
doi: 10.1007/bf02465540
Let \(F\) be a subexponential distribution function (i.e. \((1-F^{*n} (x))/(1-F(x)) \to n\) as \(x\to\infty\) for all \(n\geq 2)\), assume that \(F\) has a density \(f\), and define the hazard rate function \(q(x)=f(x)/r(x)\), where \(r(x)=1 -F(x)\), \(x>0\). Further, set \(R(x)=\int^x_0r(y)dy\) and \(U(x)=r^2 (x/2)/f (x)R(x)\), \(x>0\). A typical result of this paper, dealing with the asymptotic behaviour of the difference \(R_n(x)= 1-F^{*n}(x)- n(1-F(x))\), is as follows: Assume that \(\lim\sup_{x\to\infty} q(xy)/q(x) 0\), that \(\int^x_0 r(x-y)r(y)dy =O(1)r(x) R(x)\) as \(x\to\infty\), and that \(U(x)=O(1)\) as \(x\to\infty\). Then, for \(n\geq 2\), \(R_n(x)=O(1)f(x)R(x)\) as \(x\to\infty\). The authors discuss also applications of their results to the distance between the distribution functions of sums and maxima of i.i.d. random variables, to transient renewal theory, and to the relation between the tail of an infinitely divisible distribution function and its Lévy measure.
infinitely divisible distribution function, Probability distributions: general theory, Lévy measure, transient renewal theory
infinitely divisible distribution function, Probability distributions: general theory, Lévy measure, transient renewal theory
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