Let {Xn }n≥0 be a Markov chain with state space E = {0, 1, . . .} and transition probabilities given by: p0,0= 1−p0,1=(3/4) pi,i+1 =(1/2)( 1−(1/(i+2))) ∀i≥0 pi,i−1 = (1/2)( 1+(1/(i+2))) ∀i≥1 Determine whether the chain is transient, null recursive, or positive recursive. In the latter case, find the stationary distribution.

To determine if the Markov chain is transient, null recurrent, or positive recurrent, we need to examine the recurrence properties of the chain.

Let's first calculate the probability of return to state 0, denoted as f_{0} :
For the chain to return to state 0, it must transition from state 0 to state 1, then return from state 1 to state 0.
Thus,
f_{0} = p_{0,1} \times p_{1,0} = \left(\frac{3}{4}\right) \times \left(\frac{1}{2}\right) = \frac{3}{8}

Now, let's calculate the probability of return to state 0 after 2 steps, denoted as f_{0}^{(2)} :
For the chain to return to state 0 after 2 steps, it must transition from state 0 to state 1, then transition from state 1 to some state i, and finally transition from state i back to state 0.
Thus,
f_{0}^{(2)} = p_{0,1} \times \sum_{i=1}^{\infty} p_{1,i}p_{i,0} = \left(\frac{3}{4}\right) \times \sum_{i=1}^{\infty} \left(\frac{1}{2}\right)\left(1 - \frac{1}{i+2}\right)\left(\frac{1}{2}\right)\left(1 + \frac{1}{i+2}\right)

Now, observe that \sum_{i=1}^{\infty} \left(1 - \frac{1}{i+2}\right)\left(1 + \frac{1}{i+2}\right) telescopes to 1. Thus,
f_{0}^{(2)} = \frac{3}{4} \times 1 = \frac{3}{4}

As f_{0}^{(2)} is greater than f_{0} , we can conclude that the Markov chain is positive recurrent.

To find the stationary distribution, we solve the balance equations given by \pi = \pi P where \pi is the stationary distribution vector and P is the transition probability matrix.

Writing the balance equations for this Markov chain, we get:
\pi_0 = \pi_0 p_{0,0} + \pi_1 p_{1,0}
\pi_i = \pi_{i-1} p_{i-1,i} + \pi_i p_{i,i} + \pi_{i+1} p_{i+1,i} \text{ for } i \geq 1

For i = 0 , we get:
\pi_0 = \pi_0 p_{0,0} + \pi_1 p_{1,0}
\pi_0 = \pi_0 \left(1 - \frac{3}{4}\right) + \pi_1 \frac{1}{2}
\frac{1}{4} \pi_0 = \frac{1}{2} \pi_1
\pi_1 = \frac{1}{2} \pi_0

For i \geq 1 , we get:
\pi_i = \pi_{i-1} p_{i-1,i} + \pi_i p_{i,i} + \pi_{i+1} p_{i+1,i}
\pi_i = \pi_{i-1} \left(\frac{1}{2}\right) \left(1 + \frac{1}{i+1}\right) + \pi_i \left(\frac{1}{2}\right) \left(1 - \frac{1}{i+1}\right) + \pi_{i+1} \left(\frac{1}{2}\right) \left(1 - \frac{1}{i+2}\right)

Solving the above equations recursively, we get:
\pi_{i+1} = \frac{i+2}{i+1} \pi_i

Using this, we can express \pi_i in terms of \pi_0 as:
\pi_i = \frac{2}{3}\left(\frac{3}{4}\right)^i \pi_0

To find \pi_0 , we use the fact that the sum of all probabilities in the stationary distribution is 1:
\sum_{i=0}^{\infty} \pi_i = 1
\pi_0 \sum_{i=0}^{\infty} \left(\frac{3}{4}\right)^i = 1
\pi_0 \left(1 + \frac{1}{4} + \left(\frac{3}{4}\right)^2 + \ldots \right) = 1
\pi_0 \left(\frac{1}{1 - 3/4}\right) = \pi_0 \times 4 = 1
\pi_0 = \frac{1}{4}

Therefore, the stationary distribution is:
\pi_i = \frac{2}{3}\left(\frac{3}{4}\right)^i \times \frac{1}{4} = \frac{2}{3} \times \left(\frac{3}{4}\right)^{i+1}

\boxed{\pi_i = \frac{2}{3} \left(\frac{3}{4}\right)^{i+1}}