Find P(T$\geq$ 3) for Markov Chain

Question

Given the Markov chain $(X_n)_{n\geq 0}$ with state space $\mathcal{S}=\lbrace 0, 1,2 \rbrace$ and transition probability matrix $\mathbf{P} = \left[\begin{array}{ccc} {\textstyle\frac{5}{17}} &{\textstyle\frac{5}{17}} &{\textstyle\frac{7}{17}}\cr {\textstyle\frac{5}{23}} &{\textstyle\frac{12}{23}} &{\textstyle\frac{6}{23}}\cr 0 &0 &1 \end{array}\right]$

Define by T the time the process first reaches the absorbing state 2

Find P(T $\geq$ 3) given that P($X_0$ = $0$) = $0.3$

So far I have attempted

$P(T \geq 3) = P(X_0 = 0)*P(T \geq 3|X_0 = 0)\\= 0.3 * (1- P(T \leq 2|X_0 = 0))\\ = 0.3 (1- \frac{4375}{6647}) $

to no avail using the fact that normally in probability that $P(X \geq n) = 1- P(X \leq n-1)$

Please help me solve this question and guide me in the right direction

Edit: The question was updated to include the fact that $P(X_0 = 1) = 0.7$

Answer 1

Firstly, note that the event $T = n$ implies that $X_n = 2$ and $X_i \neq 2$ for all $i = 0,1,2,\dots,n-1$.

Let us begin with evaluating $\Pr(T = 0)$, $\Pr(T = 1)$ and $\Pr(T = 2)$. Clearly, $\Pr(T = 0) = \Pr(X_0 = 2)$. For $n \geq 1$, $\Pr(T = n|X_0 =2) = 0$ by definition. For $\Pr(T = 1)$, using law of total probability, we have, \begin{align*} \Pr(T = 1) & = \sum_{i = 0}^{2} \Pr(T = 1|X_0 = i) \Pr(X_0 = i) \\ & = \Pr(X_1 = 2|X_0 = 0) \Pr(X_0 = 0) + \Pr(X_1 = 2|X_0 = 1) \Pr(X_0 = 1) + 0 \\ & = \frac{7}{17} \cdot \Pr(X_0 = 0) + \frac{6}{23} \cdot \Pr(X_0 = 1). \end{align*}

For $T = 2$, we first evaluate $\Pr(T = 2|X_0 = 0)$. We have, \begin{align*} \Pr(T = 2|X_0 = 0) & = \Pr(X_2 = 2, X_1 \neq 2|X_0 = 0) \\ & = \Pr(X_2 = 2, X_1 = 0|X_0 = 0) + \Pr(X_2 = 2, X_1 = 1|X_0 = 0)\\ & = \Pr(X_2 = 2|X_1 = 0) \Pr(X_1 = 0|X_0 = 0) + \Pr(X_2 = 2|X_1 = 1) \Pr(X_1 = 1|X_0 = 0)\\ & = \frac{7}{17} \cdot \frac{5}{17} + \frac{6}{23} \cdot \frac{5}{17}. \end{align*} Using a similar approach, we can conclude, \begin{align*} \Pr(T = 2|X_0 = 1) = \frac{7}{17} \cdot \frac{5}{23} + \frac{6}{23} \cdot \frac{12}{23} \end{align*} Consequently, $\Pr(T = 2) = \Pr(T=2|X_0 = 0) \Pr(X_0 = 0) + \Pr(T=2|X_0 = 1) \Pr(X_0 = 1)$. You can plug in the values and compute that.

Finally, $\Pr(T \geq 3) = 1 - (\Pr(T = 0) + \Pr(T = 1) + \Pr(T = 2))$.