Ruin Probability-Based Initial Capital of the Discrete-Time Surplus Process

1. Introduction

In recent years, risk models have attracted much attention in the insurance business, in connection with the possible insolvency and the capital reserves of insurance companies. The main interest from the point of view of an insurance company is claim arrival and claim size, which affect the capital of the company.

In this paper, we assume that all the processes are defined in a probability space (Ω, F, Pr). Claims happen at the times T_i, satisfying 0 = T₀ ≤ T₁ ≤ T₂ ≤ . . . . The nth claim arriving at time T_n causes the claim size X_n. Now let the constant c represent the premium rate for one unit time; the random variable cT_n describes the inflow of capital into the business by time T_n, and $\sum_{i=1}^n X_i$ describes the outflow of capital due to payments for claims occurring in [0, T_n]. Therefore, the quantity

$$U_{0}=u, U_{n}=u+c T_{n}-\sum_{i=1}^{n} X_{i} \tag{1.1}$$

is the insurer’s balance (or surplus) at time T_n, n = 1, 2, 3, . . . , with the constant U₀ = u ≥ 0 as the initial capital.

We consider the discrete-time surplus process (1.1) in the situation that the possible insolvency (ruin) can occur only at claim arrival times T_n = n, n = 1, 2, 3, . . . . Thus, the model 1.1 becomes

$$U_{0}=u, U_{n}=u+c n-\sum_{i=1}^{n} X_{i} \tag{1.2}$$

for all n = 1, 2, 3, . . . .

The general approach for studying ruin probability in the discrete-time surplus process is through the so-called Gerber-Shiu discounted penalty function; for example, Pavlova and Willmot (2004), Dickson (2005), and Li (2005b, 2005a). All of these articles study (or calculate) the ruin probability as a function the initial capital u ≥ 0. In this paper, we shall work in the opposite direction, i.e., we study the initial capital for the discrete-time surplus process as a function of ruin probabilities.

2. Main results

Let {U_n, n ≥ 0} be a surplus process (as in Section 1) that is driven by the claim process {X_n, n ≥ 1}. We consider the finite-time ruin probabilities of the discrete-time surplus process in Equation (1.2) with the independent and identically distributed (i.i.d.) claim process {X_n, n ≥ 1}. We let $F_{X_1}(x)$ be the distribution function of X₁, i.e.,

$$F_{X_{1}}(x)=\operatorname{Pr}\left\{X_{1} \leq x\right\} \tag{2.1}$$

The premium rate c is calculated by the expected value principle, i.e.,

$$c=(1+\theta) E\left[X_{1}\right] \tag{2.2}$$

where θ > 0 which is the safety loading of insurer.

Let u ≥ 0 be an initial capital. For each n = 1, 2, 3, . . . , we let

$$\varphi_{n}(u):=\operatorname{Pr}\left\{U_{1} \geq 0, U_{2} \geq 0, U_{3} \geq 0, \ldots, U_{n} \geq 0 \mid U_{0}=u\right\} \tag{2.3}$$

denote the survival probability at the times n. Thus, the ruin probability at one of the time 1, 2, 3, . . . , n is denoted by

$$\Phi_{n}(u)=1-\varphi_{n}(u) \tag{2.4}$$

Definition 2.1.

Let {U_n, n ≥ 0} be a surplus process which is driven by the claim process {X_n, n ≥ 1} and c > 0 be a premium rate. Given α ∈ (0, 1) and N ∈ {1, 2, 3, . . .}. Let an initial capital u ≥ 0, if $\boldsymbol{\Phi}_N(u) \leq \alpha$ then u is called an acceptable initial capital corresponding to (α, N, c, {X_n, n ≥ 1}). Particularly, if

$$u^{*}=\min _{u \geq 0}\left\{u: \Phi_{N}(u) \leq \alpha\right\} \tag{2.5}$$

exists, u* is called the minimum initial capital corresponding to (α, N, c, {X_n, n ≥ 1}) and is written as

$$u^{*}:=\operatorname{MIC}\left(\alpha, N, c,\left\{X_{n}, n \geq 1\right\}\right) \tag{2.6}$$

2.1. Ruin and survival probability

We define the total claim process by

$$S_{n}:=X_{1}+X_{2}+\cdots+X_{n} \tag{2.7}$$

for all n = 1, 2, 3, . . . .

