Positive Recurrence and Aperiodicity

Drop the term ∑ i1,...in 6=k λinpinin−1 · · · pi1j we obtain λj > Pk(X1 = j, Rk > 1) + Pk(X2 = j, Rk > 2) + · · ·+ Pk(Xn = j, Rk > n) → γ j as n→∞ (where Rk is the first return time of k) So we have λ > γ(k). If in addition P is invariant, then we proved earlier that γ(k) is invariant, so μ = λ − γ(k) is also invariant with μ > 0 and μk = λk − γ (k) k = 0. Since P is irreducible, for any i ∈ I, there is some n such that p ik > 0. Now we have 0 = μk = ∑ j∈I μjp (n) jk > μip (n) ik , so μi = 0, and λ = γ(k). Recall, i ∈ I is recurrent ⇔ Pi(Xn = i infinity often) = 1 ⇔ Pi(Ri < ∞) = 1. Let mi be the expected first return time: mi = Ei(Ri)