On sequential maxima of exponential sample means, with an application to ruin probability

TL;DR

This paper derives the distribution of the maximum average of i.i.d. exponential variables, revealing a simple inverse distribution, and applies this to analyze ruin probabilities in risk models.

Contribution

It provides a closed-form expression for the inverse distribution of the maximal average of exponential samples and applies it to ruin probability analysis.

Findings

Distribution of maximal average derived explicitly

Inverse distribution admits a simple closed form

Application to ruin probability in risk models

Abstract

We obtain the distribution of the maximal average in a sequence of independent identically distributed exponential random variables. Surprisingly enough, it turns out that the inverse distribution admits a simple closed form. An application to ruin probability in a risk-theoretic model is also given.

Equations

$$Z_{\infty }:=\sup \{{\bar{X}}_i:i\in {\mathbb{N}}^{+}\}$$

$$F_{\infty }(x)=1-\sum _{k=1}^{\infty }\frac{k^{k-1}}{k!}{x}^{k-1}{e}^{-kx}$$

$$F_{\infty }^{-1}(u)=\frac{-\log (1-u)}{u}\text{ for all }u\in (0,1).$$

$$F_{\infty }(x)=1+\frac{1}{x}{W}_0(-x{e}^{-x})$$

$$M_n:=\sup \{{\bar{X}}_i:i\ge n\}$$

$$F_{M_2}(x)=F_{\infty }(x)+e^{-2x}\frac{F_{\infty }(x)}{1-F_{\infty }(x)},x\ge 0.$$

$$\sqrt{n}{M}_n\Rightarrow |Z|$$

$$\underset{n\to \infty }{\mathrm{limsup}}\frac{\sqrt{n}}{\sqrt{2\log \log n}}{M}_n=1.$$

$$Z_n:=\max \left\{{\bar{X}}_1,{\bar{X}}_2,\dots ,{\bar{X}}_n\right\}$$

$$F_{n+1}(x)$$

$$={\int }_0^x{\int }_0^{2x-y_1}\cdots {\int }_0^{(n+1)x-(y_1+y_2+\cdots +y_n)}{e}^{-(y_1+y_2+\cdots +y_{n+1})}d{\mathbf{y}}_{n+1}$$

$$\displaystyle=\int_{0}^{x}\int_{0}^{2x-y_{1}}\cdots\int_{0}^{nx-(y_{1}+y_{2}+\cdots+y_{n-1})}\Big{\{}e^{-(y_{1}+y_{2}+\cdots+y_{n})}-e^{-(n+1)x}\Big{\}}d{\mathbf{y}}_{n}$$

$$=F_n(x)-e^{-(n+1)x}\mathrm{Vol}(K_n(x))$$

$$K_n(x):=\{(y_1,y_2,\dots ,y_n)\in {\mathbb{R}}_{+}^n:0\le y_1+\cdots +y_i\le ix,i=1,2,\dots ,n\}.$$

$$F_n(x)=1-\sum _{k=1}^n\frac{k^{k-1}}{k!}{x}^{k-1}{e}^{-kx},\text{ for all }x\ge 0,n\in {\mathbb{N}}^{+}.$$

$$f_{\infty }(x):={\text{1}}_{x>1}\sum _{k=1}^{\infty }\frac{k^{k-1}}{k!}\left(k-\frac{k-1}{x}\right)x^{k-1}{e}^{-kx}.$$

$$\sum _{k=1}^{\infty }\frac{k^{k-1}}{k!}{x}^{k-1}{e}^{-kx}=1\text{ for all }x\in [0,1).$$

$$\underset{x\ge 0}{\sup }\frac{k^{k-1}}{k!}{x}^{k-1}{e}^{-kx}=\frac{(k-1{)}^{k-1}}{k!}{e}^{-(k-1)}\sim \frac{1}{k^{3/2}\sqrt{2\pi }},$$

$$\sum _{k=1}^{\infty }\frac{k^{k-1}}{k!}(x{e}^{-x}{)}^k=x\text{ for all }x\in [0,1].$$

