# The Exponential Estimate of the Ultimate Ruin Probability for the Non-Homogeneous Renewal Risk Model

^{*}

## Abstract

**:**

## 1. Introduction

**Theorem**

**1.**

## 2. Results

**Theorem**

**2.**

**Theorem**

**3.**

**Theorem**

**4.**

## 3. Proofs

**Proof**

**of**

**Theorem**

**2.**

**Proof**

**of**

**Theorem**

**3.**

**Proof**

**of**

**Theorem**

**4.**

## 4. Numerical Examples

**Example**

**1.**

${Z}_{k}$ | 0 | 1 | k |

$\mathbb{P}$ | $1-1/3k-1/3{k}^{2}$ | $1/3k$ | $1/3{k}^{2}$ |

- First of all, we find rough estimate of ruin probability according to Theorem 3. After some calculations, we get that$$\begin{array}{c}\underset{k\in \mathbb{N}}{sup}\mathbb{E}({Z}_{k}-p{\theta}_{k})=\underset{k\in \mathbb{N}}{sup}\left(\frac{2}{3k}-1\right)=-\frac{1}{3},\hfill \\ \underset{k\in \mathbb{N}}{sup}\mathbb{E}\left({\mathrm{e}}^{\gamma ({Z}_{k}-p{\theta}_{k})}1\phantom{\rule{-2.9pt}{0ex}}{\mathrm{I}}_{\{{Z}_{k}-p{\theta}_{k}>0\}}\right)=max\left\{0,\frac{{\mathrm{e}}^{\gamma}}{12},\frac{{\mathrm{e}}^{2\gamma}}{27},\frac{{\mathrm{e}}^{3\gamma}}{48},\frac{{\mathrm{e}}^{4\gamma}}{75}\right\}=\frac{{\mathrm{e}}^{\gamma}}{12}\phantom{\rule{4.pt}{0ex}}\mathrm{for}\phantom{\rule{4.pt}{0ex}}\mathrm{all}\phantom{\rule{4.pt}{0ex}}\gamma \in \left(0,61/100\right]\phantom{\rule{-0.166667em}{0ex}},\hfill \\ \underset{k\in \mathbb{N}}{sup}\mathbb{E}\left({\theta}_{k}1\phantom{\rule{-2.9pt}{0ex}}{\mathrm{I}}_{\left\{{\theta}_{k}>1\right\}}\right)=0.\hfill \end{array}$$According to the above estimates, we obtain that conditions of Theorem 3 hold with parameters $\alpha =1/3,\phantom{\rule{4pt}{0ex}}\gamma =3/5,\phantom{\rule{4pt}{0ex}}\beta ={\mathrm{e}}^{3/5}/12,\phantom{\rule{4pt}{0ex}}\varkappa =1$ and $\epsilon =0$. Substituting the obtained constants into expression (2), we get that the last parameter $\delta \in \left(0,41/100\right]$. Choosing $\delta =2/5$, we get from the assertion of Theorem 3 that$$\psi \left(u\right)\u2a7d{\psi}_{1}\left(u\right):={\mathrm{e}}^{-6u/25}$$The derived estimate is exponential but quite conservative. For instance, it follows from the obtained inequality that $\psi \left(u\right)\u2a7d0.02$, if $u\u2a7e17$.
- Now, we find more precise upper bound of ruin probability on the basis of the Theorem 4. All theorem’s conditions hold due to the derived above estimates and$$\begin{array}{cc}\hfill \underset{k\in \mathbb{N}}{sup}\mathbb{E}\phantom{\rule{0.166667em}{0ex}}{\mathrm{e}}^{\gamma {Z}_{k}}& =max\left\{\frac{2{\mathrm{e}}^{\gamma}+1}{3},\frac{{\mathrm{e}}^{2\gamma}+2{\mathrm{e}}^{\gamma}+9}{12},\frac{{\mathrm{e}}^{3\gamma}+3{\mathrm{e}}^{\gamma}+23}{27},\frac{{\mathrm{e}}^{4\gamma}+4{\mathrm{e}}^{\gamma}+43}{48},\frac{{\mathrm{e}}^{5\gamma}+5{\mathrm{e}}^{\gamma}+69}{75}\right\}\hfill \\ & =\frac{2{\mathrm{e}}^{\gamma}+1}{3}\phantom{\rule{4.pt}{0ex}}\mathrm{for}\phantom{\rule{4.pt}{0ex}}\mathrm{all}\phantom{\rule{4.pt}{0ex}}\gamma \in \left(0,4/5\right]\phantom{\rule{-0.166667em}{0ex}}.