Markov chains: recurrent state to acessible state (incomplete)
Dependencies:
Let $X = [X_0, X_1, \ldots]$ be a markov chain with transformation function $P$. Let $i$ be a recurrent state and let state $j$ be accessible from state $i$. Then the markov chain will eventually reach $j$ with probability 1 if it starts from $i$. Formally, let $R_j = \bigvee_{t=1}^{\infty} (X_t = j)$. Then $\Pr(R_j \mid X_0 = i) = 1$.
Proof
Since $i$ is recurrent, the markov chain will visit $i$ an infinite number of times. Let $T_k$ be the $k^{\textrm{th}}$ time (for $t \ge 1$) when the markov chain visits $i$. Since $j$ is accessible from $i$, there exists a path of states $Q = [k_0, k_1, k_2, \ldots, k_r]$ such that $i = k_0$, $j = k_r$ and $P(k_{t-1}, k_t) > 0$ for all $1 \le t \le r$ and $k_t \not\in \{i, j\}$ for all $1 \le t < r$. Let $q = \prod_{t=1}^r P(k_{t-1}, k_t)$. Then $q > 0$.
Let $E$ be the event that $j$ is never visited via $Q$. We will show that $\Pr(E \mid X_0 = i, T_1 = t_1, T_2 = t_2, \ldots) = 0$, which will prove that $j$ will be eventually visited (and that too via $Q$).
(Non-rigorous proof ahead)
Since $i$ is visited an infinite number of times, we have an infinite number of opportunities to visit $j$ via $Q$, and each such opportunity will succeed with probability $q > 0$. Hence, we will succeed eventually, so $j$ will be reached eventually from $i$ via $Q$.
Dependency for: None
Info:
- Depth: 9
- Number of transitive dependencies: 20
Transitive dependencies:
- /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
- Markov chains: recurrent iff infinite visits