When doing the reproduction please pay attention to the following: . You should submit a printed hardcopy by the time gi
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When doing the reproduction please pay attention to the following: . You should submit a printed hardcopy by the time gi
• If there are numbered theorems, sections, equations etc. then the numbering in your copy could be different from the original. But you must use numbered environments (when numbering is needed) and the numbering should be done by LaTeX. If you can figure out how to make LATEX produce numbers exactly as in the original documents, you will get bonus for that. • If there are references to numbered items within the page, you must use the label/ref system to make the references. Do not enter reference numbers by hand. . If there are referecences to items outside the page you are given, or citations of sources, then you may enter the reference numbers by hand (because there is no way to create those objects inside your document).
246 Random walks, Markov chains and capacity Proof of Theorem &.2.4 The right hand inequality in (8.2.1) follows from an entrance time decomposition. Lett be the first hitting time of A and let v be the hitting measure V(x) = PolX=rfor XE A. Note that v may be defective, i.c., of total mass less than 1. In fact, V(A) = Pp En 20:X, EA. (8.2.4) Now for all yEΛ: G(x,y)dv(x) = {P.!X==xG(x,y) =G{p.y). Thus (x,y) dv(x) = 1 for every y E A. Consequently 6x (cm) - S, GP) VA G(x,y) dv(s)dv= v(A) **&*(v)=v(A), () so that Cap: (A) V(A). By (8.2.4), this proves the right half of (8.2.1). To establish the left hand inequality in (8.2.1) we use the second moment method. Given a probability measure on A, consider the random variable 2- G(p.x) 'Exy) du(y) By Fubini's Theorem and the definition of G. EZ=E/G(0.9) ["ixuj dulu) = (10.)" ] P(X. = y.X – p}dt (3) G(0.3) 'G(P-9)du (v) =1 Thus by Cauchy-Schwarz 1 =(EZ)?<E,(?),(lz>o), and hence Pp[En 20:X, EA! P(Z >0) > E(2) Now we bound the second moment: EZ = E, JG(P,») "'G(px) { **-**=} du(x)du (v) < 2E, JJ, G1p.v)"G(0.4)** £ xx=du(e) du(). OSRO