# The linear combinations i1 xi aij when viewed as

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Unformatted text preview: f we only allow one individual to die and be replaced by its o↵spring on each step. If the o↵spring distributions is pk and the generating function is then the random walk increments have P (Xi = k 1) P pk . Let Sk = 1 + X1 + . . . + Xn and = T0 = min{n : Sn = 0}. Suppose µ = k kpk &gt; 1. Use Example 5.12 to show that P (T0 &lt; 1) = ⇢, the solution &lt; 1 of (⇢) = ⇢. 5.15. LetP n be a branching process with o↵spring distribution pk with p0 &gt; 0 Z P1 and µ = k kpk &gt; 1. Let (✓) = k=0 pk ✓k . (a) Show that E (✓Zn+1 |Zn ) = (✓)Zn . (b) Let ⇢ be the solution &lt; 1 of (⇢) = ⇢ and conclude that Pk (T0 &lt; 1) = ⇢k 5.16. Hitting probabilities. Consider a Markov chain with ﬁnite state space S . Let a and b be two points in S , let ⌧ = Va ^ Vb , and let C = S {a, b}. Suppose h(a) = 1, h(b) = 0, and for x 2 C we have X h(x) = p(x, y )h(y ) y (a) Show that h(Xn ) is a martingale. (b) Conclude that if Px (⌧ &lt; 1) &gt; 0 for all x 2 C , then h(x) = Px (Va &lt; Vb ) giving a proof of...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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