Properties and generation of representative points of the exponential distribution

Abstract

It is known that the exponential distribution has many nice properties. Graf and Luschgy (2000) pointed out that the mean squared error of the set of representative points of the exponential distribution is fully determined by the smallest representative point. In this paper we concern with the representative points of the exponential distribution and find a number of new interesting properties. A new algorithm is proposed to effectively generate representative points of the exponential distribution. In addition, the performance of representative points of the exponential distribution is evaluated.

Appendix

In this appendix we provide proofs for all theorems (excluding Theorem 1 and Theorem 7), corollaries and lemmas listed in Sect. 3.

Proof of Corollary 1

Combined Theorem 1 with Lemma 2, it is easy to obtain

$$\begin{aligned} a_1^2&= E(X^2)-E(Y_{MSE}^2)\\&= E(X^2)-(E(X))^2-(E(Y_{MSE}^2)-(E(Y_{MSE}))^2)\\&= \text{ Var }(X)-\text{ Var }(Y_{MSE}). \end{aligned}$$

Thus,

$$\begin{aligned} \text{ Var }(Y_{MSE}) = \text{ Var }(X)-a_1^2 = \frac{1}{\lambda ^2}-a_1^2, \end{aligned}$$

which completes the proof.

Proof of Theorem 2

The last equation in (6) gives us the following equation

$$\begin{aligned} \int _{(a_{n-1}^{(n)}+a_n^{(n)})/2}^{\infty }(x-a_n^{(n)})p(x)dx=0. \end{aligned}$$

(10)

Then,

$$\begin{aligned}&\int _{(a_{n-1}^{(n)}+a_n^{(n)})/2}^{\infty }(x-a_n^{(n)})p(x)dx = \int _{(a_{n-1}^{(n)}+a_n^{(n)})/2}^{\infty }x\lambda e^{-\lambda x}dx-a_n^{(n)}\int _{(a_{n-1}^{(n)}+a_n^{(n)})/2}^{\infty }\lambda e^{-\lambda x}dx\\&\quad = -xe^{-\lambda x}|_{(a_{n-1}^{(n)}+a_n^{(n)})/2}^{\infty }+\frac{1}{\lambda }\int _{(a_{n-1}^{(n)}+a_n^{(n)})/2}^{\infty }\lambda e^{-\lambda x}dx-a_n^{(n)}\int _{(a_{n-1}^{(n)}+a_n^{(n)})/2}^{\infty }\lambda e^{-\lambda x}dx\\&\quad =\frac{1}{2}(a_{n-1}^{(n)}+a_n^{(n)})e^{-\frac{\lambda }{2}(a_{n-1}^{(n)}+a_n^{(n)})}+(\frac{1}{\lambda }-a_n^{(n)})\int _{(a_{n-1}^{(n)}+a_n^{(n)})/2}^{\infty }\lambda e^{-\lambda x}dx\\&\quad =\frac{1}{2}(a_{n-1}^{(n)}+a_n^{(n)})e^{-\frac{\lambda }{2}(a_{n-1}^{(n)}+a_n^{(n)})}+(\frac{1}{\lambda }-a_n^{(n)})[-e^{-\lambda x}]_{(a_{n-1}^{(n)}+a_n^{(n)})/2}^{\infty }\\&\quad =\frac{1}{2}(a_{n-1}^{(n)}+a_n^{(n)})e^{-\frac{\lambda }{2}(a_{n-1}^{(n)}+a_n^{(n)})}+(\frac{1}{\lambda }-a_n^{(n)})e^{-\frac{\lambda }{2}(a_{n-1}^{(n)}+a_n^{(n)})}\\&\quad =(\frac{a_{n-1}^{(n)}+a_n^{(n)}}{2}+\frac{1}{\lambda }-a_n^{(n)})e^{-\frac{\lambda }{2}(a_{n-1}^{(n)}+a_n^{(n)})}\\&\quad =\frac{1}{2}(a_{n-1}^{(n)}-a_n^{(n)}+\frac{2}{\lambda })e^{-\frac{\lambda }{2}(a_{n-1}^{(n)}+a_n^{(n)})}. \end{aligned}$$

Since $e^{-\frac{\lambda }{2}(a_{n-1}^{(n)}+a_n^{(n)})}>0$, it follows that $a_{n-1}^{(n)}-a_n^{(n)}+\frac{2}{\lambda }=0$, which completes the proof.

