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On the absolute ruin in a map risk model with debit interest. (English) Zbl 1229.91171
A Markov-additive risk model \(\{U_t,J_t\}\) is considered, where \(U_t\) is the surplus and \(J_t\) is the state of the Markov process. If the surplus becomes negative, interest at rate \(r\) has to be paid for the deficit. The time of absolute ruin \(T\) is the first time where the payments for interest are larger than the premium income. The quantity of interest is the discounted penalty function \[ \Phi_{i j}(u) = E_{(u,i)} [e^{-\delta T} w(U_{T-}-c/r, c/r- U_T) I_{J_T = j}]\;, \] where \(c\) is the premium rate, \(w\) is a bounded measurable function, and \(I\) is the indicator function. The usual integro-differential equations are proved. In a complicated way, the boundary condition \(\Phi_{i j}(-c/r)\) is found. It would have been simpler to observe that \(\Phi\) is continuous in \(-c/r\) and that starting in \(u = -c/r\) means that absolute ruin occurs at the first claim time with \(U_{T-} = -c/r\).
Heavy-tailed claim sizes are then considered. Four classes are introduced: Subexponential and long-tailed distributions as well subexponential and long-tailed density functions. The asymptotic behaviour of \(\Phi\) for the case of subexponential distributions and subexponential densities is calculated.
Unfortunately, the paper contains some errors. For example, in the proof of Lemma 4 it is claimed that \(F\) long-tailed implies that also the density \(f\) of \(F\) is long-tailed. The following simple example gives a long-tailed distribution with an infinitely often differentiable density. Let \(h(x) = C \exp\{-1/(1-x^2)\}I_{|x| < 1}\) with \(C\) chosen such that \(\int_{-1}^1 h(x) \text{d} x = 1\). Then \[ f(x) = \text{\({1\over 2}\)} e^{-x} + \sum_{n=1}^\infty n h(2 n^2(n+1)(x-n^2))\;. \] Since \(e^x\) is not heavy-tailed, it does not matter asymptotically. The weight close to \(n^2\) is approximately \(1\over 2n(n+1)\). Thus the tail of \(F\) is \[ \sum_{n=x}^\infty {1\over 2n(n+1)} \sim \int_x^\infty {1\over 2y(y+1)} \text{d}y = {1\over 2} \log {x+1 \over x} \sim {1\over 2 x}\;. \] This proves that \(F\) is long-tailed. But for all \(y \neq 0\) \[ \limsup_{x \to \infty} {f(x+y) \over f(x)} = \limsup_{x \to \infty} {f(x)\over f(x+y)} = \limsup_{x \to \infty} f(x) = \infty\;. \]

91B30 Risk theory, insurance (MSC2010)
60J28 Applications of continuous-time Markov processes on discrete state spaces
91B70 Stochastic models in economics
Full Text: DOI
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