Cramer-Lundberg risk model

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What do the sample paths of process (11) look like?

Example 3. On some probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}) consider two sequences of nonnegative random variables {ξj,j∈N}\left\{ \xi_{j}, j \in \mathbb {N}\right\} and {ηj,j∈N}\left\{ \eta_{j}, j \in \mathbb {N}\right\} such that {ξj,ηj,j∈N}\left\{ \xi_{j}, \eta_{j}, j \in \mathbb {N}\right\} are jointly independent and Law⁡(ξj)=Law⁡(ξ1)\operatorname {Law}\left(\xi_{j}\right) = \operatorname {Law}\left(\xi_{1}\right), Law⁡(ηj)=Law⁡(η1)\operatorname {Law}\left(\eta_{j}\right) = \operatorname {Law}\left(\eta_{1}\right), j∈Nj \in \mathbb {N}. For positive constants y0y_{0} and cc define the process

Yt(ω)=y0+ct−∑j=1Xt(ω)ηj(ω),t⩾0,(11) Y_{t}(\omega ) = y_{0}+c t-\sum _{j = 1}^{X_{t}(\omega )} \eta _{j}(\omega ), \quad t \geqslant 0, \tag {11}

where Xt(ω)X_{t}(\omega ) is given by formula (10).

Example 2. For a sequence ξ1,ξ2,…\xi_{1}, \xi_{2}, \ldots of independent identically distributed nondegenerate nonnegative random variables, the renewal process is defined by the formula

X0(ω)=0,Xt(ω)=sup⁡{n:∑j⩽nξj(ω)⩽t},t>0.(10) X_{0}(\omega ) = 0, \quad X_{t}(\omega ) = \sup \left\{ n: \sum _{j \leqslant n} \xi _{j}(\omega ) \leqslant t\right\} , \quad t > 0. \tag {10}

Throughout, the sum over the empty set is taken to be zero, so Xt(ω)=0X_{t}(\omega ) = 0 if ξ1(ω)>t\xi_{1}(\omega ) > t.

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Let r(y)r(y) be the ruin function of the insurance problem (8.10.7). In the notation of that problem, show that the Laplace-Stieltjes transforms r∗(θ)r^{*}(\theta ) and F∗(θ)F^{*}(\theta ) are related by λF∗=r∗(θ−λ+λF∗)\lambda F^{*} = r^{*}\left(\theta -\lambda +\lambda F^{*}\right).

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An insurance company pays independent and identically distributed claims {Kn:n≥1}\left\{ K_{n}: n \geq 1\right\} at the instants of a Poisson process with intensity λ\lambda, where λE[K1]<1\lambda \mathbb {E}\left[K_{1}\right] < 1. Premiums are received at constant rate 1. Show that the maximum deficit MM the company will ever accumulate has moment generating function

E[eθM]=(1−ρ)θλ+θ−λE[eθK] \mathbb {E}\left[e^{\theta M}\right] = \frac{(1-\rho ) \theta }{\lambda +\theta -\lambda \mathbb {E}\left[e^{\theta K}\right]}
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Let £Yn£ Y_{n} be the assets of an insurance company after nn years of trading. During each year it receives a total (fixed) income of £P£ P in premiums. During the nnth year it pays out a total of £Cn£ C_{n} in claims. Thus Yn+1=Yn+P−Cn+1Y_{n+1} = Y_{n}+P-C_{n+1}. Suppose that C1,C2,…C_{1}, C_{2}, \ldots are independent N(μ,σ2)N\left(\mu , \sigma^{2}\right) variables and show that the probability of ultimate bankruptcy satisfies

P(Yn≤0 for some n)≤exp⁡{−2(P−μ)Y0σ2} \mathbb {P}\left(Y_{n} \leq 0 \text{ for some } n\right) \leq \exp \left\{ -\frac{2(P-\mu ) Y_{0}}{\sigma ^{2}}\right\}
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An insurance company receives premiums (net of costs) at rate ρ\rho per unit time. Claims X1,X2,…X_{1}, X_{2}, \ldots are independent random variables with the exponential distribution with parameter μ\mu, and they arrive at the times of a Poisson process of rate λ\lambda (assume the usual independence, as well as λ,μ,ρ>0)\lambda , \mu , \rho > 0). Let Y(t)Y(t) be the assets of the company at time tt where Y(0)=y>0Y(0) = y > 0. Show that

P(Y(t)≤0 for some t>0)={(1−θμ)e−θy if θ>01 otherwise  \mathbb {P}\left(Y(t) \leq 0 \text{ for some } t > 0\right) = \begin{cases} \left(1-\frac{\theta }{\mu }\right) e^{-\theta y} & \text{ if } \theta > 0 \\ 1 & \text{ otherwise }\end{cases}

where θ=μ−(λ/ρ)\theta = \mu -(\lambda / \rho ).

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References

[БуШи05] Булинский А.В., Ширяев А.Н. Теория случайных процессов. ФИЗМАТЛИТ, 2005.
[GrSt20] Grimmett G.R., Stirzaker D.R. Probability and Random Processes. 4th ed., Oxford University Press, 2020.