Remark 1

Let $y=x-a_n^{(n)}$ in (10), then we can obtain

$$\begin{aligned} \int _{(a_{n-1}^{(n)}-a_n^{(n)})/2}^{\infty }yp(y+a_n^{(n)})dy=0. \end{aligned}$$

(11)

From the memoryless property of the exponential distribution, we have

$$\begin{aligned} p(y+a_n^{(n)})=\lambda e^{-\lambda (y+a_n^{(n)})}=e^{-\lambda a_n^{(n)}}\lambda e^{-\lambda y}=e^{-\lambda a_n^{(n)}}p(y). \end{aligned}$$

Thus, equation (11) becomes

$$\begin{aligned} e^{-\lambda a_n^{(n)}}\int _{(a_{n-1}^{(n)}-a_n^{(n)})/2}^{\infty }yp(y)dy=0. \end{aligned}$$

Since $e^{-\lambda a_n^{(n)}}>0$, it follows that

$$\begin{aligned} \int _{(a_{n-1}^{(n)}-a_n^{(n)})/2}^{\infty }yp(y)dy=0. \end{aligned}$$

(12)

By using a similar way to analyse (12), the proof is completed.

For proving Theorem 3 we need the following lemma.

Lemma 5

If $0<\theta _1\le \frac{1}{\lambda }$ and $\theta _2>0$ satisfy the following equation

$$\begin{aligned} e^{\lambda \theta _2}(\frac{1}{\lambda }-\theta _2)=(\theta _1+\frac{1}{\lambda })e^{-\lambda \theta _1}, \end{aligned}$$

(13)

then $0<\theta _2<\theta _1\le \frac{1}{\lambda }$. Further, if $\theta _1$ is given, then $\theta _2$ must be unique, namely, there is a unique solution in the following equation

$$\begin{aligned} e^{\lambda x}(\frac{1}{\lambda }-x)=(\theta _1+\frac{1}{\lambda })e^{-\lambda \theta _1}, \quad \text{ when } 0<x<\theta _1. \end{aligned}$$

Proof of Lemma 5

Let

$$\begin{aligned} f(x)=(x+\frac{1}{\lambda })e^{-\lambda x},x>0, \end{aligned}$$

then equation (13) becomes $f(\theta _1)=f(-\theta _2)$. Define

$$\begin{aligned} h(x)&= f(x)-f(-x)\\&=(x+\frac{1}{\lambda })e^{-\lambda x}-(-x+\frac{1}{\lambda })e^{\lambda x},x>0. \end{aligned}$$

First, we show the monotonicity of h(x). The derivative of h(x) is given as follows

$$\begin{aligned} h'(x)&= e^{-\lambda x}+(x+\frac{1}{\lambda })(-\lambda )e^{-\lambda x}-(-e^{\lambda x}+(-x+\frac{1}{\lambda })(\lambda )e^{\lambda x})\\&= -\lambda xe^{-\lambda x}+\lambda xe^{\lambda x}= \lambda xe^{-\lambda x}(-1+e^{2\lambda x}). \end{aligned}$$

It follows that h(x) is a strict increasing function on $(0,+\infty )$ as $h'(x)>0$ when $x>0$. Thus, $h(x)>h(0)=0$, namely, $f(x)>f(-x)$ when $x>0$. As a consequence, $f(\theta _1)<f(-\theta _1)$ when $0<\theta _1\le \frac{1}{\lambda }$. Under the condition $f(-\theta _2)=f(\theta _1)$, it is clear that $f(-\theta _2)=f(\theta _1)>f(-\theta _1)$. Now, since $f'(x)=e^{-\lambda x}+(x+\frac{1}{\lambda })(-\lambda )e^{-\lambda x}=-\lambda xe^{-\lambda x}>0$ when $x<0$, it follows that f(x) is a strict increasing function on $(-\infty ,0)$. Thus, $f(-\theta _2)>f(-\theta _1)\Leftrightarrow -\theta _2>-\theta _1$, namely $\theta _1>\theta _2$.

