Both Uk and Vk are functions of the Markov process {Yk : k = 0,1,...}, so in particular, both {Un} and {Vn} depend on the initial value Y0. For clarity one should note that Uk has mean zero when Y0 0 and Vk has mean zero when Y0 follows the ≡ stationary distribution ν. The random variables and differ by a that depends on and by Uk Vk constant k, (14) the constant is not larger than O(δk). Thus, by the representation (10), we have . Now, we expand the sum in (17) and write . To estimate the first sum An of (18), we apply (13) just as we did in the derivation of (14), and this time we find . Summation then gives us . To deal with the double sum Bn, we first need a lemma to help us estimate some of the summands of Bn. Lemma 7. For any initial distribution µ one has = O(δj) for all i,j 0. Eµ[ViVi+j] ≥ Proof. To exploit the Markov property for the chain {Yn : n = 0,1,...}, we first condition on Yi−1 and Yi and note that = |Yi−1,Yi]] = |Yi]] Eµ[ViVi+j] Eµ[ViEµ[Vi+j Eµ[ViEµ[Vi+j = Eµ[ViEYi[Vj]]. If we use (13) as before, then we see that for all x X we have ∈ = Ex[Vn] Eν[Vn] + O(δn), and the implied constant does not depend on x. When we insert this into (21) and recall that the definition of Vn gives us Eν[Vn] = 0, the proof of the lemma is complete. Lemma 7 helps us deal with cross terms with large j, but we also need a relation that deals with arbitrary j. Here, we again use (13) to get for all j 0 that ≥ E0[ViVi+j] = Eν[ViVi+j] + O(δi) = Eν[V1V1+j] + O(δi).
Both Uk and Vk are functions of the Markov process {Yk : k = 0,1,...}, so in particular, both {Un} and {Vn} depend on the initial value Y0. For clarity one should note that Uk has mean zero when Y0 ≡ 0 and Vk has mean zero when Y0 follows the stationary distribution ν. The random variables Uk and Vk differ by a constant that depends on k, and by (14) the constant is not larger than O(δk). Thus, by the representation (10), we have . Now, we expand the sum in (17) and write . To estimate the first sum An of (18), we apply (13) just as we did in the derivation of (14), and this time we find . Summation then gives us . To deal with the double sum Bn, we first need a lemma to help us estimate some of the summands of Bn. Lemma 7. For any initial distribution µ one has Eµ[ViVi+j] = O(δj) for all i,j ≥ 0. Proof. To exploit the Markov property for the chain {Yn : n = 0,1,...}, we first condition on Yi−1 and Yi and note that Eµ[ViVi+j] = Eµ[ViEµ[Vi+j |Yi−1,Yi]] = Eµ[ViEµ[Vi+j |Yi]] = Eµ[ViEYi[Vj]]. If we use (13) as before, then we see that for all x ∈ X we have Ex[Vn] = Eν[Vn] + O(δn), and the implied constant does not depend on x. When we insert this into (21) and recall that the definition of Vn gives us Eν[Vn] = 0, the proof of the lemma is complete. Lemma 7 helps us deal with cross terms with large j, but we also need a relation that deals with arbitrary j. Here, we again use (13) to get for all j ≥ 0 that E0[ViVi+j] = Eν[ViVi+j] + O(δi) = Eν[V1V1+j] + O(δi).
Peng-Steele-Analogs.docx