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Let ${X_t}$ be an ergodic Markov chain such that $E(X_{t+1}-X_t)=-\epsilon$ for $X_t\in [2,n-1]$, $E(X_{t+1}-X_t)<-\epsilon$ for $X_t=n$, and $E(X_{t+1}-X_t)=\beta$ for $0\leq X_t\leq 1$, where $\epsilon$ is a small positive number, $\beta>\epsilon$, and $n$ is a big positive constant.

Intuitively I feel like $\lim_{t\to\infty}E(X_t)\leq 1 + \beta$. Or how can we show that $\lim_{t\to\infty}E(X_t)/n\rightarrow 0$, as $n\rightarrow\infty$?

Is there some theorem that says something similar to this?

John
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  • Is $n$ fixed? If so, I'm not too sure what "big" means? Do you mean big relative to $\beta,\epsilon$? – mathworker21 Apr 13 '20 at 17:14
  • Yes, it's fixed. Yes, it's very big relative to $\beta$ and $\epsilon$ (I guess we could say $\beta, \epsilon < 2$ and $n>1000$). I feel like there must be some theorem about this. I mean, if a process has negative expected increase for all states except the lowest ones. As time passes, the expected state must be near these lowest states. – John Apr 13 '20 at 18:41

1 Answers1

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This is a bit too long for a comment, you could try to go this way :

I assume that $X_t$ takes values in $\{0,\dots,n\}$.

I feel like you mean statements like $\mathbb E[X_{t+1}-X_t|X_t\in [2,n-1]]=-\varepsilon$, and similarly for the last ones. Now by ergodicity you know that \begin{align*} 0&=\lim_{t\to\infty}\mathbb E[X_{t+1}-X_t]\\ &\leq\lim_{t\to\infty} \mathbb P[X_t\in \{0,1\}] \beta - (1-\mathbb P[X_t\in \{0,1\}])\varepsilon\\ &= -\varepsilon +(\varepsilon+\beta) \lim_{t\to \infty} \mathbb P[X_t\in \{0,1\}] \end{align*}

P. Quinton
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  • Thank you for your answer, your assumptions are correct. But how can we from that say anything about $\lim_{t\to\infty}\mathbb E[X_{t+1}]$? I also added a more general version to my question: how can we show that this expectation is independent of $n$? – John Apr 15 '20 at 10:19