1 as b 1 giving another solution of exercise 946 from

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Unformatted text preview: lity. (The denominator is the same, while the numerator is rearranged.) Since there are n ways to choose the j draws on which we get j red, ✓ ◆ j 1 P Xn = = for 1 j n + 1 n+2 n+1 and it follows that the distribution of the limit X1 is uniform. Example 5.15. Branching Processes. In this system introduced in Example 1.8, Zn is the number of individuals in generation n and each gives rise to an independent and identically distributed number of individuals with mean 0 &lt; µ &lt; 1 in generation n + 1. If p(x, y ) is the transition probability of the Markov chain X 1X µx p(x, y )f (y, n + 1) = n+1 p(x, y )y = n+1 = h(x, n) µ µ y y so using Theorem 5.5 we see that Wn = Zn /µn is a martingale. Using this we can rederive some of the facts proved in Example 1.52, and prove at least one new one. Subcritical. If µ &lt; 1 then P (Zn &gt; 0) µn EZ0 ! 0 as n ! 1 Proof. Since Zn /µn is a martingale, EZn = µn EZ0 . Using this with P (Zn 1) EZn gives the desired result. Critical. Let pk be the probability an individual h...
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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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