Markov Chain: Using one-step analysis to calculate the probability to reach state $5$ before reaching state $0$ when starting in state$ 3.$

Question

I have an alternating random walk question where the stochastic process ${X_n : n = 0, 1, 2, . . .}$ is a Markov chain on the integers with $X_0 = 0$. If $|X_n|$ is even, the conditional probabilities $P(X{n+1} = i + 1|X_n = i) = 3/4$ and $P(X_{n+1} = i − 1|X_n = i) = 1/4 $ are given, whereas if $|X_n|$ is odd, the conditional probabilities are$ P(X_n+1 = i + 1|X_n = i) = 1/4$ and $P(X_{n+1}= i − 1|X_n = i) = 3/4.$ I have been asked to use a one-step analysis to calculate the probability to reach state $5$ before reaching state $0$ when starting in state $3$.

I thought of this question as a Gambler's Ruin problem, but i'm not sure if i'm correct. I defined $A_j$ as the event of reaching state $5$ starting in state $j=3$ before reaching state $N=0$. So there are two absorbing states $5$ and $0$. So $P(A_3)= \theta_3$ with $\theta_5 =1$ (since probability of reaching 5 when starting at 5 is equal to 1) and using one step analysis i have that $\theta_3 = 3/4\theta_4 +1/4\theta_2 $ from $\theta_j = p\theta_{j+1} +q\theta_{j-1} $ since they're even and from the question i have $p=3/4$ and $q=1/4$. How should i find $\theta_2, \theta_4$? Would greatly appreciate the help

Answer 1

You would find $\theta_2$ and $\theta_4$ in the same way as you found $\theta_3$, so in this case $\theta_2 = 3/4 \theta_3 + 1/4 \theta_1$ and $\theta_4 = 3/4 \theta_5 + 1/4 \theta_1$. This will give you 4 equations in the 4 unknowns $\theta_1, \theta_2, \theta_3, \theta_4$, which you can then solve with linear algebra.