$$t(x):=g_r^{-1}(x{e}^{-x})=h(x{e}^{-x}),x\ge 0.$$

$$F_{\infty}(x)=\left\{\begin{array}[]{lll}0,&\mbox{if}&x\leq 1,\\ 1-\frac{t(x)}{x},&\mbox{if}&x\geq 1.\end{array}\right.$$

$$e^{xu}=\frac{e^{-t(x)}}{e^{-x}}=\frac{x}{t(x)}=\frac{1}{1-u}.$$

$$\text{mbox}\mathrm{IE}{Z}_n^{\alpha }=\alpha \sum _{k=1}^n\frac{\Gamma (\alpha +k-1)}{k^{\alpha }k!}.$$

$$\text{mbox}\mathrm{IE}{Z}_n=\sum _{k=1}^n\frac{1}{k^2}$$

$$K_n(x,t):=\{(y_1,y_2,\dots ,y_n)\in {\mathbb{R}}_{+}^n:y_1+\cdots +y_i\le ix+t\text{ for all }i=1,2,\dots ,n\}.$$

$$V_n(x,t):=\mathrm{Vol}(K_n(x,t))=\frac{1}{n!}(x+t)((n+1)x+t{)}^{n-1},n=1,2,\dots ,$$

$$V_{n+1}(x,t)$$

$$={\int }_0^{x+t}{\int }_0^{x+(x+t-y_1)}\cdots {\int }_0^{nx+(x+t-y_1)-(y_2+\cdots +y_n)}d{\mathbf{y}}_{n+1}$$

$$={\int }_0^{x+t}{V}_n(x,x+t-y_1)d{y}_1.$$

$$\underset{n\to \infty }{\lim }\sqrt{n}\left(M_n-\underset{k\in \mathbb{N},k\ge n}{\sup }\frac{W_k}{k}\right)=0$$

$$\underset{n\to \infty }{\lim }\sqrt{n}\left(\underset{k\in \mathbb{N},k\ge n}{\sup }\frac{W_k}{k}-\underset{s\ge n}{\sup }\frac{W_s}{s}\right)=0.$$

$$\sqrt{n}\underset{s\ge n}{\sup }\frac{W_s}{s}\stackrel{d}{=}\underset{s\ge 1}{\sup }\frac{W_s}{s}\stackrel{d}{=}\underset{s\in [0,1]}{\sup }{W}_s\stackrel{d}{=}|W_1|,$$

$$Z_{n;\lambda }:=\max \left\{\frac{X_1}{1+\lambda },\frac{X_1+X_2}{2+\lambda },\dots ,\frac{X_1+X_2+\cdots +X_n}{n+\lambda }\right\}$$

Full text

60E05,

60F05,

exponential distribution,

maximal average,

Lambert $W$ function,

ruin probability,

and

1 Introduction

Consider a sequence $(X_{i})_{i\geq 1}$ of independent identically distributed (i.i.d.) random variables, each having exponential distribution with mean 1. For each $i\in\mathbb{N}^{+}$ define the sample mean of the first $i$ variables as $\overline{X}_{i}:=(X_{1}+X_{2}+\cdots+X_{i})/i$. The supremum of this sequence

[TABLE]

is finite because the sequence converges to 1 with probability 1.

In this note we compute the distribution function, $F_{\infty}$, of $Z_{\infty}$. In fact, what has nice form is the inverse of this distribution function. Our main result is the following.

Theorem 1**.**

(a)* $Z_{\infty}$ has distribution function*

[TABLE]

for $x>0$, and density which is continuous on \mathbb{R}{\raise 1.29167pt\hbox{\scriptstyle\setminus}}\{1\}, positive on $(1,\infty)$, and zero on $(-\infty,1)$.

(b)* The restriction of $F_{\infty}$ on $(1,\infty)$ is one to one and onto $(0,1)$ with inverse*

[TABLE]

Remark 1**.**

(a) For $F_{\infty}$ we have the alternative expression

[TABLE]

where $W_{0}$ is the principal branch of the Lambert $W$ function, that is, the inverse function of $x\mapsto xe^{x},x\geq 1$; see [2]. Indeed, the power series $\sum_{k=1}^{\infty}\frac{k^{k-1}}{k!}y^{k}$ has interval of convergence $[-1/e,1/e]$ and equals $-W_{0}(-y)$.

