By Nicolas Privault

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**Extra resources for Notes on Markov chains**

**Example text**

6) and the change of variable (k, n) → (k, l) with l = n + k. Hence Gτ0 (s) = 1 − 1 = 1 − (1 − 4pqs2 )1/2 , H(s) 4pqs2 < 1. 10) n=0 can now be used to determine the probability distribution P(τ0 = n) of τ0 , as follows: ∞ 2 We used the formula (1 + x)α = k=0 xk α(α − 1) × · · · × (α − (k − 1)), cf. 4). k! html Notes on Markov Chains Gτ0 (s) = 1 − (1 − 4pqs2 )1/2 ∞ = 1− k=0 ∞ s2k = k=1 1 − 1 × ··· × 2 1 1 (−4pqs2 )k k! 2 (4pq)k 1 k! 4) for α = 1/2. P(τ0 = 2k) = (4pq)k 1 k! 2 (4pq)k = 2k! 10) we find 1− 1 2 k−1 m− m=1 × ··· × k − 1 − 1 2 , 1 2 k ≥ 1, while P(τ0 = 2k + 1) = 0, k ∈ N.

Discrete Distributions Next, let X : Ω → N be a discrete random variable. The expectation IE[X] of X is defined as the sum ∞ IE[X] = kP(X = k), k=0 in which the possible values k ∈ N of X are weighted by their probabilities. More generally we have ∞ IE[φ(X)] = φ(k)P(X = k), k=0 for all sufficiently summable functions φ : N → R. The expectation of the indicator function X = 1A can be recovered as IE[1A ] = 0 × P(Ω \ A) + 1 × P(A) = P(A). e. 7) provided IE[|X|] + IE[|Y |] < ∞. html N. 8) where we used the relation ∞ X= k1{X=k} k=0 which holds since X takes only integer values.

1). html Notes on Markov Chains ∞ Gτ0 (s) := IE[sτ0 1{τ0 <∞} ] = ∞ sn P(τ0 = n) = n=0 sn g(n), −1 ≤ s ≤ 1. 8) Computing Gτ0 (s) provides a number of informations on τ0 , such as P(τ0 < ∞) = IE[1{τ0 <∞} ] = Gτ0 (1) and ∞ nP(τ0 = n) = Gτ0 (1− ). IE[τ0 1{τ0 <∞} ] = n=1 Our aim is now to compute Gτ0 (s) for all s ∈ [−1, 1]. 6) implies that Gτ0 (s)H(s) = H(s) − 1, s ∈ [−1, 1]. html N. Privault ∞ (−4pqs2 )k = k=0 (−1/2) × (−3/2) × · · · (3/2 − k) × (−1/2 − (k − 1)) k! 6) and the change of variable (k, n) → (k, l) with l = n + k.