\hfill \end{array}$$For positive h, we have that$$\begin{array}{c}\hfill \phantom{\rule{-8.53581pt}{0ex}}\underset{k\in \mathbb{N}}{sup}\mathbb{E}{\mathrm{e}}^{h({Z}_{k}-p{\theta}_{k})}=max\left\{\frac{2{\mathrm{e}}^{h}+1}{3{e}^{h}},\frac{{\mathrm{e}}^{2h}+2{\mathrm{e}}^{h}+9}{12{\mathrm{e}}^{h}},\frac{{\mathrm{e}}^{3h}+3{\mathrm{e}}^{h}+23}{27{\mathrm{e}}^{h}},\frac{{\mathrm{e}}^{4h}+4{\mathrm{e}}^{h}+43}{48{\mathrm{e}}^{h}},\frac{{\mathrm{e}}^{5h}+5{\mathrm{e}}^{h}+69}{75{\mathrm{e}}^{h}}\right\}\phantom{\rule{-0.166667em}{0ex}}.\end{array}$$Consequently, $\underset{k\in \mathbb{N}}{sup}\mathbb{E}{\mathrm{e}}^{h({Z}_{k}-p{\theta}_{k})}\u2a7d1$ for all $h\in (0,47/50]$, and Theorem 4 implies that$$\psi \left(u\right)\u2a7d\underset{h\in (0,47/50]}{inf}\left\{{\mathrm{e}}^{-hu}\underset{k\in \mathbb{N}}{sup}\mathbb{E}{\mathrm{e}}^{h({Z}_{k}-p{\theta}_{k})}\right\}\u2a7d{\psi}_{2}\left(u\right):={\mathrm{e}}^{-47u/50}$$The derived estimate is almost four times accurate with respect to the previous one. For instance, it follows from the obtained inequality that $\psi \left(u\right)\u2a7d0.02$, if $u\u2a7e5$.
- Finally, we use the Monte Carlo simulations to get approximate values of ruin probability $\psi \left(u\right)$. For this we use the fact that $\widehat{\psi}(u,N)\approx \psi \left(u\right)$ for fixed u and for sufficiently large N. We consider the case when $u\in \{0,1,2,\dots ,10\}$ and $N=1000$. For each u, we simulate ${10}^{7}$ trajectories of the renewal risk process with N random claims and we calculate how many times on average they fall below zero in order to get values of $\widehat{\psi}(u,N)$. According to the obtained values of ruin probability, we get that $\psi \left(u\right)\u2a7d0.02$, even if the initial surplus is especially small, i.e., $u\u2a7e1$.
- After the above calculations, for $u\in \{0,1,2,\dots ,10\}$, we can compare approximate values of ruin probability $\psi \left(u\right)$ with their conservative estimate ${\psi}_{1}\left(u\right)$ and with their sharp estimate ${\psi}_{2}\left(u\right)$. All results are presented in Table 1 and Figure 1. In fact, the conservative exponential upper bound ${\psi}_{1}$ is quite rough, but it is very easy to obtain an expression of this function using Theorem 3. The values of the sharp exponential upper bound ${\psi}_{2}$ are much closer to the simulated values of the ruin probability, however due to Theorem 4, more deeper analysis of the model elements is needed to obtain this bound. Values of $\psi \left(u\right)$, which are obtained by the Monte Carlo method, are sufficiently accurate, but such a procedure takes a lot of time and resources, because we need to generate a lot of process trajectories in order to get these values.

