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Unformatted text preview: April 14, 2007 HW5 Problem Solutions #4.3.5. Note that in this problem with α ( i ) = δ i, , we have v n = f n , ρ = ∑ k ≥ 1 f k , and u n satisfies the discrete renewal equation with the assumed nonlattice probability vector { f k } . When ρ = 1, we have a proper renewal equation for u n , and the discrete renewal theorem implies that the limit of u n is 1 / ∑ k ≥ 1 kf k . For the rest of the problem, note that: ρ = k ≥ 1 e β θ k 1 ( k 1)! = e θ β , k ≥ 1 e β γ k θ k 1 ( k 1)! = γ e γθ β Since γ e γθ β is monotone increasing in γ , there is a unique value γ such that γ e γθ β = 1, and it is obvious that γ is respectively > 1 , = 1 , < 1 if θ < β , respectively θ = β, θ > β . We saw above that u n has the limit 1 / ∑ k ≥ 1 kf k when θ = β . When θ < β , we find by means of Theorem 3.5 (since v k = f k ) that lim n γ n u n = ∑ ∞ k =0 γ k f k ∑ ∞ k =0 k γ k f k = ∑ k ≥ 1 e β γ k θ k 1 / ( k 1)! ∑ k ≥ 1 e β (1 + ( k 1)) γ k θ k 1 / ( k 1)! The summations in this last ratio can be found analytically, giving the result that γ n u n → 1 / (1 + γθ ), i.e. u n decays geometrically with ratio 1 /γ . In the remaining case, where ρ > 1 or equivalently θ > β , we can also reach a conclusion. Using exactly the same trick as in Theorem 3.5, i.e....
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This note was uploaded on 09/09/2009 for the course STAT 650 taught by Professor Slud during the Spring '09 term at Maryland.
 Spring '09
 SLUD
 Probability

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