(b) Clearly, the results of the theorem extend immediately to the case that the $X_{i}$’s are i.i.d. and $X_{1}=aY+b$ with $a>0$, $b\in\mathbb{R}$ and $Y\sim{\mbox{E}xp}(1)$. However, we were not able to find an explicit formula for the distribution of $Z_{\infty}$ for any other distribution of the $X_{i}$’s.

(c) Although it is intuitively clear that $F_{\infty}(x)>0$ for $x>1$, it is not entirely obvious how to verify it by direct calculations. However, this fact is evident from Theorem 1.

(d) Formula (1) enables the explicit calculation of the percentiles of $F_{\infty}$. Therefore, the result is useful for the following kind of problems: Suppose that a quality control machine calculates subsequent averages, and alarms if some average $\bar{X}_{n}$ is greater than $c$, where $c$ is a predetermined constant such that the probability of false alarm is small, say $\alpha$. For $\alpha\in(0,1)$, the upper percentage point of $F_{\infty}$ (that is, the point $c_{\alpha}$ with $F_{\infty}(c_{\alpha})=1-\alpha$) is given by $c_{\alpha}=\frac{-\log\alpha}{1-\alpha}$, and thus the proper value of $c$ is $c=c_{\alpha}$.

If in the definition of $Z_{\infty}$ we discard the first $n-1$ values of $\bar{X_{i}}$, we obtain the random variable

[TABLE]

for which, however, (for $n\geq 2$) the distribution function is quite complicated even for the exponential case. For instance, the distribution of $M_{2}$ is given by (we omit the details)

[TABLE]

What we can compute is the asymptotic distribution of $\sqrt{n}(M_{n}-1)$ as $n\to\infty$. This distribution is the same for a large class of distributions of the $X_{i}$’s, as the following theorem shows.

Theorem 2**.**

Assume that the $(X_{i})_{i\geq 1}$ are i.i.d. with mean 0, variance 1, and there is $p>2$ with $\mbox{\rm\hskip 0.86108ptIE\hskip 0.86108pt}|X_{1}|^{p}<\infty$. Let $M_{n}:=\sup\{\bar{X}_{i}:i\geq n\}$ for all $n\in\mathbb{N}^{+}$. Then,

[TABLE]

where $Z\sim N(0,1)$ is a standard normal random variable.

It is easy to see that under the assumptions of Theorem 2, by the law of the iterated logarithm, it holds

[TABLE]

2 Proofs

Proof of Theorem 1.

(a) For each $n\in\mathbb{N}^{+}$ consider the random variable

[TABLE]

and call $F_{n}$ its distribution function. The sequence $(Z_{n})_{n\geq 1}$ is increasing and $Z_{\infty}=\lim_{n\to\infty}Z_{n}$, $F_{\infty}(x)=\lim_{n\to\infty}F_{n}(x)$. We will compute $F_{n}$ recursively.

For $n\in\mathbb{N}^{+}$ and $x\geq 0$ we have

[TABLE]

where $d{\mathbf{y}}_{k}=dy_{k}\cdots dy_{2}dy_{1}$ and

[TABLE]

Note that $F_{1}(x)=1-e^{-x}$ and introduce the convention $\operatorname{Vol}(K_{0}(x))=1$. It follows that $F_{n}(x)=1-\sum_{k=1}^{n}\operatorname{Vol}(K_{k-1}(x))e^{-kx}$ and from Lemma 1, below, we get the explicit form

[TABLE]

This implies the first formula for $F_{\infty}$. By the law of large numbers, we get that $F_{\infty}(x)=0$ for all $x\in(-\infty,1)$, and thus, the derivative of $F_{\infty}$ in \mathbb{R}{\raise 1.29167pt\hbox{\scriptstyle\setminus}}\{1\} is

[TABLE]