Next, we will show the uniqueness of the solution of $g(x)=0$ on $(0,\theta _1)$, where $0<\theta _1\le \frac{1}{\lambda }$ and

$$\begin{aligned} g(x)=e^{\lambda x}(\frac{1}{\lambda }-x)-(\theta _1+\frac{1}{\lambda })e^{-\lambda \theta _1}. \end{aligned}$$

Since $g'(x) = \lambda e^{\lambda x}(\frac{1}{\lambda }-x)+e^{\lambda x}(-1)=-\lambda x e^{\lambda x}<0$ when $0<x<\theta _1\le \frac{1}{\lambda }$, it follows that g(x) is a strict decreasing function on $(0,\theta _1)$. Then, the sign of g(0) can be determined as follows

$$\begin{aligned} g(0)&= \frac{1}{\lambda }-(\theta _1+\frac{1}{\lambda })e^{-\lambda \theta _1}\\&= \frac{1}{\lambda }e^{-\lambda \theta _1}(e^{\lambda \theta _1}-(\lambda \theta _1+1))\\&>0 \quad (\text{ since } e^x>x+1 \text{ for } x\in {\mathbb {R}}^+). \end{aligned}$$

Because g(x) is a strict decreasing function on $(0,\theta _1)$ and $g(0)>0$, if we could prove $g(\theta _1)<0$, then g(x) must have a unique solution on $(0,\theta _1)$ based on intermediate value theorem. The function $g(\theta _1)$ can be expressed as

$$\begin{aligned} g(\theta _1)&= e^{\lambda \theta _1}(\frac{1}{\lambda }-\theta _1)-(\theta _1+\frac{1}{\lambda })e^{-\lambda \theta _1}\\&=\frac{1}{\lambda }e^{-\lambda \theta _1}((1-\lambda \theta _1)e^{2\lambda \theta _1}-(\lambda \theta _1+1)). \end{aligned}$$

In order to prove $g(\theta _1)<0$, it is equivalent to prove the following inequality:

$$\begin{aligned} (1-\lambda \theta _1)e^{2\lambda \theta _1}-(\lambda \theta _1+1)<0,\quad 0<\theta _1\le \frac{1}{\lambda }. \end{aligned}$$

(14)

Let $k(x)=(1-x)e^{2x}-(x+1),0<x\le 1$. It is easy to calculate that

$$\begin{aligned} k'(x)&= -e^{2x}+2(1-x)e^{2x}-1= e^{2x}-2xe^{2x}-1,\\ k''(x)&= 2e^{2x}-2(e^{2x}+2xe^{2x})= -4xe^{2x}<0. \end{aligned}$$

Since $k''(x)<0$ when $0<x\le 1$, it follows that $k'(x)$ is a strict decreasing function on (0, 1] and we can obtain $k'(x)<k'(0)=0$ when $0<x\le 1$. It leads that k(x) is a strict decreasing function on (0, 1], so $k(x)<k(0)=0$ when $0<x\le 1$ and equation (14) holds, i.e., $g(\theta _1)<0$. Based on intermediate value theorem, we finish the proof.