Lemma 2.1.

Let N ∈ {1, 2, 3, . . .} and c > 0 be given. If {X_n, n ≥ 1} is an i.i.d. claim process, then $\Phi_N(u)$ is increasing and right continuous and $\Phi_N(u)$ is decreasing and right continuous in u.

Proof. The survival probability at the time N as mentioned in (2.3) can be expressed as follows.

$$\begin{aligned} \varphi_{N}(u)= & \operatorname{Pr}\left\{S_{1} \leq u+c, S_{2} \leq u\right. \\ & \left.+2 c, \ldots, S_{N} \leq u+N c\right\} \\ = & \operatorname{Pr}\left(\bigcap_{k=1}^{N}\left\{S_{k}-k c-u \leq 0\right\}\right) \\ = & E\left[\bigcap_{k=1}^{\mathbb{I}_{N k-k c-u \leq 0\}}}\right] \\ = & E\left[\prod_{k=1}^{N} \mathbb{I}_{\left\{S_{k}-k c-u \leq 0\right\}}\right] \end{aligned} \tag{2.8}$$

where

$$\mathbb{I}_{A}(x)=\left\{\begin{array}{l} 1, x \in A \\ 0, x \notin A \end{array}\right.$$

for all A  ℝ. Since $\mathbb{I}_{\left\{S_k-k c-u \leq 0\right\}}(\omega)=\mathbb{I}_{(-, 0]}\left(S_k(\omega)-k c-u\right)$ for all ω ∈ ,

$$\varphi_{N}(u)=E\left[\prod_{k=1}^{N} \mathbb{I}_{(-\infty, 0]}\left(S_{k}-k c-u\right)\right] \tag{2.9}$$

For each a ∈ ℝ and u ≥ 0, we obtain

$$\mathbb{I}_{(-\infty, 0]}(a-u)=\left\{\begin{array}{l} 1, {u} \geq {a} \\ 0, {u}<{a} \end{array}\right.$$

then 𝕀_{(−∞, 0]}(a − u) is increasing and right continuous in u. This implies that $\prod_{i=1}^N \mathbb{I}_{(-\infty, 0]}\left(a_i-u\right)$ is also increasing and right continuous in u, moreover, this bounding function is identically equal to 1, where a_k ∈ ℝ, k = 1, 2, 3, . . . , N. Therefore, by the monotone convergence theorem, we have

$$\begin{aligned} \lim _{v \rightarrow u^{+}} \varphi_{N}(v) & =\lim _{v \rightarrow u^{+}} E\left[\prod_{k=1}^{N} \mathbb{I}_{(-\infty, 0]}\left(S_{k}-k c-v\right)\right] \\ & =E\left[\lim _{v \rightarrow u^{+}} \prod_{k=1}^{N} \mathbb{I}_{(-\infty, 0]}\left(S_{k}-k c-v\right)\right] \\ & =E\left[\prod_{k=1}^{N} \mathbb{I}_{(-\infty, 0]}\left(S_{k}-k c-u\right)\right] \\ & =\varphi_{N}(u) . \end{aligned} \tag{2.10}$$

Therefore, $\varphi_N(u)$ is increasing and right continuous. Moreover, we can conclude that $\Phi_N(u)=1- \varphi_N(u)$) is decreasing and also right continuous.

