# Calculations with a Markov chain

Let $$(X_n)_{n \in N_0}$$ be a Markov chain with the following graph:

(a) Find its probability transition matrix.

(b) Find all stationary (invariant) distributions of the Markov chain.

(c) Let the initial distribution be $$X_0 = (0.5, 0.5, 0)$$. Calculate the distribution of $$X_2$$ and $$P(X_0·X_2 = 1)$$.

Here's my work:

(a) $$P = \begin{bmatrix}0 & 0.5 & 0.5 \\0.5 & 0.5 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$$

(b) Since $$3$$ is the only absorbing state, the stationary distribution has to be $$(0, 0, 1).$$

(c) The distribution of $$X_2 = X_0 \cdot P^2 = (\dfrac{1}{4}, \dfrac{3}{6}, \dfrac{3}{8}).$$

I'm not sure how to solve for $$P(X_0 \cdot X_2 = 1)$$. Is the question looking for $$P(X_0 = 1, X_2 = 1)?$$

The statement of the question is a little muddled. If the Markov chain is $$\ \big(X_n\big)_{n\in N_0}\$$ with the transitions illustrated in the given graph and initial distribution $$\ \pi_0=(0.5, 0.5,0)\$$, then $$\ X_0\$$ must be the state at time $$0$$, a random variable which assumes the value $$0$$ with probability $$0.5$$ and the value $$1$$ with probability $$0.5$$. It cannot be its own distribution, so using the same symbol "$$X_0$$" to represent that distribution is a confusing misuse of notation, which makes it a little puzzling as to what $$\ P\big(X_0\cdot X_2=1\big)\$$ is supposed to mean.
However, your interpretation of it is the only one I can see that makes any sense. Taking $$\ X_0\$$ to be the state of the chain at time $$0$$ (rather than the distribution of that state), and $$\ X_2\$$ to be its state at time $$2$$, the only way $$\ X_0\cdot X_2\$$ can be $$1$$ is if $$\ X_0\$$ and $$\ X_2\$$ are both $$1$$.