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I am trying to understand a proof of the following claim:

If $P$ is irreducible and $\pi$ any stationary distribution, then $\pi_i = \frac{1}{m_i}$, where $m_i$ is the mean return time to state $i$.


Here's the proof:

Suppose $\pi$ is stationary for $P$, and let $X$ be a Markov chain with initial distribution $\pi$ and transition matrix $P$. Let $V_i(n)$ be the number of visits to state $i$ before time $n$. Then, by stationarity, $\mathbb{P}(X_n=i) = \pi_i$ for all $n$, and $$ \frac{\mathbb{E}[V_i(n)]}{n} = \frac{1}{n} \sum_{r=0}^{n-1}\mathbb{E}[1\{X_r=i\}] = \frac{1}{n} \sum_{r=0}^{n-1}\mathbb{P}(X_r=i) = \pi_i \tag{$1$} \label{ref1} $$ From the ergodic theorem, for any $\epsilon$, $$ \mathbb{P}\left(\left|\frac{V_i(n)}{n} - \frac{1}{m_i} \right| > \epsilon\right) < \epsilon \tag{$2$} \label{ref2} $$ for large enough $n$ (since almost sure convergence implies convergence in probability).

But since $\frac{V_i(n)}{n}$ is bounded between $0$ and $1$, it follows from \eqref{ref2} that $$ \boxed{\frac{\mathbb{E}[V_i(n)]}{n} \to \frac{1}{m_i} \quad \text{as} \quad n \to \infty} $$ Comparing to \eqref{ref1}, we obtain $\pi_i = \frac{1}{m_i}$.


Why does the boxed statement follow from \eqref{ref2}?

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1 Answer 1

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You can read (2) as

$$\mathbb{P}\left(-\epsilon \le \frac{V_i(n)}{n}-\frac{1}{m_i} \le \epsilon\right ) > 1- \epsilon$$

and this means with probability $1-\epsilon<1$ the absolute difference is less than $\epsilon$ and with the other probability $\epsilon$ the absolute difference is less than $1$ (since $\frac{V_i}{n}$ is bounded between $0$ and $1$), so there is a probability at least $1−ϵ$ that $\frac{V_i(n)}{n}-\frac{1}{m_i}$ is between $−ϵ$ and $+ϵ$; otherwise it is between $−1$ and $+1$. So its expectation $\mathbb{E}\left[\frac{V_i(n)}{n}-\frac{1}{m_i}\right]=\frac{\mathbb{E}[V_i(n)]}{n}-\frac{1}{m_i}$ is bounded below by $(1−ϵ)×(−ϵ)+ϵ×(−1)>−2ϵ$ and bounded above by $(1−ϵ)×(+ϵ)+ϵ×(+1)<+2ϵ$, i.e.

$$-2\epsilon < \frac{\mathbb{E}[V_i(n)]}{n}-\frac{1}{m_i} < 2\epsilon$$ for large enough $n$ (which depends on $\epsilon$). You can then take $\epsilon$ as small as you like to conclude $$\frac{\mathbb{E}[V_i(n)]}{n} \to \frac{1}{m_i} \quad \text{as} \quad n \to \infty.$$

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  • $\begingroup$ Thanks. Why is the absolute difference is less than one, though? $\endgroup$
    – Bob Zorro
    Commented Apr 4, 2023 at 11:33
  • $\begingroup$ I'm not really sure how you got the second inequality. How do you go from probability to expectation? $\endgroup$
    – Bob Zorro
    Commented Apr 4, 2023 at 11:49
  • $\begingroup$ $\frac{V_i(n)}{n}$ is between $0$ and $1$ and $\frac{1}{m_i}$ is between $0$ and $1$ so their absolute difference is between $0$ and $1$ $\endgroup$
    – Henry
    Commented Apr 4, 2023 at 12:40
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    $\begingroup$ There is a probability at least $1-\epsilon$ that $\frac{V_i(n)}{n}-\frac{1}{m_i}$ is between $-\epsilon$ and $+\epsilon$; otherwise it is between $-1$ and $+1$. So its expectation $\mathbb{E}\left[\frac{V_i(n)}{n}-\frac{1}{m_i}\right]=\frac{\mathbb{E}[V_i(n)]}{n}-\frac{1}{m_i}$ is bounded below by $(1-\epsilon)\times(-\epsilon)+\epsilon\times (-1) > -2\epsilon$ and bounded above by $(1-\epsilon)\times(+\epsilon)+\epsilon\times (+1) <+2\epsilon$ $\endgroup$
    – Henry
    Commented Apr 4, 2023 at 12:47
  • $\begingroup$ Ah thank you, that makes more sense. Perhaps you could add this to the answer to make it clearer? $\endgroup$
    – Bob Zorro
    Commented Apr 4, 2023 at 14:16

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