Show that i.i.d. process is ergodic
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
asked Jan 4 at 15:00
0xbadf00d0xbadf00d
1,80641534
1,80641534
add a comment |
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1 Answer
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$begingroup$
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.
$endgroup$
add a comment |
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$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Jan 4 at 15:40
sazsaz
82.1k862131
82.1k862131
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