Markov Chain (Simple Random Walk) distribution Let $\{X_n\}_{n = 0,1,\ldots}$  be the Markov Chain with state space $\mathbb{Z}$ and the transition probability  
$P_{n,n+1} = p \qquad P_{n,n-1} = 1-p$
where $p \gt \frac{1}{2}$. Assume $X_0 = 0$
Let $Y  = min\{X_0,X_1,\ldots\}$. What is the distribution of $Y$?
I can only get that for all $k \in \mathbb{Z}_{> 1}, P(Y=k) = 0$  since the chain is already in state $0$. And I think the meaning of event $Y = -k$ means that this chain have across state $-1,-2,\ldots,-k$ but never reach state $-k-1$. However I still cannot solve this problem, can some one help me? Thanks!
 A: Let the chance for $\{X_n\}$ to ever reach a negative integer $-k\,$ from $\,0\,$ be $P_{-k}$. Then we have
$$P_{-1}=P_0\,(1-p)+P_0\,p\,P_{-1}^2.$$
Since $P_0=1$, because the particle was initially at $X_0=0$,  we have
$$P_{-1}=\frac{1-p}{p}\mbox{ or }\,1.$$
For $\,p>1/2$, the random walk drifts to the positive side. So the chances to reach negative numbers are less than $1$. The random walk no longer returns to the origin infinitely often because of the drift. So we take the first root $P_{-1}=(1-p)/p$. Then the chance to reach $-k\,$ is
$$P_{-k}=(P_{-1})^k=\left(\frac{1-p}{p}\right)^k.$$
This is because to reach $-k$, the walker has to first reach $-1$, which happens with probability $P_{-1}$, and then it has to reach $-2\,$ from $-1$, which also happens with probability $P_{-1}$ independently of its history, and so on. Finally, notice that
$$\mathrm{Pr}(Y\equiv\mbox{minimum}\leq-k)=P_{-k}.$$
The expected value of the minimum $Y$ is given by
$$\langle Y\rangle=-\sum_{k=1}^\infty\,\mathrm{Pr}(Y\leq-k)=-\frac{1-p}{2p-1}.$$
