By Hermann Thorisson
It is a booklet on coupling, together with self-contained remedies of stationarity and regeneration. Coupling is the principal subject within the first half the publication, after which enters as a device within the latter part. the 10 chapters are grouped into 4 components.
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Extra resources for Coupling, Stationarity, and Regeneration
D > xn ) Moreover, ℙ(X0 = x0 , … , Xn = xn , Xn+1 = 0) = ℙ(X0 = x0 , … , Xn = xn ) ℙ(D = xn + 1) ℙ(D > xn ) is obtained similarly, or by complement to 1. Hence, the only thing that matters is the age of the component in function, and (Xn )n≥0 is a Markov chain on ℕ with matrix P and graph given by P(x, x + 1) = ℙ(D > x + 1) = ℙ(D > x + 1|D > x) ∶= px , ℙ(D > x) P(x, 0) = ℙ(D = x + 1|D > x) = 1 − px , 1 – p0 p0 0 1 – p1 p1 1 1 – p2 p2 2 x ∈ ℕ, px – 1 ··· px x ··· . 5) FIRST STEPS 35 This Markov chain is irreducible if and only if ℙ(D > k) > 0 for every k ∈ ℕ and ℙ(D = ∞) < 1.
Prove that if Ln reaches ∅ (the empty word), then the gambler wins the sum S. b) Let Xn be the length of the list (or word) Ln for n ≥ 0. Prove that (Xn )n≥0 is a Markov chain on ℕ and give its matrix P and its graph. 5 Three-card Monte Three playing cards are lined face down on a cardboard box at time n = 0. At times n ≥ 1, the middle card is exchanged with probability p > 0 with the card on the right and with probability q = 1 − p > 0 with the one on the left. a) Represent the evolution of the three cards by a Markov chain (Yn )n≥0 .
The game stops when k = 0, and hence, the sum S has been won. ) a) Represent the list evolution by a Markov chain (Ln )n≥0 on the set = ⋃ ℕk k≥0 of words of the form n1 · · · nk . Describe its transition matrix Q and its graph. Prove that if Ln reaches ∅ (the empty word), then the gambler wins the sum S. b) Let Xn be the length of the list (or word) Ln for n ≥ 0. Prove that (Xn )n≥0 is a Markov chain on ℕ and give its matrix P and its graph. 5 Three-card Monte Three playing cards are lined face down on a cardboard box at time n = 0.