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world:non-geometric [2024/03/04 14:39]
rdouc [Proof]
world:non-geometric [2024/03/05 10:07]
rdouc
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 ==== Proof ==== ==== Proof ====
-The proof works by contradiction. ​If $P$ is geometrically ergodic, then there exists a $m$-small set $C$ such that $\sup_{x\in C} \PE_x[\beta^{\sigma_C}]<​\infty$ for some constant $\beta>​1$. ​+The proof works by contradiction. ​Assume that $P$ is geometrically ergodic, then there exists a $m$-small set $C$ such that $\sup_{x\in C} \PE_x[\beta^{\sigma_C}]<​\infty$ for some constant $\beta>​1$. ​
  
 Now, for any $\eta<​1$,​ define $A_\eta=\set{x \in \Xset}{P(x,​\{x\})\geq \eta}$. We can assume that $\pi(A_\eta)>​0$ (since in the assumptions,​ the esssup is taken wrt $\pi$). Then, if $x\in A_\eta$, ​ Now, for any $\eta<​1$,​ define $A_\eta=\set{x \in \Xset}{P(x,​\{x\})\geq \eta}$. We can assume that $\pi(A_\eta)>​0$ (since in the assumptions,​ the esssup is taken wrt $\pi$). Then, if $x\in A_\eta$, ​
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 \PP_x(X_1=\ldots=X_j=x) \geq \eta^j \PP_x(X_1=\ldots=X_j=x) \geq \eta^j
 \end{equation*} \end{equation*}
-Using that $C$ is a small set, we can easily show that $\sup_{x \in C} P(x,​\{x\})<​1$ (indeed, if $C\cap A_\eta$ contains two distincts elements $x,x'$ for $\eta$ sufficiently small, then $\epsilon \nu(\{x\}^c) \leq P^m(x,​\{x\}^c) \leq 1-\eta^m$ showing that $\nu(\{x\}^c) + \nu(\{x'​\}^c)$ is arbitrary small which is not possible since this sum is bounded from below by $\nu(\Xset)$). This allows to choose $B=A_\eta$ with $\eta$ chosen sufficiently close to $1$ so that $B \cap A_\eta=\emptyset$. Then, there exists $w_0\in C$ and $k \in \nset$ such that $\PP_{w_0}(X_k \in B, \sigma_C >​k)>​0$ (which can be easily seen by contradiction). ​+Moreover,  
 +  * Using that $C$ is a small set, we can easily show that $\sup_{x \in C} P(x,​\{x\})<​1$ (indeed, if $C\cap A_\eta$ contains two distincts elements $x,x'$ for $\eta$ sufficiently small, then $\epsilon \nu(\{x\}^c) \leq P^m(x,​\{x\}^c) \leq 1-\eta^m$ showing that $\nu(\{x\}^c) + \nu(\{x'​\}^c)$ is arbitrary small which is not possible since this sum is bounded from below by $\nu(\Xset)$). ​ 
 +  * This allows to choose $B=A_\eta$ with $\eta$ chosen sufficiently close to $1$ so that $B \cap A_\eta=\emptyset$. Then, there exists $w_0\in C$ and $k \in \nset$ such that $\PP_{w_0}(X_k \in B, \sigma_C >​k)>​0$ (which can be easily seen by contradiction). ​
  
 Now, write for any $\beta>​1$, ​ Now, write for any $\beta>​1$, ​
 \begin{align*} \begin{align*}
         \sup_{x\in C} \PE_x[\beta^{\sigma_C}]&​\geq \PE_{w_0}[\beta^{\sigma_C}-1]+1=(\beta-1) \sum_{i=0}^{\infty} \beta^i \PP_{w_0}(\sigma_C > i ) +1\\         \sup_{x\in C} \PE_x[\beta^{\sigma_C}]&​\geq \PE_{w_0}[\beta^{\sigma_C}-1]+1=(\beta-1) \sum_{i=0}^{\infty} \beta^i \PP_{w_0}(\sigma_C > i ) +1\\
-        & \geq (\beta-1) \sum_{i=0}^{\infty} \beta^{k+j} \PP_{w_0}(X_k\in B,\sigma_C > k+j ) +1 \\  +        & \geq (\beta-1) \sum_{j=0}^{\infty} \beta^{k+j} \PP_{w_0}(X_k\in B,\sigma_C > k+j ) +1 \\  
-        &  \geq (\beta-1) \sum_{i=0}^{\infty} \beta^{k+j} \PP_{w_0}(X_k\in B,\sigma_C > k, X_k=X_{k+1}= \ldots=X_{k+j} ) +1 \\  +        &  \geq (\beta-1) \sum_{j=0}^{\infty} \beta^{k+j} \PP_{w_0}(X_k\in B,\sigma_C > k, X_k=X_{k+1}= \ldots=X_{k+j} ) +1 \\  
-        & \geq (\beta-1) \sum_{i=0}^{\infty} \beta^{k+j} \PP_{w_0}(X_k\in B,\sigma_C > k) \eta^{j} +1 +        & \geq (\beta-1) \sum_{j=0}^{\infty} \beta^{k+j} \PP_{w_0}(X_k\in B,\sigma_C > k) \eta^{j} +1 
 \end{align*} \end{align*}
 which is divergent for $\eta$ sufficiently close to 1.  which is divergent for $\eta$ sufficiently close to 1. 
  
world/non-geometric.txt · Last modified: 2024/03/27 17:27 by rdouc