**Example**

**2.**

- Firstly, we start with the rough estimate of ruin probability, which can be derived using Theorem 3. In the case, we get that$$\begin{array}{c}\underset{k\in \mathbb{N}}{sup}\mathbb{E}({Z}_{k}-p{\theta}_{k})=\underset{k\in \mathbb{N}}{sup}\left(\frac{1}{3+cosk}-\frac{11}{10}\right)\u2a7d-\frac{3}{5},\hfill \\ \underset{k\in \mathbb{N}}{sup}\mathbb{E}\left({\theta}_{k}1\phantom{\rule{-2.9pt}{0ex}}{\mathrm{I}}_{\left\{{\theta}_{k}>4\right\}}\right)=\underset{k\in \mathbb{N}}{sup}\phantom{\rule{0.166667em}{0ex}}\underset{4}{\overset{\infty}{\int}}x\phantom{\rule{0.166667em}{0ex}}\frac{{k}^{k}{x}^{k-1}{\mathrm{e}}^{-kx}}{\mathsf{\Gamma}\left(k\right)}\mathrm{d}x=\underset{4}{\overset{\infty}{\int}}x{\mathrm{e}}^{-x}\mathrm{d}x=\frac{5}{{\mathrm{e}}^{4}}.\hfill \end{array}$$In addition, we have that$$\begin{array}{ccc}\hfill \mathbb{P}\left({Z}_{k}-p{\theta}_{k}\u2a7dx\right)& =& \underset{-\infty}{\overset{\infty}{\int}}\mathbb{P}\left({Z}_{k}-\frac{11}{10}y\u2a7dx\right)\mathrm{d}\mathbb{P}\left({\theta}_{k}\u2a7dy\right)\hfill \\ & =& \frac{{k}^{k}}{\mathsf{\Gamma}\left(k\right)}\underset{0}{\overset{\infty}{\int}}\mathbb{P}\left({Z}_{k}-\frac{11}{10}y\u2a7dx\right){y}^{k-1}{\mathrm{e}}^{-ky}\mathrm{d}y\hfill \\ & =& \frac{{k}^{k}}{\mathsf{\Gamma}\left(k\right)}\underset{0}{\overset{\infty}{\int}}\left(1-{e}^{-(3+cosk)(x+11y/10)}\right)1\phantom{\rule{-2.9pt}{0ex}}{\mathrm{I}}_{\left[0,\infty \right)}\left(x+\frac{11}{10}y\right){y}^{k-1}{\mathrm{e}}^{-ky}\mathrm{d}y\hfill \\ & =& \frac{{k}^{k}}{\mathsf{\Gamma}\left(k\right)}\underset{max\left\{0,-10x/11\right\}}{\overset{\infty}{\int}}\left(1-{e}^{-(3+cosk)(x+11y/10)}\right){y}^{k-1}{\mathrm{e}}^{-ky}\mathrm{d}y.\hfill \end{array}$$Since random values ${Z}_{k}$ and ${\theta}_{k}$ are absolutely continuous, so that random value ${Z}_{k}-p{\theta}_{k}$ has density, which for all $x\u2a7e0$ has the following expression$$\begin{array}{c}\hfill {f}_{{Z}_{k}-p{\theta}_{k}}\left(x\right)=\frac{\left(3+cosk\right){k}^{k}}{\mathsf{\Gamma}\left(k\right)}{\mathrm{e}}^{-\left(3+cosk\right)x}\underset{0}{\overset{\infty}{\int}}{y}^{k-1}{\mathrm{e}}^{-\left(k+33/10+11(cosk)/10\right)y}\mathrm{d}y.\end{array}$$Hence, we get that$$\begin{array}{ccc}\hfill \underset{k\in \mathbb{N}}{sup}\mathbb{E}\left({\mathrm{e}}^{\gamma ({Z}_{k}-p{\theta}_{k})}1\phantom{\rule{-2.9pt}{0ex}}{\mathrm{I}}_{\{{Z}_{k}-p{\theta}_{k}>0\}}\right)& =& \underset{k\in \mathbb{N}}{sup}\underset{0}{\overset{\infty}{\int}}{\mathrm{e}}^{\gamma x}{f}_{{Z}_{k}-p{\theta}_{k}}\left(x\right)\mathrm{d}x\hfill \\ & & \phantom{\rule{-85.35826pt}{0ex}}=\underset{k\in \mathbb{N}}{sup}\left(\frac{\left(3+cosk\right){k}^{k}}{\mathsf{\Gamma}\left(k\right)}\underset{0}{\overset{\infty}{\int}}{y}^{k-1}{\mathrm{e}}^{-\left(k+33/10+11(cosk)/10\right)y}\mathrm{d}y\underset{0}{\overset{\infty}{\int}}{\mathrm{e}}^{\left(\gamma -3-cosk\right)x}\mathrm{d}x\right)\hfill \\ & & \phantom{\rule{-85.35826pt}{0ex}}\u2a7d\underset{k\in \mathbb{N}}{sup}\left(\frac{\left(3+cosk\right){k}^{k}}{(3+cosk-\gamma )\mathsf{\Gamma}\left(k\right)}\underset{0}{\overset{\infty}{\int}}{y}^{k-1}{\mathrm{e}}^{-\left(k+22/10\right)y}\mathrm{d}y\right)\hfill \\ & & \phantom{\rule{-85.35826pt}{0ex}}=\underset{k\in \mathbb{N}}{sup}\left(\frac{\left(3+cosk\right){k}^{k}}{(3+cosk-\gamma )\mathsf{\Gamma}\left(k\right)}\frac{\mathsf{\Gamma}\left(k\right)}{{\left(k+\frac{11}{5}\right)}^{k}}\right)\hfill \\ & & \phantom{\rule{-85.35826pt}{0ex}}\u2a7d\underset{k\in \mathbb{N}}{sup}\left(\frac{3+cosk}{3+cosk-\gamma}\right)\underset{k\in \mathbb{N}}{sup}{\left(\frac{k}{k+\frac{11}{5}}\right)}^{k}\u2a7d\frac{5}{8\left(2-\gamma \right)}\phantom{\rule{4.pt}{0ex}}\mathrm{for}\phantom{\rule{4.pt}{0ex}}\mathrm{all}\phantom{\rule{4.pt}{0ex}}\gamma \in \left(0,2\right)\phantom{\rule{-0.166667em}{0ex}}.