Theorem 2.2. Let N ∈ {1, 2, 3, . . .} and c 0 be given. If {X_n, n ≥ 1} is an i.i.d. claim process, then

$$\lim _{u \rightarrow \infty} \varphi_{N}(u)=1 \text { and } \lim _{u \rightarrow \infty} \Phi_{N}(u)=0 \text {. } \tag{2.11}$$

Proof. First, we will show the following properties

$$\bigcap_{i=1}^{N}\left\{X_{i} \leq u+c\right\} \subset \bigcap_{i=1}^{N}\left\{S_{i} \leq N u+i c\right\} . \tag{2.12}$$

Let ω $\in \bigcap^{N}${X_i ≤ u + c} be given. For each i ∈ {1, 2, 3, . . . , N}, we have X_i(ω) ≤ u + c and

$$S_{i}(\omega)=\sum_{k=1}^{i} X_{k}(\omega) \leq i u+i c \leq N u+i c \tag{2.13}$$

That is, ω ∈ {S_i ≤ Nu + ic}. Therefore, (2.12) follows. Next, since the process {X_n, n ≥ 1} is i.i.d., then

$$\operatorname{Pr}\left(\bigcap_{i=1}^{N}\left\{X_{i} \leq u+c\right\}\right)=\prod_{i=1}^{N} \operatorname{Pr}\left\{X_{i} \leq u+c\right\}=(F(u+c))^{N} . \tag{2.14}$$

By Equation (2.8), we have

$$\varphi_{N}(N u)=\operatorname{Pr}\left(\bigcap_{i=1}^{N}\left\{S_{i} \leq N u+i c\right\}\right) \tag{2.15}$$

By (2.12), (2.14) and (2.15), we obtain

$$(F(u+c))^{N} \leq \varphi_{N}(N u) \leq 1 \tag{2.16}$$

Since (F(u + c))^N → 1 as u → ∞, then $\varphi_N(N u)$ → 1 as u → ∞. Thus, we conclude that $\varphi_N(u)$ → 1, and $\Phi_N(u)=1-\varphi_N(u) \rightarrow 0$ as u → ∞. This is the proof.

Corollary 2.3. Let α ∈ (0, 1), N ∈ {1, 2, 3, . . .} and c 0 be given. If {X_n, n ≥ 1} is an i.i.d. claim process, then there exists ũ ≥ 0 such that, for all u ≥ ũ, u is an acceptable initial capital corresponding to (α, N, c, {X_n, n ≥ 1}).

Proof. We consider by case. Case 1: $\Phi_N(0) \leq \alpha$. Since $\Phi_N(u)$ is decreasing, then $\Phi_N(u) \leq \Phi_N(0) \leq 0$ for all $u \geq 0$. Case 2: $\Phi_N(0)>\alpha$. By Theorem 2.2 , we have $\Phi_N(u) \rightarrow 0$ as $u \rightarrow \infty$. Thus, there exists $\tilde{u}>0$ such that $\Phi_N(\tilde{a})<\alpha$. Since $\Phi_N(u)$ is decreasing, we conclude that $\Phi_N(u) \leq \Phi_N(\tilde{u})<\alpha$ for all $u \geq \tilde{u}$.

2.2. Recursive formula of ruin probabilities

From Theorem 2.2 and Corollary 2.3, we know that the small ruin probability can be controlled by a large initial capital. In this part, we shall describe the upper bound of ruin probability with negative exponential. In order to prove the following lemma, we shall use an equivalent definition of the ruin probability which is given as follows:

$$\Phi_{n}(u)=\operatorname{Pr}\left(\max _{1 \leq k \leq n}\left(\sum_{i=1}^{k} X_{i}-c k\right)>u\right) \tag{2.17}$$

Theorem 2.4. Let N ∈ {1, 2, 3, . . .}, c 0 and u ≥ 0 be given. If {X_n, n ≥ 1} is an i.i.d. claim process, then the ruin probability at one of the times 1, 2, 3, . . . , N satisfies the following equation

$$\Phi_{N}(u)=\Phi_{1}(u)+\int_{-\infty}^{u+c} \Phi_{N-1}(u+c-x) d F_{X_{1}}(x) \tag{2.18}$$

where Φ₀(u) = 0.

