The problem that I'm working on is given below:
A Markov chain $X_0,X_1,X_2,\ldots$ has the transition probability matrix: $$P=\begin{Vmatrix} 0.3 & 0.2 & 0.5\\ 0.5 & 0.1 & 0.4 \\ 0 & 0 & 1 \end{Vmatrix}$$ and is known to start in state $X_0=0$. Eventually, the process will end up in state 2. What is the probability that the time $T=\text{min}\{n\geq 0;X_n=2\}$ is an odd number?
The way I've tried to approach the problem is this: lets define $p_{i,o}$ as the probability that a Markov chain at state $i$ would end up at state $2$, given that it is on an odd-valued step. Let $p_{i,e}$ denote the probability that a Markov chain at state $i$ would end up at state $2$, given that it is on an even-valued step. Then we have the following system of equations: $$p_{0,o}=0.3p_{0,e} + 0.2p_{1,e} + 0.5$$ $$p_{1,o}=0.5p_{0,e} + 0.1p_{1,e} + 0.4$$ $$p_{0,e}=0.3p_{0,o} + 0.2p_{1,o}$$ $$p_{1,e}=0.5p_{0,o} + 0.1p_{1,o}$$ The issue is that when I solve this system of equations, it outputs the incorrect answer. I'm not sure if this is due to a computational error on my part, or if it's due to a fundamental flaw in how I'm doing the first-step analysis. How would you guys approach this problem?
Here is how I would approach this problem. From the total law of probability, $$\begin{eqnarray*}\mathbb{P}(T-\text{odd}|X_0=0) &=& \frac{3}{10}\mathbb{P}(T-\text{odd}|X_0=0,X_1=0)+\frac{1}{5}\mathbb{P}(T-\text{odd}|X_0=0,X_1=1)+\frac{1}{2}\mathbb{P}(T-\text{odd}|X_0=0,X_1=2)\end{eqnarray*}$$ We can write this as $$\mathbb{P}(T-\text{odd}|X_0=0)=\frac{3}{10}\mathbb{P}(T-\text{even}|X_0=0)+\frac{1}{5}\mathbb{P}(T-\text{even}|X_0=1)+\frac{1}{2}$$ Since $\mathbb{P}(T-\text{odd}|X_0=i)+\mathbb{P}(T-\text{even}|X_0=i)=1$ for any $i\in\{0,1,2\}$, the above equation reduces to $$\frac{13}{10}\mathbb{P}(T-\text{odd}|X_0=0)+\frac{1}{5}\mathbb{P}(T-\text{odd}|X_0=1)=1$$ Executing the same logic on $\mathbb{P}(T-\text{odd}|X_0=1)$ yields $$\frac{1}{2}\mathbb{P}(T-\text{odd}|X_0=0)+\frac{11}{10}\mathbb{P}(T-\text{odd}|X_0=1)=1$$ Solving this system yields $\mathbb{P}(T-\text{odd}|X_0=0)=\frac{90}{133}$.