Mean number of visits till absorbed by closed class

Question

Let $\{X_n\}$ be a markov chain, with transition probability matrix: $$\displaystyle P=\left( \begin{array}{ccccccc} 1/2 & 1/2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1/2 & 1/2 & 0 & 0 & 0 & 0 \\ 1/4 & 0 & 0 & 1/2 & 1/4 & 0 & 0 \\ 0 & 0 & 0& 1/2 & 1/2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1/2 & 1/4 & 1/4 \\ 0 & 0 & 0 & 0 & 1/4 & 1/2 & 1/4 \\ 0 & 0 & 0 & 0 & 1/2 & 1/2 & 0 \\ \end{array} \right).$$ If $X_0=1$, find $$\mathbb{E}\bigg[\sum_{k=1}^{T}\mathbb{I}_{\{X_k=3\}}\bigg],$$ where $T=\inf\{k\geq 0: ~X_k\in \{5,6,7\}\}.$

($\mathbb{I}_A$ denotes the index function on set $A$)

Attempt. There are three communication classes, the open $\{1,2,3\},~\{4\}$ and the closed $\{5,6,7\}$ (whose invariant distribution is $\pi =(2/5,2/5,1/5)$). I thought of a one step analysis, but i don't think it would get me anywhere.

Thank you in advance!

Answer 1

Answer: $4/3$.

Sketch: The only way from $1$ to $\{5,6,7\}$ (in fact, here we will always stop in $5$) lies through $3$. Each time when we visit $3$, a coin is flipped, and we either (with probability $1/4$) get heads and return to $1$ or (with probability $3/4$) get tails and go either to $5$ or to $4$, which also eventually leads to $5$. Thus, the number of visits of the state $3$ is the number of attempts to get tails when flipping a coin with $\mathbb{P}(\text{tails}) = 3/4$. So this number follows a geometric distribution with success probability $3/4$, hence the answer.

Answer 2

From state $1$ the process transitions to $1$ or $2$. From state $2$ the process transitions to $2$ or $3$. From state $3$ the process transitions to $1$ or $4$ or $5$. For any higher state, the process never returns to $3$.

Let $v_{i}$ be value of state $i=1,\ldots,7$. Then $$\begin{aligned} v_{1}&=\tfrac{1}{2}v_{1}+\tfrac{1}{2}v_{2}\\ v_{2}&=\tfrac{1}{2}v_{2}+\tfrac{1}{2}v_{3}\\ v_{3}&=1+\tfrac{1}{4}v_{1}+\tfrac{3}{4}(v_{4}=0\text{ or }v_{5}=0)\\ \end{aligned}$$ Solving this system gives $v_{1}=v_{2}=v_{3}=\frac{4}{3}$.