Random Walks- Stochastic Processes

Question

Suppose a random walk on $S = \{0, 1, 2, \dots , 5\}$ defined as follows. If $1 ≤ k ≤ 4$ and $X_n = k$, then $X_{n+1}$ is $k ± 1$ each with probability 1/2. If $X_n = 0$, then $X_{n+1} = 0$ (0 is absorbing). If $X_n = 5$, then $X_{n+1}$ is either 5 or 4, each with probability 1/2 (5 is "retaining"). Let $T = \min\{n ≥ 0 : X_n = 0\}$ denote the time that the walk gets absorbed at 0. For each k with 0 ≤ k ≤ 5, determine $τ(k) = E[T|\ X_0 = k]$. (E.g., τ(0) = 0.)

Answer 1

Let the probability of moving to the right be $p$ and the probability of moving to the left be $1-p$ (in the case of state $5$, the probability of remaining in state $5$ is $p$). We have the system of linear equations \begin{align} \tau(0) &= 0\\ \tau(1) &= 1 + (1-p)\tau(0) + p\tau(2)\\ \tau(2) &= 1 + (1-p)\tau(1) + p\tau(3)\\ \tau(3) &= 1 + (1-p)\tau(2) + p\tau(4)\\ \tau(4) &= 1 + (1-p)\tau(3) + p\tau(5)\\ \tau(5) &= 1 + (1-p)\tau(4) + p\tau(5) \end{align} which yields the unique solution \begin{align} \tau(0) &= 0\\ \tau(1) &= \frac{p^4-2 p^3+4 p^2-3 p+1}{(1-p)^5}\\ \tau(2) &= \frac{p^4-4 p^3+8 p^2-6 p+2}{(1-p)^5}\\ \tau(3) &= \frac{2 p^4-7 p^3+12 p^2-9 p+3}{(1-p)^5}\\ \tau(4) &= \frac{2 p^4-8 p^3+15 p^2-12 p+4}{(1-p)^5}\\ \tau(5) &= \frac{3 p^4-12 p^3+21 p^2-16 p+5}{(1-p)^5}. \end{align}