By Menzies A.W.C., Lacoss D.A.

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**Example text**

Let (Tn , n ≥ 1) be the sequence of partial sums of the variables Un . Then the map ∞ Φ: ω∈Ω→ δ(Tn (ω),Xn (ω)),Yn (ω)) , n=1 26 Renewal Reward Processes where δ(x,y,z) is the Dirac measure in (x, y, z), defines a Poisson point process on E = [0, ∞) × [0, ∞) × [0, ∞) with intensity measure ν(dtdxdy) = dtdH(x, y), where H is the joint cdf of X1 and Y1 . Let Mp (E) be the set of all point measures on E. We will denote the distribution of Φ over Mp (E) by Pν . Define for t ≥ 0 the functionals AX (t) and AY (t) on Mp (E) by AX (t)(µ) = 1[0,t) (s)xµ(dsdxdy) E and AY (t)(µ) = E 1[0,t) (s)yµ(dsdxdy).

Then √ converges in probability to 0. ν(t) Assume that the function φ is non-negative and non-decreasing, or bounded. Let Yn := φ(Xn ). Then N (t) N (t) Yn = n=1 N (t)+1 φ(Xn ) ≤ Rφ (t) ≤ n=1 N (t)+1 φ(Xn ) = n=1 Yn . 27) n=1 2 Assume that σX and σY2 are finite. Then using the Central Limit Theorem for random sums, see Embrechts et al. 28) µX µX µ X n=1 where Var Y1 − µY X1 µX t µX 2 − 2µX µY σXY µ2X σY2 + µ2Y σX µ3X = Now we will consider the limiting distribution of C= Note that N (t)+1 n=1 N (t)+1 n=1 t.

Pν×G = P ◦ Φ−1 . 39) where L(t)(µ) = E yh(t − s)µ(dsdy). 1, we obtain ˜ R(t) e− = M+ ([0,∞)) = R E Pν×G (dν)Π(dν) 1 − e−αyh(t−s)1[0,t] (s) ν × G(dsdy) Π(dν) exp − M+ ([0,∞)) = αyh(t−s)1[0,t] (s)µ(dsdy) Mp (E) E t 1 − G∗ (αh(t − s)) ν(ds) Π(dν). exp − 0 M+ ([0,∞)) The last equality follows from the independence assumption between (Sn ) and (Yn ). As an example let M+ ([0, ∞)) = {µ : µ(dt) = λdt, λ ∈ [0, ∞)} and Π be a probability distribution of an exponential random variable with parameter η on M+ ([0, ∞)).

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