an:05883114 Zbl 1229.91171 Zhang, Zhimin; Yang, Hailiang; Yang, Hu On the absolute ruin in a map risk model with debit interest EN Adv. Appl. Probab. 43, No. 1, 77-96 (2011). 00277675 2011
j
91B30 60J28 91B70 Markov additive processes; absolute ruin; discounted penalty function; matrix renewal equation; asymtotics; heavy-tailed distribution 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 $$y \neq 0$$ $\limsup_{x \to \infty} {f(x+y) \over f(x)} = \limsup_{x \to \infty} {f(x)\over f(x+y)} = \limsup_{x \to \infty} f(x) = \infty\;.$ Hanspeter Schmidli (K??ln)