Proof. We will prove (2.18) by induction. We start with n = 1. Since Φ₀(u) = 0 for all u ≥ 0, then

$$\int_{-\infty}^{u+c} \Phi_{0}(u+c-x) d F_{X_{1}}(x)=0 \tag{2.19}$$

This proves (2.18) for n = 1. Now assume that (2.18) holds for n = k ≥ 1. Then

$$\begin{aligned} \Phi_{k+1}(u)= & \operatorname{Pr}\left(\max _{1 \leq s \leq k+1}\left(\sum_{i=1}^{n} X_{i}-c n\right)>u\right) \\ = & \operatorname{Pr}\left(X_{1}-c>u\right) \\ & +\operatorname{Pr}\left(\max _{2 s n s k+1}\left(X_{1}+\sum_{i=2}^{n} X_{i}-c n\right)>u, X_{1} \leq u+c\right) \\ = & \Phi_{1}(u)+\int_{-\infty}^{u+c} \operatorname{Pr}\left(\max _{1 s n \leq k}\left(x+\sum_{i=2}^{n} X_{i}-c n\right)>u\right) \\ & d F_{X_{1}}(x) \\ = & \Phi_{1}(u)+\int_{-\infty}^{u+c} \operatorname{Pr}\left(\max _{2 s n s k+1}\left(\sum_{i=2}^{n} X_{i}-c(n-1)\right)\right. \\ & >u+c-x) d F_{X_{1}}(x) \\ = & \Phi_{1}(u)+\int_{-\infty}^{u+c} \operatorname{Pr}\left(\max _{2 s n s k+1}\left(\sum_{i=1}^{n-1} X_{i}-c(n-1)\right)\right. \\ & >u+c-x) d F_{X_{1}}(x) \\ = & \Phi_{1}(u)+\int_{-\infty}^{u+c} \operatorname{Pr}\left(\max _{1 \leq n s k}\left(\sum_{i=1}^{n} X_{i}-c n\right)\right. \\ & >u+c-x) d F_{X_{1}}(x) \\ = & \Phi_{1}(u)+\int_{-\infty}^{u+c} \Phi_{k}(u+c-x) d F_{X_{1}}(x) . \end{aligned}$$

which proves (2.18) for n = k + 1 and concludes the proof.

Corollary 2.5. Let N ∈ {1, 2, 3, . . .}, c 0 and u ≥ 0 be given. If {X_n, n ≥ 1} is an i.i.d. claim process, then the ruin probability at one of the times 1, 2, 3, . . . , N satisfies the following equation:

$$\begin{aligned} \Phi_{0}(u) & =0, \Phi_{1}(u)=1-\operatorname{Pr}(X \leq u+c), \Phi_{N}(u) \\ & =\Phi_{N-1}(u)+\Theta_{N}(u) \end{aligned} \tag{2.20}$$

where

$$\Theta_{N}(u)=\int_{-\infty}^{u+c}\left(\int_{-\infty}^{u+c-x} \Phi_{N-2}(u+2 c-x-v) d F_{X_{1}}(v)\right) d F_{X_{1}}(x)$$

for all n = 2, 3, 4, . . . .

Proof. Let N ≥ 2, by Theorem 2.4, we obtain

$$\begin{aligned} \Phi_{N}(u) & =\Phi_{1}(u)+\int_{-\infty}^{u+c} \Phi_{N-1}(u+c-x) d F_{X_{1}}(x) \\ = & \Phi_{1}(u)+\int_{-\infty}^{u+c}\left(\Phi_{N-2}(u+c-x)\right. \\ & \left.+\int_{-\infty}^{u+2 c-x} \Phi_{N-2}(u+2 c-x-v) d F_{X_{1}}(v)\right) d F_{X_{1}}(x) \\ = & \Phi_{1}(u)+\int_{-\infty}^{u+c}\left(\Phi_{N-2}(u+c-x) d F_{X_{1}}(x)\right. \\ & +\int_{-\infty}^{u+c}\left(\int_{-\infty}^{u+2 c-x} \Phi_{N-2}(u+2 c-x-v) d F_{X_{1}}(v)\right) d F_{X_{1}}(x) \\ = & \Phi_{N-1}(u)+\int_{-\infty}^{u+c}\left(\int_{-\infty}^{u+2 c-x} \Phi_{N-2}(u+2 c-x-v) d F_{X_{1}}(v)\right) \\ & d F_{X_{1}}(x) . \end{aligned}$$

This completes the proof.