\hfill \end{array}$$According to the derived estimates, conditions of Theorem 3 hold with such parameters $\alpha =3/5,\phantom{\rule{4pt}{0ex}}\gamma =13/10,\phantom{\rule{4pt}{0ex}}\beta =25/28,\phantom{\rule{4pt}{0ex}}\varkappa =44/10$ and $\epsilon =5/{\mathrm{e}}^{4}$. Substituting the obtained constants into expression (2), we get that the parameter $\delta $ should be selected from the interval $\left(0,0.02855\right]$. If $\delta =7/250$, then Theorem 3 implies that$$\psi \left(u\right)\u2a7d{\psi}_{1}\left(u\right):={\mathrm{e}}^{-91u/2500}$$The derived estimate is extremely conservative. For instance, it follows from the obtained inequality that $\psi \left(u\right)\u2a7d0.02$, if $u\u2a7e108$.
- We can use Theorem 4 to obtain the more sharper upper bound for the ruin probability. Conditions of this theorem are satisfied due to the above estimates and$$\begin{array}{c}\hfill \underset{k\in \mathbb{N}}{sup}\mathbb{E}\phantom{\rule{0.166667em}{0ex}}{\mathrm{e}}^{\gamma {Z}_{k}}=\underset{k\in \mathbb{N}}{sup}\left(\frac{3+cosk}{3+cosk-\gamma}\right)\u2a7d\frac{2}{2-\gamma}\phantom{\rule{4.pt}{0ex}}\mathrm{for}\phantom{\rule{4.pt}{0ex}}\mathrm{all}\phantom{\rule{4.pt}{0ex}}\gamma \in \left(0,2\right)\phantom{\rule{-0.166667em}{0ex}}.\end{array}$$In addition, we have that$$\begin{array}{c}\hfill \underset{k\in \mathbb{N}}{sup}\mathbb{E}{\mathrm{e}}^{h({Z}_{k}-p{\theta}_{k})}=\underset{k\in \mathbb{N}}{sup}\left(\frac{3+cosk}{3+cosk-h}\frac{1}{{\left(1+\frac{11h}{10k}\right)}^{k}}\right)\u2a7d\frac{2}{2-h}\phantom{\rule{0.166667em}{0ex}}\underset{k\in \mathbb{N}}{sup}\left(\frac{1}{{\left(1+\frac{11h}{10k}\right)}^{k}}\right)\u2a7d1\end{array}$$Accordingly, Theorem 4 together with the remark presented after this theorem imply that$$\psi \left(u\right)\u2a7d\underset{h\in (0,12/11]}{inf}\left\{{\mathrm{e}}^{-hu}\underset{k\in \mathbb{N}}{sup}\mathbb{E}{\mathrm{e}}^{h({Z}_{k}-p{\theta}_{k})}\right\}\u2a7d{\psi}_{2}\left(u\right):={\mathrm{e}}^{-12u/11}$$The derived estimate is exponential and almost 30 times more accurate than the previous one. For instance, it follows from the obtained inequality that $\psi \left(u\right)\u2a7d0.02$, if $u\u2a7e4$.
- After all these calculations, we again apply the Monte Carlo method in order to get approximate values of ruin probability $\psi \left(u\right)$ as in the previous example. We also analyze the same way, while $u\in \{0,1,2,\dots ,10\}$ and $N=1000$, and we simulate ${10}^{7}$ trajectories of the renewal risk process with N random claims and with N random inter-arrival times for each u. Although this example is sufficiently erratic, on the basis of the received values of ruin probability $\widehat{\psi}(u,N)\approx \psi \left(u\right)$, we get that $\psi \left(u\right)\u2a7d0.02$, even if the initial surplus is relatively small, i.e., $u\u2a7e2$.
- Ultimately, we can make a comparison of approximate values of ruin probability $\psi \left(u\right)$ to their conservative exponential upper bound ${\psi}_{1}\left(u\right)$ and to their sharp exponential upper bound ${\psi}_{2}\left(u\right)$ for $u\in \{0,1,2,\dots ,10\}$. All results are presented in Table 2 and Figure 2. Actually, the conservative estimate ${\psi}_{1}$ is too rough. The values of sharp estimate ${\psi}_{2}$ are more accurate than ${\psi}_{1}$. It should be noted that the Monte Carlo method takes even more time in this example than in Example 1, because we need to generate values of two random variables ${Z}_{k}$ and ${\theta}_{k}$, which are non-identically distributed for each $k=1,2,\dots ,N$.

