I am very new to Markov chains, and during my basic practice I've encountered a few problems that I need clarification of.
I am given a time-homogenous Markov chain $\{X_n\}$ with state space $\{0,1,2,3\}$ with a $4\times4$ transition matrix P, and told that $X_0 \sim$ Bin$(3,0.5)$.
One of the questions is to find the probability of $P(X_0=0 \mid X_1=1, X_2 = 3)$.
In the Markov chains I have come into contact thus far, it has always been such that the final state succeeds the conditional state. That is to say, in $P(X_i=0 \mid X_j=1)$, $i \gt j$ so that I interpret a succeeding state given a preceding state. So how it is possible in this case to find a preceding state only given the succeeding state? This seems to defy intuition for me.
In addition, what role exactly does the Binomial distribution play in this? Does this mean I need to multiply the binomial probability into each state, since that is the probability of that state occurring at that step?
Thank you in advance!
We can use the Bayes theorem,
\begin{align} &P(X_0 = 0|X_1 = 1, X_2=3) \\&= \frac{P(X_1=1, X_2=3|X_0=0)P(X_0=0)}{P(X_1=1, X_2=3)}\\ &=\frac{P(X_1=1, X_2=3|X_0=0)P(X_0=0)}{P(X_1=1, X_2=3|X_0=0)P(X_0=0) +P(X_1=1, X_2=3|X_0=1)P(X_0=1) }\\ &=\frac{P(X_2=3|X_1=1)P(X_1=1|X_0=0)P(X_0=0)}{P(X_2=3|X_1=1)P(X_1=1|X_0=0)P(X_0=0)+P(X_2=3|X_1=1)P(X_1=1|X_0=1)P(X_0=1)}\\ &=\frac{P(X_1=1|X_0=0)P(X_0=0)}{P(X_1=1|X_0=0)P(X_0=0)+P(X_1=1|X_0=1)P(X_0=1)} \end{align}
You are told the the transition probabilities and the distribution of $X_0$. Hence you should have enough unformation to compute the above quantity.