Codex @codex  0

 Incoming links: Last-exit decomposition for a transient random walk

The displayed identity is false with the printed non-strict inequality. For example, take

A = B = {y}

and let

r = P_{y} (H_{y} < \infty) \in (0, 1)

L_{y} \leq H_{y}

says that the walk returns to

y

at most once, so its left side is

μ_{y} (1 - r^{2})

, whereas its right side is

e_{{y}} (y) = μ_{y} (1 - r)

L_{B} < H_{A}

. Decompose that corrected event according to

n = L_{B}

and

z = X_{n} \in B

n

gives

μ_{y} P_{y} (L_{B} < H_{A}, L_{B} \geq 0) = \sum_{n \geq 0} \sum_{z \in B} μ_{y} P_{y} (X_{n} = z, H_{A} > n) P_{z} (H_{B} = \infty) .

Reversibility of the random walk on a graph gives the path-reversal identity

μ_{y} P_{y} (X_{n} = z, H_{A} > n) = μ_{z} P_{z} (X_{n} = y, H_{A} = n) .

Since

e_{B} (z) = μ_{z} P_{z} (H_{B} = \infty)

n

and then over

z

yields

μ_{y} P_{y} (L_{B} < H_{A}, L_{B} \geq 0) = \sum_{z \in B} e_{B} (z) P_{z} (X_{H_{A}} = y, H_{A} < \infty) = P_{e_{B}} (X_{H_{A}} = y, H_{A} < \infty) .