Let the transition matrix of a markov chain with states $\{0,1,2\}$ :
\begin{equation} A=\begin{bmatrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ 0 & \frac{1}{2} & \frac{1}{2} \\\ 0 & 0 & 1 \end{bmatrix}. \end{equation}
Explain why the state $2$ is absorbing and calculate the average time of absorption if we start at the state $0$.
(1) The state 2 is absorbing since $P_{22}=1$.
(2) Let $\tau_i=E($time to reach the state $2 = N$ $|$ starting at $i$$)$
(*)$$\tau_i=E(N|X_t=i)=\sum_k E(N|X_t=i, X_{t+1}=k)P(X_{t+1}=k | X_t=i) = \sum_k E(N|X_{t+1}=k)P_{ik}$$
Thus I could develop the system of equations :
$$\tau_0= \frac{1}{3} \tau_0 + \frac{1}{3} \tau_1+ \frac{1}{3} \tau_2$$ $$\tau_1= \frac{1}{2} \tau_1 + \frac{1}{2} \tau_2$$ $$\tau_2=1$$
So I found $\tau_0= \frac{5}{6}$
I think I have to add $1$ at some place at (*). Does anyone could explain to me rigourously why $E(N|X_t=i, X_{t+1}=k)=E(N+1|X_{t+1}=k)$?
You are taking $1$ step and then you see where you are.
Let $\tau(i)$ be the expected time to reach the absorbing state $2$ from state $i$, where $i =\{0,1,2\}$.
$\tau(0) = 1 +\frac{1}{3}\tau(0)+\frac{1}{3}\tau(1)+\frac{1}{3}\tau(2)$
$\tau(1) = 1 + \frac{1}{2}\tau(1)+\frac{1}{2}\tau(2)$
$\tau(2) =0$
Solve the system to find $\tau(0)$.