Corollary 2.6. Let N ∈ {1, 2, 3, . . .} and u ≥ 0. Assume that {X_n, n ≥ 1} is a sequence of exponential distribution with intensity λ 0, i.e., X₁ has the probability density function $f(x)=\lambda e^{-\lambda x}$. The obtained ruin probability is in the following recursive form

$$\begin{aligned} \Phi_{0}(u)= & 0, \Phi_{n}(u)=\Phi_{n-1}(u) \\ & +\frac{(u+c) \lambda^{n-1}(u+n c)^{n-2}}{(n-1) !} e^{-\lambda(u+n c)} \end{aligned} \tag{2.21}$$

for all n = 1, 2, 3, . . . , where the initial capital u ≥ 0 and premium rate c E[X₁] = 1/λ.

Proof. We will prove (2.21) by induction. We start with n = 1, Φ₁(u) = 1 − Pr {X ≤ u + c} = 1 − (1 − $e^{-\lambda(u+c)}$ = $e^{-\lambda(u+c)}$.

This proves (2.21) for n = 1. Next we assume that (2.21) holds for n = k ≥ 1. From Theorem 2.2, we have

$$\begin{aligned} \Phi_{k+1}(u)= & \Phi_{1}(u)+\int_{0}^{u+c} \Phi_{k}(u+c-x) d F_{X_{1}}(x) \\ = & \Phi_{1}(u)+\int_{0}^{u+c}\left(\Phi_{k-1}(u+c-x)\right. \\ & +\frac{(u+2 c-x) \lambda^{k-1}(u+(k+1) c-x)^{k-2}}{(k-1) !} \\ & \left.e^{-\lambda(u+(k+1) c-x)}\right) d F_{X_{1}}(x) \end{aligned}$$

$$\begin{aligned} = & \Phi_{k}(u)+\int_{0}^{u+c} \frac{(u+2 c-x) \lambda^{k-1}(u+(k+1) c-x)^{k-2}}{(k-1) !} \\ & e^{-\lambda(u+(k+1) c-x)} d F_{X_{1}}(x) \end{aligned} \tag{2.22}$$

and

$$\begin{aligned} \Theta_{k+1}(u)= & \int_{0}^{u+c} \frac{(u+2 c-x) \lambda^{k-1}(u+(k+1) c-x)^{k-2}}{(k-1) !} \\ = & \frac{e^{-\lambda(u+(k+1) c-x)} e^{-\lambda x} d x}{(k-1) !} \int_{0}^{-\lambda(u+(k+1) c)} \int^{u+c}(u+(k+1) c-x)^{k-2} \\ & (u+(k+1) c-x-(k-1) c) d x \\ = & \frac{\lambda^{k} e^{-\lambda(u+(k+1) c)} \int_{0}^{u+c}}{(k-1) !}\left((u+(k+1) c-x)^{k-1}\right. \\ & \left.+(k-1) c(u+(k+1) c-x)^{k-2}\right) d x \\ = & \frac{(u+c) \lambda^{k}(u+(k+1) c)^{k-1}}{k !} e^{-\lambda(u+(k+1) c)}, \end{aligned}$$

which proves (2.21) for n = k + 1 and completes the proof.

2.3. Existence of minimum initial capital

A quantity α, which was discussed in the previous section, can be described as the most acceptable probability that the insurance company will become insolvent. As a result of Corollary 2.3, we obtain that $\left\{u \geq 0: \Phi_N(u) \leq \alpha\right\}$ is a non-empty set. This means that we can always choose an initial capital which makes the value of ruin probability not exceed α. Since the set $\left\{u \geq 0: \Phi_N(u) \leq \alpha\right\}$ is an infinite set, then there are many acceptable initial capital corresponding to (α, N, c, {X_n, n ≥ 1}). In this section, we will prove the existence of

$$\operatorname{MIC}\left(\alpha, N, c,\left\{X_{n}, n \geq 1\right\}\right)=\min _{u \geq 0}\left\{u: \Phi_{N}(u) \leq \alpha\right\} \tag{2.23}$$

