Discussion
Suppose \(f_{ij}^{(n)}\) denotes the probability of the first visit to j from i in n steps. Suppose the Markov chain is periodic with period 2. The following are three statements :
(I) \(f_{ii}^{(2)} = p_{ii}^{(2)} - p_{ii}^{(1)} f_{ii}^{(1)}\)
(II) \(f_{ii}^{(4)} = p_{ii}^{(4)} - p_{ii}^{(2)} f_{ii}^{(2)}\)
(III) \(f_{ii}^{(7)} = 0\)
Which of the following is true ?
Answer: D
All three are true.
The first-passage decomposition gives \(p_{ii}^{(n)} = \sum_{k=1}^{n} f_{ii}^{(k)} p_{ii}^{(n-k)}\), with \(p_{ii}^{(0)} = 1\).
For \(n = 2\): \(p_{ii}^{(2)} = f_{ii}^{(2)} + f_{ii}^{(1)}p_{ii}^{(1)}\), which rearranges to (I). This holds in any chain.
For \(n = 4\):
\[ \begin{aligned} p_{ii}^{(4)} = &f_{ii}^{(4)} + f_{ii}^{(3)}p_{ii}^{(1)} \\ &+ f_{ii}^{(2)}p_{ii}^{(2)} + f_{ii}^{(1)}p_{ii}^{(3)} \end{aligned} \]
With period 2, returns are possible only at even times, so \(p_{ii}^{(1)} = p_{ii}^{(3)} = 0\) and \(f_{ii}^{(1)} = f_{ii}^{(3)} = 0\), leaving (II).
For the same reason a first return at the odd time 7 is impossible, so \(f_{ii}^{(7)} = 0\), which is (III).
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