# Find probability of markov chain ended in state $0$.

Given transition matrix probability $$\begin{bmatrix} 1&0&0&0\\ 0.1&0.6&0.1&0.2\\ 0.2&0.3&0.4&0.1\\ 0&0&0&1 \end{bmatrix}$$ with state = $$\{0,1,2,3\}$$. Start from state $$1$$, find probability of markov chain ended in state $$0$$.

I have found state transition diagram. But I don't know to find probability of Markov chain ended in state $$0$$. What formula can be used to answer this problem? Can anyone help me?

Note that $$X_n$$ of the given transition probability is eventually absorbed at $$\{0,3\}$$ with probability $$1$$. For $$i=1,2$$, let $$p_i=\Bbb P(X_n\text{ is absorbed at }0\mid X_0=i)$$. Then we can see that \begin{align*}\require{cancel} p_1&=\sum_{k=0}^3\Bbb P(X_n\text{ is absorbed at }0, X_1=k\mid X_0=1) \\&=\sum_{k=0}^3\Bbb P(X_n\text{ is absorbed at }0\mid X_1=k\cancel{ X_0=1})\Bbb P(X_1=k\mid X_0=1) \\&=1\cdot (0.1)+p_1\cdot (0.6)+p_2\cdot (0.1)+0\cdot (0.2). \end{align*} In the same way, $$p_2 = (0.2)+p_1\cdot (0.3)+p_2\cdot (0.4).$$ holds. Solving those 2 equations, we obtain $$p_1= \frac8{21}, \ \ \ p_2=\frac{11}{21}.$$
Note: The vector $$\pi_0=(1,p_1,p_2,0)$$ of probability being absorbed at $$0$$ satisfies the equilibrium equation $$P\pi_0=\pi_0$$where $$P$$ is the transition matrix, because $$\pi_0 = \lim_{n\to\infty} P^n (0,1,0,0).$$
• What you want is the probability of absorption at $0$ starting from $\color{red}1$. So the answer is $p_1=\frac8{21}$. $p_2$ is only need to solve the equation. – Song Mar 3 at 13:42