Lemma 2.2. Let a, b and α be real numbers such that a ≤ b. If f is decreasing and right continuous on [a, b] and α ∈ [f(b), f(a)], then there exists d ∈ [a, b] such that

$$d=\min \{x \in[a, b]: f(x) \leq \alpha\} . \tag{2.24}$$

Proof. Let

$$S:=\{x \in[a, b]: f(x) \leq \alpha\}$$

Since α ∈ [f(b), f(a)], i.e., f(b) ∈ α ≤ f(a), then b ∈ S. That is, S is a non-empty set. Since S is a subset of closed and bounded interval [a, b], then there exists d ∈ [a, b] such that d = inf S. Next, we consider by case.

Case 1: d = b. We know that b ∈ S, thus b = min S.

Case 2: a ≤ d b. Since d = inf S, then there exists d_n ∈ S such that

$${d} \leq {d}_{n}<{d}+1 / {n}$$

for all n ∈ ℕ. For each n 2/(b − d), we have

$${d}<{d}+1 / {n}<{d}+\frac{b-d}{2}=\frac{b+d}{2}<{b} .$$

This means that d + 1/n ∈ (d, b) ⊂ [a, b] for all n 2/(b − d). Since f is decreasing and d_n ∈ S, we get

$$f(d+1 / n) \leq f\left(d_{n}\right) \leq \alpha$$

i.e., d + 1/n ∈ S for all n 2/(b − d). Since f is right continuous at d, we have

$$f(d)=\lim _{n \rightarrow \infty} f(d+1 / n) \leq \alpha$$

Therefore, d ∈ S, i.e., d = min S. This completes the proof.

Theorem 2.7. Let α ∈ (0, 1), N ∈ {1, 2, 3, . . .}, and c 0. Then there exist u* ≥ 0 such that

$$u^{*}=\operatorname{MIC}\left(\alpha, N, c,\left\{X_{n}, n \geq 1\right\}\right)$$

Proof. We consider by case. Case 1: $\Phi_N(0) \leq \alpha$, we have

$$\operatorname{MIC}\left(\alpha, N, c,\left\{X_{n}, n \geq 1\right\}\right)=0$$

Case 2: $\Phi_N(0)>\alpha$, by Corollary 2.3 , there exists $\tilde{a}>0$ such that $\Phi_N(\tilde{a})<\alpha$, i.e., $\alpha \in\left[\Phi_N(\tilde{a}), \Phi_N(0)\right]$. Since $\Phi_N(u)$ is decreasing and right continuous, by Lemma 2.2 there exists $u^* \in[0, \tilde{u}]$ such that

$$u^{*}=\min _{u \in 0, \bar{i}]}\left\{u: \Phi_{N}(u) \leq \alpha\right\}=\min _{u \in 0, \infty)}\left\{u: \Phi_{N}(u) \leq \alpha\right\}$$

That is,

$$u^{*}=\operatorname{MIC}\left(\alpha, N, c,\left\{X_{n}, n \geq 1\right\}\right)$$

Next, we will approximate the minimum initial capital MIC(α, N, c, {X_n, n ≥ 1}) by applying the bisection technique for the decreasing and right continuous function.

Theorem 2.8. Let α ∈ (0, 1), N ∈ {1, 2, 3, . . .}, and v₀, u₀ ≥ 0 such that v₀ u₀. Let $\left\{u_n\right\}_{n=1}^{\infty}$ and $\left\{v_n\right\}_{n=1}^{\infty}$ be a real sequence defined by

$$\left\{\begin{array}{ll} v_{k}=v_{k-1} & \text { and } u_{k}=\frac{u_{k-1}+v_{k-1}}{2}, \\ & \text { if } \Phi_{N}\left(\frac{u_{k-1}+v_{k-1}}{2}\right) \leq \alpha \\ v_{k}=\frac{v_{k-1}+u_{k-1}}{2} & \text { and } u_{k}=u_{k-1}, \\ & \text { if } \Phi_{N}\left(\frac{u_{k-1}+v_{k-1}}{2}\right)>\alpha \end{array}\right.$$