Proof of Theorem 3

From equation (6), we can obtain

$$\begin{aligned} (\frac{1}{\lambda }+\frac{a_{i+1}^{(n)}-a_i^{(n)}}{2})e^{-\lambda \frac{a_{i+1}^{(n)}-a_i^{(n)}}{2}} =(\frac{1}{\lambda }-\frac{a_i^{(n)}-a_{i-1}^{(n)}}{2})e^{\lambda \frac{a_i^{(n)}-a_{i-1}^{(n)}}{2}} \end{aligned}$$

(15)

for $i=2,...,n-1$. From Theorem 2, we know the final gap $a_{n}^{(n)}-a_{n-1}^{(n)}$ is constant $\frac{2}{\lambda }$. By the mathematical induction, if we set $\theta _1=(a_{i+1}^{(n)}-a_i^{(n)})/2$ and $\theta _2=(a_i^{(n)}-a_{i-1}^{(n)})/2$ in Lemma 5 and the condition is satisfied for each $i=2,...,n-1$, then we can obtain

$$\begin{aligned} \frac{a_{i+1}^{(n)}-a_i^{(n)}}{2}>\frac{a_i^{(n)}-a_{i-1}^{(n)}}{2},\quad i=2,...,n-1, \end{aligned}$$

which completes the proof.

Proof of Theorem 4

According to equation (6) and Theorem 2, we can obtain the following two system of equations. The first one is for n MSE RPs and the second is for $n+1$ case.

$$\begin{aligned}&\left\{ \begin{array}{lr} \displaystyle {(\frac{1}{\lambda }+\frac{a_{i+1}^{(n)}-a_i^{(n)}}{2})e^{-\lambda \frac{a_{i+1}^{(n)}-a_i^{(n)}}{2}} =(\frac{1}{\lambda }-\frac{a_i^{(n)}-a_{i-1}^{(n)}}{2})e^{\lambda \frac{a_i^{(n)}-a_{i-1}^{(n)}}{2}}, i=2,..,n-1}\\ \displaystyle {a_{n}^{(n)}-a_{n-1}^{(n)}=\frac{2}{\lambda }}. \end{array} \right. \\&\left\{ \begin{array}{lr} \displaystyle {(\frac{1}{\lambda }+\frac{a_{j+1}^{(n+1)}-a_j^{(n+1)}}{2})e^{-\lambda \frac{a_{j+1}^{(n+1)}-a_j^{(n+1)}}{2}} =(\frac{1}{\lambda }-\frac{a_j^{(n+1)}-a_{j-1}^{(n+1)}}{2})e^{\lambda \frac{a_j^{(n+1)}-a_{j-1}^{(n+1)}}{2}}, j=3,...,n-1}\\ \displaystyle {a_{n+1}^{(n+1)}-a_{n}^{(n+1)}=\frac{2}{\lambda }}. \end{array} \right. \end{aligned}$$

If we compare these two system equations for each pair i and j, the proof is completed by using Theorem 3 and Lemma 5 directly.

Proof of Theorem 5

Set $\delta =(a_1^{(n+1)}+a_2^{(n+1)})/2$ and $\rho =1$ in Lemma 4, then $X+\delta$ has has the density function

$$\begin{aligned} q(x)=\frac{\lambda e^{-\lambda x}}{e^{-\lambda \delta }}{\mathbb {I}}_{(x\ge \delta )}, \end{aligned}$$

and its RPs are

$$\begin{aligned} \beta _i^{(n)}=a_i^{(n)}+\frac{a_1^{(n+1)}+a_2^{(n+1)}}{2}. \end{aligned}$$

(16)

Consider the $n+1$ MSE RPs of $EP(\lambda )$ and denote them by $a_1^{(n+1)},...,a_{n+1}^{(n+1)}$ with $a_1^{(n+1)}<...<a_{n+1}^{(n+1)}$. And those points must satisfy

$$\begin{aligned} \left\{ \begin{array}{lr} \displaystyle {\int _{0}^{(a_1^{(n+1)}+a_2^{(n+1)})/2}(x-a_1^{(n+1)})p(x)dx=0}\\ \displaystyle {\int _{(a_{i-1}^{(n+1)}+a_i^{(n+1)})/2}^{(a_i^{(n+1)}+a_{i+1}^{(n+1)})/2}(x-a_i^{(n+1)})p(x)dx=0,\quad i=2,...,n,}\\ \displaystyle {\int _{(a_{n}^{(n+1)}+a_{n+1}^{(n+1)})/2}^{\infty }(x-a_{n+1}^{(n+1)})p(x)dx=0} \end{array} \right. \end{aligned}$$

