mu$-preserving, i.e. the distribution of $tau$ under $operatorname P$ is $operatorname P$.




      How can we show that $operatorname P_mu$ is $tau$-ergodic, i.e. $$mathcal I:=left{Binmathcal E^{otimes I}:tau^{-1}(B)=Bright}$$ is $operatorname P_mu$-trivial (each event has $operatorname P_mu$-measure $0$ or $1$)?




      Seems to be an easy task, but I don't see where we can make use of the i.i.d.-property of $X$.






      probability-theory stochastic-processes markov-chains markov-process ergodic-theory







      share|cite|improve this question













      Let





      • $(Omega,mathcal A,operatorname P)$ be a probability space


      • $(E,mathcal E)$ be a measurable space


      • $(X_n)_{ninmathbb N_0}$ be an $(E,mathcal E)$-valued i.i.d. process on $(Omega,mathcal A,operatorname P)$


      • $mu$ denote the distribution of $X_0$ under $operatorname P$


      Assume that there is a Markov kernel $pi$ with source $(E,mathcal E)$ and target $left(E^{mathbb N_0},mathcal E^{otimesmathbb N_0}right)$ with $$operatorname Pleft[Xin Bmid X_0right]=pi(X_0,B);;;text{almost surely for all }Binmathcal E^{otimesmathbb N_0}tag1.$$ Now, let $$tau:E^{mathbb N_0}to E^{mathbb N_0};,;;;(x_n)_{ninmathbb N_0}mapsto(x_{n+1})_{ninmathbb N_0}$$ and $operatorname P_mu:=mupi$ denote the composition of $mu$ and $pi$, i.e. $$operatorname P_mu[B]=intmu({rm d}x_0)pi(x_0,B);;;text{for all }(x_0,B)in Etimesmathcal E^{otimesmathbb N_0}tag2.$$



      Note that $tau$ is $operatorname P_mu$-preserving, i.e. the distribution of $tau$ under $operatorname P$ is $operatorname P$.




      How can we show that $operatorname P_mu$ is $tau$-ergodic, i.e. $$mathcal I:=left{Binmathcal E^{otimes I}:tau^{-1}(B)=Bright}$$ is $operatorname P_mu$-trivial (each event has $operatorname P_mu$-measure $0$ or $1$)?




      Seems to be an easy task, but I don't see where we can make use of the i.i.d.-property of $X$.






      probability-theory stochastic-processes markov-chains markov-process ergodic-theory




      probability-theory stochastic-processes markov-chains markov-process ergodic-theory






      share|cite|improve this question













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      asked Jan 4 at 15:00









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          Set $A:={X in B}$. Since $B=tau^{-1}(B)$ we have



          $$A = {X in tau^{-1}(B)} = {tau(X) in B}.$$



          Iterating the procedure we get



          $$A = {tau^n(X) in B},$$



          for any $n in mathbb{N}$ and therefore



          $$A in sigma(X_n,X_{n+1},ldots), qquad n in mathbb{N}.$$



          Since the sequence $(X_n)_{n in mathbb{N}}$ is iid, it follows from Kolmogorov's 0-1 law that $P(A) in {0,1}$. As



          $$P(A) = P(X in B) = int_E P(X in B mid X_0 = x_0) , mu(dx_0) = int_E pi(x_0,B) , mu(dx_0) = P_{mu}(B)$$



          this proves the assertion.






          share|cite|improve this answer





















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            Set $A:={X in B}$. Since $B=tau^{-1}(B)$ we have



            $$A = {X in tau^{-1}(B)} = {tau(X) in B}.$$



            Iterating the procedure we get



            $$A = {tau^n(X) in B},$$



            for any $n in mathbb{N}$ and therefore



            $$A in sigma(X_n,X_{n+1},ldots), qquad n in mathbb{N}.$$



            Since the sequence $(X_n)_{n in mathbb{N}}$ is iid, it follows from Kolmogorov's 0-1 law that $P(A) in {0,1}$. As



            $$P(A) = P(X in B) = int_E P(X in B mid X_0 = x_0) , mu(dx_0) = int_E pi(x_0,B) , mu(dx_0) = P_{mu}(B)$$



            this proves the assertion.






            share|cite|improve this answer


























              1














              Set $A:={X in B}$. Since $B=tau^{-1}(B)$ we have



              $$A = {X in tau^{-1}(B)} = {tau(X) in B}.$$



              Iterating the procedure we get



              $$A = {tau^n(X) in B},$$



              for any $n in mathbb{N}$ and therefore



              $$A in sigma(X_n,X_{n+1},ldots), qquad n in mathbb{N}.$$



              Since the sequence $(X_n)_{n in mathbb{N}}$ is iid, it follows from Kolmogorov's 0-1 law that $P(A) in {0,1}$. As



              $$P(A) = P(X in B) = int_E P(X in B mid X_0 = x_0) , mu(dx_0) = int_E pi(x_0,B) , mu(dx_0) = P_{mu}(B)$$



              this proves the assertion.






              share|cite|improve this answer
























                1












                1








                1






                Set $A:={X in B}$. Since $B=tau^{-1}(B)$ we have



                $$A = {X in tau^{-1}(B)} = {tau(X) in B}.$$



                Iterating the procedure we get



                $$A = {tau^n(X) in B},$$



                for any $n in mathbb{N}$ and therefore



                $$A in sigma(X_n,X_{n+1},ldots), qquad n in mathbb{N}.$$



                Since the sequence $(X_n)_{n in mathbb{N}}$ is iid, it follows from Kolmogorov's 0-1 law that $P(A) in {0,1}$. As



                $$P(A) = P(X in B) = int_E P(X in B mid X_0 = x_0) , mu(dx_0) = int_E pi(x_0,B) , mu(dx_0) = P_{mu}(B)$$



                this proves the assertion.






                share|cite|improve this answer












                Set $A:={X in B}$. Since $B=tau^{-1}(B)$ we have



                $$A = {X in tau^{-1}(B)} = {tau(X) in B}.$$



                Iterating the procedure we get



                $$A = {tau^n(X) in B},$$



                for any $n in mathbb{N}$ and therefore



                $$A in sigma(X_n,X_{n+1},ldots), qquad n in mathbb{N}.$$



                Since the sequence $(X_n)_{n in mathbb{N}}$ is iid, it follows from Kolmogorov's 0-1 law that $P(A) in {0,1}$. As



                $$P(A) = P(X in B) = int_E P(X in B mid X_0 = x_0) , mu(dx_0) = int_E pi(x_0,B) , mu(dx_0) = P_{mu}(B)$$



                this proves the assertion.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 4 at 15:40









                sazsaz

                78.5k758123




                78.5k758123






























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