for all k = 1, 2, 3, . . . . If $\Phi_N\left(u_0\right) \leq \alpha<\Phi_N\left(v_0\right)$, then

$$\lim _{k \rightarrow \infty} u_{k}=\operatorname{MIC}\left(\alpha, N, c,\left\{X_{n}, n \geq 1\right\}\right) \tag{2.25}$$

and

$$0 \leq u_{k}-\operatorname{MIC}\left(\alpha, N, c,\left\{X_{n}, n \geq 1\right\}\right) \leq \frac{u_{0}-v_{0}}{2^{k}} \tag{2.26}$$

for all k = 1, 2, 3, . . . .

Proof. Obviously, $\left\{u_n\right\}_{n=1}^{\infty}$ is decreasing and $\left\{v_n\right\}_{n=1}^{\infty}$ is increasing, moreover, v_k ≤ u_k for all k = 1, 2, 3, . . . . Thus, $\left\{u_n\right\}_{n=1}^{\infty}$ and $\left\{v_n\right\}_{n=1}^{\infty}$ are convergent. Since

$$0 \leq u_{k}-v_{k}=\left(u_{0}-v_{0}\right) / 2^{k} \rightarrow 0 \text { as } k \rightarrow \infty,$$

then there exists u* ∈ [v₀, u₀] such that

$$\lim _{k \rightarrow \infty} u_{k}=\lim _{k \rightarrow \infty} v_{k}:=u^{*} \tag{2.27}$$

Since $\Phi_N(u)$ is decreasing and $\Phi_{N}\left(v_k\right)>\alpha$ for all ${k}=1,2,3, \ldots$, then $\Phi_N(u)>\alpha$ for all ${u}<{u}^*$. Since $\Phi_N(u)$ is right continuous and $\Phi_N\left(u_k\right) \leq \alpha$ for all ${k}=1$, $2,3, \ldots$, then

$$\Phi_{N}\left(u^{*}\right)=\lim _{k \rightarrow \infty} \Phi_{N}\left(u_{k}\right) \leq \alpha \tag{2.28}$$

Therefore,

$$u^{*}=\operatorname{MIC}\left(\alpha, N, c,\left\{X_{n}, n \geq 1\right\}\right) . \tag{2.29}$$

For each k = 0, 1, 2, . . . , we have v_k ≤ u* ≤ u_k. This implies that

$$\begin{array}{c} 0 \leq u_{k}-u^{*} \leq u_{k}-u^{*}+u^{*}-v_{k} \\ =u_{k}-v_{k}=\frac{u_{0}-v_{0}}{2^{k}} . \end{array} \tag{2.30}$$

This completes the proof.

2.4. Numerical results

We provide numerical illustrations of the main results. We approximate the minimum initial capital of the discrete-time surplus process (1.2) by using Theorem 2.8 in the case of {X_n, n ≥ 1} a sequence of i.i.d. exponential distribution with intensity λ = 1, by choosing model parameter combinations θ = 0.10 and 0.25, i.e., c = 1.10 and c = 1.25, respectively; and α = 0.1, 0.2, and 0.3.

Table 1 shows the approximation of MIC(α, N, c, {X_n, n ≥ 1}) with u₂₅ as mentioned in Theorem 2.8, choosing v₀ = 0 and u₀ = 20, and $\Phi_N(u)$ is computed from the recursive form (2.21).

Figure 1 shows the approximation of MIC(α, N, c, {X_n, n ≥ 1}) for the various values of α with u₂₅ as mentioned in Theorem 2.8. Here we choose v₀ = 0, u₀ = 20, and parameter combinations θ = 0.10, θ = 0.25, i.e., c = 1.10 and c = 1.25, respectively.

Figure 1.Minimum initial capital MIC(α, N, c, {$X_n$, n≥1}) in the discrete-time surplus process with exponential claims (λ = 1, N = 100)