An Introduction to Heavy-Tailed and Subexponential by Sergey Foss, Dmitry Korshunov, Stan Zachary

By Sergey Foss, Dmitry Korshunov, Stan Zachary

This monograph presents a whole and accomplished creation to the speculation of long-tailed and subexponential distributions in a single measurement. New effects are provided in an easy, coherent and systematic manner. the entire common houses of such convolutions are then got as effortless effects of those effects. The ebook makes a speciality of extra theoretical facets. A dialogue of the place the components of purposes presently stand in incorporated as is a few initial mathematical fabric. Mathematical modelers (for e.g. in finance and environmental technology) and statisticians will locate this ebook helpful.

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Extra resources for An Introduction to Heavy-Tailed and Subexponential Distributions

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We show first that the class S∗ of strong subexponential distributions is a subclass of the class L of long-tailed distributions on R. 23. Let the distribution F on R belong to S∗ . Then F is long-tailed. Proof. Since, for x ≥ 2, x 0 F(x − y)F(y)dy ≥ 2F(x) 1 0 F(y)dy + 2F(x − 1) x/2 F(y)dy, 1 the inclusion F ∈ S∗ implies (F(x − 1) − F(x)) where again m = x/2 1 1 x F(x − y)F(y)dy − F(x) 2 0 = mF(x) − mF(x) + o(F(x)) F(y)dy ≤ ∞ 0 F(y)dy. 22 that F is long-tailed. 7. 24. Let F be a distribution on R.

On the other hand, G(x) ≤ F ∗n (x) ∼ nF(x). Hence the distributions F and G are weakly tail-equivalent. 11, as x → ∞, F ∗ G(x) ≥ (1 + o(1))(F(x) + G(x)) = F(x) + G(x) + o(F(x)). Recalling that F ∗ G(x) = F ∗n (x) ∼ nF(x), we deduce the following upper bound: F ∗(n−1) (x) = G(x) ≤ (n − 1 + o(1))F(x). 6) this implies that F ∗(n−1)(x) ∼ (n − 1)F(x) as x → ∞. By induction we deduce then that F ∗2 (x) ∼ 2F(x), which completes the proof. 4 The Class S∗ of Strong Subexponential Distributions We have already observed that a heavy-tailed distribution F on R+ is subexponential if and only if it is long-tailed and its tail is sufficiently regular that limx→∞ F ∗ F(x)/F(x) exists (and that this limit is then equal to 2).

Have common mean a > 0 and finite variance. Fix ε > 0. 9 Comments 37 √ N and A such√that P{|Sn − na| ≤ A n} ≥ 1 − ε for all n ≥ N. It follows from the definition of x-insensitivity that there is n0 such that √ F(na ± A n)) − 1 ≤ ε for all n ≥ n0 . 52). To show (ii)⇒(i) assume that the independent identically distributed random variables ξ1 , ξ2 , . . have common mean a >√0 and finite variance σ 2 > 0, but that, on the contrary, the distribution F fails to be x-insensitive. Then there exists ε > 0 and an increasing sequence nk such that, for all k, F(nk a + nk σ 2 ) ≤ (1 − ε )F(nk a).

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