Markov chains: recurrent iff infinite visits
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Let $X = [X_0, X_1, \ldots]$ be a markov chain. For state $i$, define \[ N_i = \sum_{t=1}^{\infty} \begin{cases} 1 & \textrm{ if } X_t = i \\ 0 & \textrm{ otherwise} \end{cases}. \]
Then $\Pr(N_i = \infty \mid X_0 = i) = 1$ iff state $i$ is recurrent.
Proof
Define $R_i$ and $T_i$ as \[ R_i = \bigvee_{t=1}^{\infty} (X_t = i). \] \[ T_i = \max_{t \ge 0} (X_t = i). \]
Note that $R_i \iff N_i \ge 1$. Suppose $\Pr(N_i = \infty \mid X_0 = i) = 1$. Then $\Pr(R_i \mid X_0 = i) = \Pr(N_i \ge 1 \mid X_0 = i) \ge \Pr(N_i = \infty \mid X_0 = i) = 1$. Hence, $i$ is recurrent.
Now assume $i$ is recurrent. We will prove that $\Pr(N_i = \infty \mid X_0 = i) = 1$. Note that $T_i = \infty \iff N_i = \infty$. \begin{align} & \Pr(N_i \neq \infty \mid X_0 = i) \\ &= \Pr(T_i \neq \infty \mid X_0 = i) \\ &= \sum_{t=1}^{\infty} \Pr(T_i = t \mid X_0 = i) \\ &= \sum_{t=1}^{\infty} \Pr\left(X_t = i \land \bigwedge_{r=t+1}^{\infty} (X_r \neq i) \mid X_0 = i\right) \\ &= \sum_{t=1}^{\infty} \Pr\left(\bigwedge_{r=t+1}^{\infty} (X_r \neq i) \mid X_t = i, X_0 = i\right) \Pr(X_t = i \mid X_0 = i) \\ &= \sum_{t=1}^{\infty} \Pr\left(\bigwedge_{r=1}^{\infty} (X_r \neq i) \mid X_0 = i \right) \Pr(X_t = i \mid X_0 = i) \tag{$\begin{array}{ll}\textrm{by markov property} \\ \textrm{and time homogeneity}\end{array}$} \\ &= \sum_{t=1}^{\infty} \Pr\left(\overline{R_i} \mid X_0 = i \right) \Pr(X_t = i \mid X_0 = i) \\ &= \sum_{t=1}^{\infty} 0 \Pr(X_t = i \mid X_0 = i) \tag{$i$ is recurrent} \\ &= 0 \end{align} Hence, if $i$ is recurrent, then $\Pr(N_i = \infty \mid X_0 = i) = 1$.
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- /analysis/topological-space
- /sets-and-relations/countable-set
- /sets-and-relations/de-morgan-laws
- σ-algebra
- Generated σ-algebra
- Borel algebra
- Measurable function
- Generators of the real Borel algebra (incomplete)
- Measure
- σ-algebra is closed under countable intersections
- Group
- Ring
- Semiring
- Matrix
- Probability
- Conditional probability (incomplete)
- Random variable
- Markov chain
- Markov chains: recurrent states