where $p(x)=\lambda e^{-\lambda x},x\ge 0$. With some simple calculation, we can obtain the following n equations with deleting the first original equation

$$\begin{aligned} \left\{ \begin{array}{lr} \displaystyle {(\frac{a_{i-1}^{(n+1)}-a_i^{(n+1)}}{2}+\frac{1}{\lambda })e^{-\frac{\lambda }{2}(a_{i-1}^{(n+1)}+a_i^{(n+1)})}=(\frac{a_{i+1}^{(n+1)}-a_i^{(n+1)}}{2}+\frac{1}{\lambda })e^{-\frac{\lambda }{2}(a_{i+1}^{(n+1)}+a_i^{(n+1)})}, i=2,...,n}\\ \displaystyle {a_{n+1}^{(n+1)}-a_n^{(n+1)}=\frac{2}{\lambda }} \end{array} \right. \end{aligned}$$

(17)

For q(x), as we know that $\beta _1^{(n)},...,\beta _n^{(n)}$ are the n MSE RPs of q(x) where $a=(a_1^{(n+1)}+a_2^{(n+1)})/2$, then we can obtain the following n equations

$$\begin{aligned} \left\{ \begin{array}{lr} \displaystyle {(\frac{a_1^{(n+1)}-(2\beta _1^{(n)}-a_2^{(n+1)})}{2}+\frac{1}{\lambda })e^{-\frac{\lambda }{2}(a_1^{(n+1)}+a_2^{(n+1)})}=(\frac{\beta _2^{(n)}-\beta _1^{(n)}}{2}+\frac{1}{\lambda })e^{-\frac{\lambda }{2}(\beta _1^{(n)}+\beta _2^{(n)})}}\\ \displaystyle {(\frac{\beta _{i-1}^{(n)}-\beta _i^{(n)}}{2}+\frac{1}{\lambda })e^{-\frac{\lambda }{2}(\beta _{i-1}^{(n)}+\beta _i^{(n)})}=(\frac{\beta _{i+1}^{(n)}-\beta _{i}^{(n)}}{2}+\frac{1}{\lambda })e^{-\frac{\lambda }{2}(\beta _i^{(n)}+\beta _{i+1}^{(n)})},\quad i=2,...,n-1}\\ \displaystyle {\beta _{n}^{(n)}-\beta _{n-1}^{(n)}=\frac{2}{\lambda }} \end{array} \right. \end{aligned}$$

(18)

Based on Lemma 3, the solutions of equation (18) must be unique. Thus, compared equations (17) with (18), we can obtain

$$\begin{aligned} \beta _i^{(n)}=a_{i+1}^{(n+1)},\quad i=1,...,n. \end{aligned}$$

With equation (16), we conclude

$$\begin{aligned} a_i^{(n)}+\frac{a_1^{(n+1)}+a_2^{(n+1)}}{2}=a_{i+1}^{(n+1)},\quad i=1,...,n. \end{aligned}$$

Particularly,

$$\begin{aligned} a_{2}^{(n+1)}=a_1^{(n)}+\frac{a_1^{(n+1)}+a_2^{(n+1)}}{2}, \end{aligned}$$

which completes the proof.