## 5. Conclusions

## Acknowledgments

## Author Contributions

## Conflicts of Interest

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u | $\mathit{\psi}\left(\mathit{u}\right)$ | ${\mathit{\psi}}_{1}\left(\mathit{u}\right)$ | ${\mathit{\psi}}_{2}\left(\mathit{u}\right)$ |
---|---|---|---|

0 | 0.1069843 | 1.0000000 | 1.0000000 |

1 | 0.0192021 | 0.7866279 | 0.3906278 |

2 | 0.0068947 | 0.6187834 | 0.1525901 |

3 | 0.0019112 | 0.4867523 | 0.0596059 |

4 | 0.0006655 | 0.3828929 | 0.0232837 |

5 | 0.0002378 | 0.3011942 | 0.0090953 |

6 | 0.0000675 | 0.2369278 | 0.0035529 |

7 | 0.0000217 | 0.1863740 | 0.0013878 |

8 | 0.0000060 | 0.1466070 | 0.0005421 |

9 | 0.0000014 | 0.1153251 | 0.0002118 |

10 | 0.0000006 | 0.0907180 | 0.0000827 |

u | $\mathit{\psi}\left(\mathit{u}\right)$ | ${\mathit{\psi}}_{1}\left(\mathit{u}\right)$ | ${\mathit{\psi}}_{2}\left(\mathit{u}\right)$ |
---|---|---|---|

0 | 0.2628618 | 1.0000000 | 1.0000000 |

1 | 0.0262527 | 0.9642545 | 0.3359110 |

2 | 0.0035110 | 0.9297868 | 0.1128362 |

3 | 0.0005077 | 0.8965511 | 0.0379029 |

4 | 0.0000739 | 0.8645034 | 0.0127320 |

5 | 0.0000102 | 0.8336013 | 0.0042768 |

6 | 0.0000015 | 0.8038039 | 0.0014366 |

7 | 0.0000001 | 0.7750715 | 0.0004826 |

8 | 0.0000000 | 0.7473662 | 0.0001621 |

9 | 0.0000000 | 0.7206512 | 0.0000545 |

10 | 0.0000000 | 0.6948912 | 0.0000183 |

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**MDPI and ACS Style**

Kizinevič, E.; Šiaulys, J.
The Exponential Estimate of the Ultimate Ruin Probability for the Non-Homogeneous Renewal Risk Model. *Risks* **2018**, *6*, 20.
https://doi.org/10.3390/risks6010020

**AMA Style**

Kizinevič E, Šiaulys J.
The Exponential Estimate of the Ultimate Ruin Probability for the Non-Homogeneous Renewal Risk Model. *Risks*. 2018; 6(1):20.
https://doi.org/10.3390/risks6010020

**Chicago/Turabian Style**

Kizinevič, Edita, and Jonas Šiaulys.
2018. "The Exponential Estimate of the Ultimate Ruin Probability for the Non-Homogeneous Renewal Risk Model" *Risks* 6, no. 1: 20.
https://doi.org/10.3390/risks6010020