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**Example text**

Let x1 , x2 , … ∈ [0, ∞) and define ∞ ∑ ????xi , ????= i=1 where ????xi is the Dirac measure at xi ; that is, for every A ∈ [0,∞) , we have that ????xi (A) = 1 if xi ∈ A or ????xi (A) = 0, otherwise. The measure ???? is said to be a counting measure on [0, ∞). Let p ([0, ∞)) be the space of counting measures on ([0, ∞), [0,∞) ). We equip this space with the vague topology, that is, ????n → ???? in p ([0, ∞)) whenever ????n (????) → ????(????) for any continuous function ???? ∶ [0, ∞) → ℝ with compact support. A point process N on [0, ∞) is a random element of p ([0, ∞)); that is, let (X, X , ????) be a probability space, then any measurable map N ∶ X → p ([0, ∞)) is a point process on [0, ∞).

2 Absence of Clusters In the independent case, the exceedances appear scattered on the time line and the waiting times between them are (asymptotically) exponentially distributed. 1) converges in distribution to a standard Poisson process N of intensity 1. Let us consider the general stationary stochastic processes, where conditions D(un ) and D′ (un ) hold. This implies that there is no clustering of exceedances. In this case, the waiting times between exceedances assume exactly the same exponential pattern as in the independent case.

We define the first hitting time rA ∶ → ℕ ∪ {∞} by rA (x) ∶= inf{n ∈ ℕ ∶ T n (x) ∈ A}. 1) If x never enters A under iteration by f then rA (x) = ∞. When the point x ∈ A then we say that rA (x) is the first return of x to A. Kac’s theorem gives that the mean return time to A is equal to the reciprocal of ℙ(A). Hence, at least in average, rA (x) should go to ∞ as ℙ(A) goes to zero. A natural question then is if we can find a limit distribution for rA , when conveniently normalized. By Kac’s theorem the natural candidate to normalize rA would be ℙ(A).