Markov Chain Basics

A Markov chain is a stochastic process that satisfies the Markov property. This means that the future state depends only on the current state, not on the sequence of events that preceded it. In this problem, we have a two-state Markov chain with state space S = {0, 1}. The transitions between these states are governed by a transition matrix P.

The given transition matrix P is:

The entry \(P_{ij}\) in the matrix represents the probability of transitioning from state \(i\) to state \(j\) in one step.

Expected Return Time Calculation

The question asks for the expected return time to state 0, assuming the chain starts in state 0 (X0 = 0). The expected return time to a state \(i\), denoted by E[T\(_{i}\) | X\(_0\)=\(i\)], is the expected number of steps until the chain returns to state \(i\) for the first time, given it started in state \(i\).

For an irreducible and aperiodic Markov chain, the expected return time to state \(i\) is the reciprocal of the stationary probability of being in state \(i\). That is, \(E[T_i | X_0=i] = \frac{1}{\pi_i}\), where \(\pi_i\) is the stationary probability of state \(i\).

Stationary Distribution

To find the expected return time to state 0, we first need to find the stationary distribution \(\pi = [\pi_0, \pi_1]\) of the Markov chain. The stationary distribution satisfies the equation \(\pi P = \pi\) and the normalization condition \(\pi_0 + \pi_1 = 1\).

The matrix equation \(\pi P = \pi\) can be written as:

• \(\pi_0 P_{00} + \pi_1 P_{10} = \pi_0\)
• \(\pi_0 P_{01} + \pi_1 P_{11} = \pi_1\)

Substituting the values from the transition matrix P:

• \(\pi_0 \left(\frac{1}{2}\right) + \pi_1 \left(\frac{1}{3}\right) = \pi_0\)
• \(\pi_0 \left(\frac{1}{2}\right) + \pi_1 \left(\frac{2}{3}\right) = \pi_1\)

Let's use the first equation to find a relationship between \(\pi_0\) and \(\pi_1\):

\(\frac{1}{2}\pi_0 + \frac{1}{3}\pi_1 = \pi_0\)

\(\frac{1}{3}\pi_1 = \pi_0 - \frac{1}{2}\pi_0\)

\(\frac{1}{3}\pi_1 = \frac{1}{2}\pi_0\)

Multiplying by 6 to clear the denominators:

\(2\pi_1 = 3\pi_0\)

Now, we use the normalization condition \(\pi_0 + \pi_1 = 1\). Substitute \(\pi_1 = \frac{3}{2}\pi_0\) into this equation:

\(\pi_0 + \frac{3}{2}\pi_0 = 1\)

\(\frac{2\pi_0}{2} + \frac{3\pi_0}{2} = 1\)

\(\frac{5\pi_0}{2} = 1\)

\(5\pi_0 = 2\)

\(\pi_0 = \frac{2}{5}\)

With \(\pi_0 = \frac{2}{5}\), we can find \(\pi_1\):

\(\pi_1 = 1 - \pi_0 = 1 - \frac{2}{5} = \frac{3}{5}\)

So, the stationary distribution is \(\pi = \left[\frac{2}{5}, \frac{3}{5}\right]\).

Expected Return Time to State 0

Now that we have the stationary probability for state 0, \(\pi_0 = \frac{2}{5}\), we can calculate the expected return time to state 0 using the formula \(E[T_0 | X_0=0] = \frac{1}{\pi_0}\).

\(E[T_0 | X_0=0] = \frac{1}{\frac{2}{5}} = \frac{5}{2}\)

Thus, the expected return time to state 0 is \(\frac{5}{2}\).