(b) First we rewrite $F_{\infty}$ in a more convenient form. The fact that $F_{\infty}(x)=0$ for $x\in[0,1)$ implies the remarkable identity (see Fig. 1)

[TABLE]

Our aim is to compute the value of the series in the left hand side also for $x\geq 1$. The series converges uniformly for $x\in[0,\infty)$ because

[TABLE]

which is summable in $k$. Thus, by continuity, (2) holds also for $x=1$. Now we rewrite (2) in the form

[TABLE]

The power series $h(y):=\sum_{k=1}^{\infty}\frac{k^{k-1}}{k!}y^{k}$ is strictly increasing in $[0,e^{-1}]$ and thus (3) says that $h$ is the inverse function of the restriction, $g_{r}$, on $[0,1]$ of the function $g:[0,\infty)\to[0,e^{-1}]$ with $g(x)=xe^{-x}$. The function $g$ is continuous, strictly increasing in $[0,1]$, and strictly decreasing in $[1,\infty)$ with $g(0)=0,g(1)=e^{-1},g(\infty)=0$. Thus, for each $x\in[1,\infty)$, there exists a unique $t=t(x)\in(0,1]$ such that $g_{r}(t)=xe^{-x}$, i.e., $te^{-t}=xe^{-x}$; hence, we define

[TABLE]

Since $t(x)=x$ for $x\in[0,1]$, we have

[TABLE]

Now for any fixed $u\in(0,1)$, the relation $F_{\infty}(x)=u$ gives $x-t(x)=xu$ so that $t(x)=(1-u)x$. Consequently,

[TABLE]

Thus, $x=-\log(1-u)/u$ and the proof is complete. ∎

Remark 2**.**

From the well-known relation $\mbox{\rm\hskip 0.86108ptIE\hskip 0.86108pt}Z_{n}^{\alpha}=\alpha\int_{0}^{\infty}x^{\alpha-1}(1-F_{n}(x))dx$ for $\alpha>0$, we obtain a simple expression for the moments:

[TABLE]

In particular,

[TABLE]

Since $Z_{n}\nearrow Z_{\infty}$ with probability one, the above relations combined with the monotone convergence theorem give the moments of $Z_{\infty}$ and in particular that it has mean $\frac{\pi^{2}}{6}$ and variance $\frac{\pi^{2}}{6}(2-\frac{\pi^{2}}{6})$.

The next lemma is a special case of Theorem 1 in [5] (see relation (7) in that paper), however, to keep the exposition self-contained, we provide a proof.

Lemma 1**.**

For $x\geq 0$, $x+t\geq 0$, and $n\in\mathbb{N}^{+}$, define

[TABLE]

Then,

[TABLE]

and, in particular, setting $t=0$, $\operatorname{Vol}(K_{n}(x))=\frac{1}{n!}(n+1)^{n-1}x^{n}$.

Proof.

Clearly $V_{1}(x,t)=x+t$ and for $n\geq 1$

[TABLE]

The claim follows by induction on $n$. ∎

It is consistent with the recursion (7) for $V_{n}$ and (6) to define $V_{0}(x,t):=1$ so that (6) holds for all $n\in\mathbb{N}^{+}\cup\{0\}$. This agrees with the convention $\operatorname{Vol}(K_{0}(x))=1$ we made in the proof of Theorem 1(a).

Proof of Theorem 2.

By Theorem 2.2.4 in [3] we may assume that we can place $(X_{i})_{i\geq 1}$ in the same probability space with a standard Brownian motion $(W_{s})_{s\geq 0}$, so that, with probability 1, we have $|n\bar{X}_{n}-W_{n}|/n^{1/p}(\log n)^{1/2}\to 0$ as $n\to\infty$. This implies that

[TABLE]

with probability 1. On the other hand, with probability one, we have for all large $n$ the bound $\sup_{s\in[n,n+1]}|W_{s}-W_{n}|\leq 2\sqrt{\log n}$, thus

[TABLE]

Finally, by scaling and time inversion, we conclude that

[TABLE]

and the proof is complete. ∎

3 An application to ruin probability

Following the same steps as in the proof of Theorem 1(b), one can evaluate the distribution function, $F_{n;\lambda}$, of the random variable