Proof of Corollary 2

For $n+1$ MSE RPs, the first integral equation in (6) becomes

$$\begin{aligned} \int _0^{(a_1^{(n+1)}+a_2^{(n+1)})/2}(x-a_1^{(n+1)})p(x)dx=0. \end{aligned}$$

After simplification, we obtain

$$\begin{aligned} \frac{1}{\lambda }-a_1^{(n+1)}=(\frac{a_2^{(n+1)}-a_1^{(n+1)}}{2}+\frac{1}{\lambda })e^{-\frac{\lambda }{2}(a_1^{(n+1)}+a_2^{(n+1)})}. \end{aligned}$$

Multiplying $e^{\lambda a_1^{(n+1)}}$ on both sides, the above equation becomes

$$\begin{aligned} e^{\lambda a_1^{(n+1)}}(\frac{1}{\lambda }-a_1^{(n+1)})=(\frac{a_2^{(n+1)}-a_1^{(n+1)}}{2}+\frac{1}{\lambda })e^{-\frac{\lambda }{2}(a_2^{(n+1)}-a_1^{(n+1)})}. \end{aligned}$$

Substituting $a_2^{(n+1)}-a_1^{(n+1)}=2a_1^{(n)}$ based on Theorem 5, we have

$$\begin{aligned} e^{\lambda a_1^{(n+1)}}(\frac{1}{\lambda }-a_1^{(n+1)})=(a_1^{(n)}+\frac{1}{\lambda })e^{-\lambda a_1^{(n)}}, \end{aligned}$$

which completes the proof.

Proof of Theorem 6

The proof will be divided into two steps. First, in the process of proving Theorem 5, we can obtain

$$\begin{aligned} a_i^{(n)}+\frac{a_1^{(n+1)}+a_2^{(n+1)}}{2}=a_{i+1}^{(n+1)},\quad i=1,...,n. \end{aligned}$$

Thus, $a_i^{(n)}<a_{i+1}^{(n+1)}$ for $i=1,2,...,n$. Second, we consider the relationship between $a_i^{(n+1)}$ and $a_i^{(n)}$ for $i=1,2,...,n$. For $a_1^{(n+1)}$ and $a_1^{(n)}$, from Corollary 2 and Lemma 5, it is easy to obtain $a_1^{(n+1)}<a_1^{(n)}$. For $a_2^{(n+1)}$ and $a_2^{(n)}$, in the process of proving Corollary 2, we consider n MSE RPs, and obtain

$$\begin{aligned} e^{\lambda a_1^{(n)}}(\frac{1}{\lambda }-a_1^{(n)})=(\frac{a_2^{(n)}-a_1^{(n)}}{2}+\frac{1}{\lambda })e^{-\frac{\lambda }{2}(a_2^{(n)}-a_1^{(n)})}. \end{aligned}$$

So, from Lemma 5, we can obtain

$$\begin{aligned} \frac{a_2^{(n)}-a_1^{(n)}}{2}>a_1^{(n)}. \end{aligned}$$

Then,

$$\begin{aligned}&a_2^{(n)}-a_1^{(n)}>2a_1^{(n)}>a_1^{(n+1)}+a_1^{(n)}\\&\Rightarrow 2a_1^{(n)}+a_1^{(n+1)}<a_2^{(n)}\\&\Rightarrow a_2^{(n+1)}<a_2^{(n)}. \end{aligned}$$

For $a_i^{(n+1)}$ and $a_i^{(n)},i=3,...,n$, from Theorem 3 and the above results, we obtain

$$\begin{aligned}&a_1^{(n)}+a_1^{(n+1)}<a_2^{(n)}-a_1^{(n)}<a_3^{(n)}-a_2^{(n)}<...<a_i^{(n)}-a_{i-1}^{(n)},\quad i=3,...,n\\&\Rightarrow a_1^{(n)}+a_1^{(n+1)}<a_i^{(n)}-a_{i-1}^{(n)},\quad i=3,...,n\\&\Rightarrow \frac{a_1^{(n+1)}+a_2^{(n+1)}}{2}<a_i^{(n)}-a_{i-1}^{(n)},\quad i=3,...,n\\&\Rightarrow a_{i-1}^{(n)}+\frac{a_1^{(n+1)}+a_2^{(n+1)}}{2}<a_i^{(n)},\quad i=3,...,n\\&\Rightarrow a_{i}^{(n+1)}<a_i^{(n)},\quad i=3,...,n \end{aligned}$$

Thus, $a_i^{(n+1)}<a_{i}^{(n)}$ for $i=1,...,n$.