Consider a Markov Chain defined on the integers. Suppose $P_{n,n+1}=p$ and $P_{n,n-1}=1-p$, with $p > \frac{1}{2}$. Assume $X_{0}=0$.
What is the distribution of $\min\{X_{0}, X_{1}, \dots, \}$?
Consider a Markov Chain defined on the integers. Suppose $P_{n,n+1}=p$ and $P_{n,n-1}=1-p$, with $p > \frac{1}{2}$. Assume $X_{0}=0$.
What is the distribution of $\min\{X_{0}, X_{1}, \dots, \}$?
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Let $m_k=P(Y=-k)$ for $k\geq 0$. Then $m_k$ satisfies the following recurrence:
$$m_k=pm_{k+1}+(1-p)m_{k-1}$$
Solving it subject to condition $\sum_{k\geq 0} m_k=1$ yields $$m_k=\big(1-\frac{1-p}p\big)\big(\frac {1-p}p\big)^k$$
ETA: To show that the above condition is satisfied, we need to show that $Y$ is finite a.s. This follows from the fact that $X_n\to \infty$ with probability $1$.