Lectures from Markov Processes to Brownian Motion (Springer by Kai Lai Chung

By Kai Lai Chung

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Extra info for Lectures from Markov Processes to Brownian Motion (Springer Advanced Texts in Life Sciences)

Example text

D Remark. Let us change notation and put Then {At} is an increasing process, namely for each w, At(w) increases with t; and Ar. = Y x by definition of the latter. Tt! Vb 1. t -- EJA ( y i: - A t· (9) This is a simple case of the Doob-Meyer decomposition of a supermartingale into a uniformly integrable martingale minus an increasing process. 4, the supermartingale in (9) has a right continuous version. It is in fact a superpotential of dass D (exercise). 2. 1, we turn our attention to a dass of transition semigroups having nice analytic properties.

3. For a Markov chain suppose that there exists an i such that lim t10 [I Pii(t)J/t = + w. (Such astate is called instantaneous. It is not particularly easy to construct such an example; see Chung [2J, p. ) Show that for any 6 > 0 we have pi( X(t) = i for all tE [0, 6J] = O. Thus under pi, no version of the process can be right continuous at t = O. Hence (Pt) cannot be Fellerian. Prove the last assertion analytically. 4. Prove that in the definition of Feiler property, condition (2) implies condition (I).

8) Since Yü is bounded, {1";, ~} is a martingale. It is an easy proposition that for any integrable Y, the family E {Y I-;§} where -;§ ranges over all O"-fields of 48 2. , P), is uniformly integrable. Here is the proof: r JIIE(YI(1)12:/l1 IE(YI~§)ldP< - p{IE(YI~§)1 ~ r Jlllc(YIWll2:/I} IYldP', 1 11) ::;; - E(IYI) -> O. 11 Hence {r;} is uniformly integrable by (8). Finally, the first term on the right side of (7) is continuous in t, hence it is progressively measurable as a process. On the other hand since U a!

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