8H (Markov Chains)

Expand according to the hitting time and use detailed balance to reverse each finite path:

P_{π} (X_{T_{A}} = z) = n \geq 0 \sum x_{0}, \dots, x_{n - 1} \in / A \sum π (x_{0}) P (x_{0}, x_{1}) \dots P (x_{n - 1}, z) = π (z) n \geq 0 \sum P_{z} (X_{1}, \dots, X_{n} \in / A) .

The event in the final probability is

{T_{A}^{+} > n}

\sum_{n \geq 0} P_{z} (T_{A}^{+} > n) = E_{z} T_{A}^{+} .

Therefore

P_{π} (X_{T_{A}} = z) = π (z) E_{z} T_{A}^{+} .

b

Sum the identity from part (a) over

z \in A

. Since

X_{T_{A}} \in A

almost surely under positive recurrence,

1 = \sum_{z \in A} π (z) E_{z} T_{A}^{+} .

Using the conditional stationary law

π_{A} (z) = π (z) / π (A)

gives

E_{π_{A}} T_{A}^{+} = \frac{1}{π ( A )} \sum_{z \in A} π (z) E_{z} T_{A}^{+} .

E_{π_{A}} T_{A}^{+} = \frac{1}{π ( A )} .

There are no discussions about this article yet.

There are currently no matching articles.