[TABLE]

for all $\lambda>-1$ and $n\in\mathbb{N}^{+}$. Indeed, using (6) and induction on $n$ it is easily verified that for all $x\geq 0$ we have

[TABLE]

Thus, the distribution function of $Z_{\infty,\lambda}:=\lim_{n\to\infty}Z_{n;\lambda}$ equals

[TABLE]

where the function $t$ is defined by (4). To justify the equality (9), we use the same arguments that lead from (2) to (5). Similarly as in Theorem 1(b), we find that $F_{\infty;\lambda}$ is zero in $(-\infty,1]$, strictly increasing in $[1,\infty)$ with range $[0,1)$, and its distribution inverse is given by

[TABLE]

Remark 3**.**

By the law of large numbers, the series in the right hand side of (8) equals to one for all $x\in[0,1]$. Therefore, setting $x=\alpha$, $1+\lambda=\theta$ and $k\to k+1$, the function

[TABLE]

defines a probability mass function supported on $\mathbb{N}^{+}\cup\{0\}$, known (after a suitable re-parametrization) as generalized Poisson distribution with parameter $(\alpha,\theta)\in[0,1]\times(0,\infty)$; see [1] and references therein.

Consider now the following risk model. Assume that the aggregate claim at time $n$ is described by $S_{n}:=X_{1}+\cdots+X_{n}$, where the $(X_{i})_{i\geq 1}$ are i.i.d. with $\mbox{\rm\hskip 0.86108ptIE\hskip 0.86108pt}X_{1}=1$, the premium rate (per time unit) is $c=1+\theta>0$ ($\theta$ is the safety loading of the insurance), and the initial capital is $u>-(1+\theta)$, where negative initial capital is allowed for technical reasons. The risk process is defined by

[TABLE]

Clearly, the ruin probability

[TABLE]

is of fundamental importance. Our explicit formulae are useful in computing the minimum initial capital needed to ensure that $\psi(u)$ is small. The particular problem (for general claims) has been studied in [4], under the name discrete-time surplus-process model. It is well-known that $\psi(u)=1$ when $c\leq 1$, no matter how large $u$ is, because $\mbox{\rm\hskip 0.86108ptIE\hskip 0.86108pt}X_{i}=1$. Hence, the problem is meaningful only for $c>1$, i.e., $\theta>0$.

Theorem 3**.**

Assume that the i.i.d. individual claims $(X_{i})_{i\geq 1}$ are exponential random variables with mean 1, fix $\alpha\in(0,1)$ and $\theta>0$, and set $c=1+\theta$. Then,

(a)* the ruin probability (11) is given by*

[TABLE]

where the function $t$ is given by (4);

(b)* the minimum initial capital $u=u(\alpha,\theta)$ needed to ensure that $\psi(u)\leq\alpha$ is given by the unique root of the equation*

[TABLE]

Proof.

(a) For $u>-c$, we can use (9) to get

[TABLE]

which is (12). Then, the definition of $t$ shows that $\lim_{u\to-c^{+}}\psi(u)=\frac{t(c)e^{-t(c)}}{ce^{-c}}=1$, and the monotonicity of $\psi$ implies that $\psi(u)=1$ for $u\leq-c$.

(b) By the formula of part (a), the function $\psi$ is strictly decreasing in the interval $(-c,\infty)$ and maps that interval to $(0,1)$. Therefore, there is a unique $u=u(\alpha,\theta)>-c$ such that $\psi(u)=\alpha$. Let $\lambda:=u/c$, which is greater than $-1$. Then, using (10), we see that

[TABLE]

We substitute $c=1+\theta,\lambda=u/(1+\theta)$, and the above equivalences show that $u$ is the unique solution of

[TABLE]

∎

The exact values of $u$ in (13) are in perfect agreement with the numerical approximations given in the last line of Table 1 in [4]. Notice that the initial capital $u$ can be negative sometimes, e.g., $u(.5,.5)\simeq-.3107$.