Random walk with absorbing barriers

Question

Consider a random walk with absorbing barriers at $0$ and $3$. $\mathbb P(S_{n+1}-S_n=1)=0.6$ and $\mathbb P(S_{n+1}-S_n=-1)=0.4$. What is the probability of eventual absorption at $0$, given that the starting point is $S_0=1$?

Answer 1

First, form the transition matrix corresponding to the random walk with 0 and 3 as absorbing states: $$P=\begin{bmatrix} 1&0&0&0\\2/5&0&3/5&0\\0&2/5&0&3/5\\0&0&0&1 \end{bmatrix}.$$ Then, rearrange to have the absorbing states first: $$\tilde{P}=\left[ \begin{array}{c|c} I & 0 \\ \hline S & Q \end{array} \right]=\begin{bmatrix} 1&0&0&0\\0&1&0&0\\2/5&0&3/5&0\\0&2/5&0&3/5 \end{bmatrix}.$$

Then, compute $M=(I-Q)^{-1},$ and $M\cdot S$. This yields $$M\cdot S = \begin{bmatrix}10/19&9/19\\4/19&15/19 \end{bmatrix} $$ These entries are the probabilities of starting at state 1 or 2 and being absorbed at 0 or 3. So, if the chain starts at state 1, for example, the probability of eventual absorption at 0 is 10/19.

FUN with basic linear algebra!

Answer 2

Your particular query can be answered very easily by considering a 2D lattice path with each point on X-axis representing a "winning" step (Pr = 0.6), and each step on the Y-axis representing a step towards ruin (Pr = 0.4). If you take a cursory look, you'll see that ruin can only occur after an odd # of steps, and that for each such possibility, there is just one way to reach ruin.

So P(get absorbed at 0) = sum of (2/5)^k*(3/5)^(k-1) for k = 1 to oo = 10/19