Exponential and Hypoexponential Distributions: Some Characterizations

Abstract

The (general) hypoexponential distribution is the distribution of a sum of independent exponential random variables. We consider the particular case when the involved exponential variables have distinct rate parameters. We prove that the following converse result is true. If for some $n\ge 2$, $X_1,X_2,\dots ,X_n$ are independent copies of a random variable X with unknown distribution F and a specific linear combination of $X_j$’s has hypoexponential distribution, then F is exponential. Thus, we obtain new characterizations of the exponential distribution. As corollaries of the main results, we extend some previous characterizations established recently by Arnold and Villaseñor (2013) for a particular convolution of two random variables.

1. Introduction and Main Results

Sums of exponentially distributed random variables play a central role in many stochastic models of real-world phenomena. Hypoexponential distribution is the convolution of k exponential distributions each with their own rate ${\lambda }_i$, the rate of the $i^{th}$ exponential distribution. As an example, consider the distribution of the time to absorption of a finite state Markov process. If we have a $k+1$ state process, where the first k states are transient and the state $k+1$ is an absorbing state, then the time from the start of the process until the absorbing state is reached is phase-type distributed. This becomes the hypoexponential if we start in state 1 and move skip-free from state i to $i+1$ with rate ${\lambda }_i$ until state k transitions with rate ${\lambda }_k$ to the absorbing state $k+1$.

We write $Z_i\sim \mathrm{Exp}\left({\lambda }_i\right)$ for ${\lambda }_i>0$, if $Z_i$ has density

$$f_i\left(z\right)={\lambda }_i{\mathrm{e}}^{-{\lambda }_{i}z},z\ge 0\left(\mathit{exponential}\mathit{distribution}\right).$$

The distribution of the sum $S_n:=Z_1+Z_2+\dots +Z_n$, where ${\lambda }_i$ for $i=1,\dots ,n$ are not all identical, is called (general) hypoexponential distribution (see [1,2]). It is absolutely continuous and we denote by $g_n$ its density. It is called the hypoexponetial distribution as it has a coefficient of variation less than one, compared to the hyper-exponential distribution which has coefficient of variation greater than one and the exponential distribution which has coefficient of variation of one. In this paper, we deal with a particular case of the hypoexponential distribution when all ${\lambda }_i$ are distinct, i.e., ${\lambda }_i\ne {\lambda }_j$ when $i\ne j$. In this case, it is known ([3], p. 311; [4], Chapter 1, Problem 12) that

$$S_n=Z_1+Z_2+\dots +Z_n\mathrm{has}\mathrm{density}{g}_n\left(z\right):={\displaystyle \sum _{j=1}^n}{\ell }_j{f}_j\left(z\right),z\ge 0.$$

(1)

Here the weight ${\ell }_j$ is defined as

$${\ell }_j={\displaystyle \prod _{i=1,i\ne j}^n}\frac{{\lambda }_i}{{\lambda }_i-{\lambda }_j}.$$

Please note that ${\ell }_j:={\ell }_j\left(0\right)$, where ${\ell }_1\left(x\right),\dots ,{\ell }_n\left(x\right)$ are identified (see [5]) as the Lagrange basis polynomials associated with the points ${\lambda }_1,\dots ,{\lambda }_n$. The convolution density $g_n$ in (1) is the weighted average of the values of the densities of $Z_1,Z_2,\dots ,Z_n$, where the weights ${\ell }_j$ sum to 1 (see [5]). Notice, however, since the weights can be both positive or negative, $g_n$ is not a “usual” mixture of densities. If we place ${\lambda }_j$’s in increasing or decreasing order, then the corresponding coefficients ${\ell }_j$’s alternate in sign.

Consider the Laplace transforms ${\phi }_i\left(t\right):=\mathsf{E}\left[{\mathrm{e}}^{-t{Z}_i}\right]$, $t\ge 0$, $i=1,2,\dots ,n$. They are well-defined and will play a key role in the proofs of the main results.

To begin with, let us look at the case when all $Z_i$’s are identically distributed, i.e., ${\lambda }_i=\lambda$ for $i=1,2,\dots ,n$, so we can use $\phi$ for the common Laplace transform. The sum $S_n=Z_1+Z_2+\dots +Z_n$ has Erlang distribution whose Laplace transform $\tilde{\phi }$, because of the independence, is expressed as follows:

$$\tilde{\phi }\left(t\right)=\mathsf{E}\left[{\mathrm{e}}^{-t{S}_n}\right]={\phi }^n\left(t\right)={\left(\frac{\lambda }{\lambda +t}\right)}^n.$$

If we go in the opposite direction, assuming that $S_n$ has Erlang distribution with Laplace transform $\tilde{\phi }$, then we conclude that ${\phi }_i\left(t\right)=\lambda {(\lambda +t)}^{-1}$ for each $i=1,2,\dots ,n$, which in turn implies that $Z_i\sim \mathrm{Exp}\left(\lambda \right)$. By words, if $Z_i$ are independent and identically distributed random variables and their sum has Erlang distribution, then the common distribution is exponential.

Does a similar characterization hold when the rate parameters${\lambda }_i$are all different? The answer to this question is not obvious. It is our goal in this paper to show that the answer is positive.

Let ${\mu }_1,{\mu }_2,\dots ,{\mu }_n$ be positive real numbers, such that ${\lambda }_i=\lambda /{\mu }_i$. Without loss of generality suppose that ${\mu }_1>{\mu }_2>\dots >{\mu }_n>0$. Assume that $X_1,X_2,\dots ,X_n$, for fixed $n\ge 2$, are independent and identically distributed as a random variable X with density f, $f\left(x\right)=\lambda {\mathrm{e}}^{-\lambda x}$, $x>0$. Then (1) is equivalent to the following:

$$S_n:={\mu }_1{X}_1+{\mu }_2{X}_2+\cdots +{\mu }_n{X}_n\mathrm{has}\mathrm{density}{g}_n\left(x\right)={\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}f\left(\frac{x}{{\mu }_j}\right),x\ge 0.$$

(2)

Here the coefficients/weights are given as follows:

$${\ell }_j={\displaystyle \prod _{i=1,i\ne j}^n}\frac{{\mu }_i^{-1}}{{\mu }_i^{-1}-{\mu }_j^{-1}}={\displaystyle \prod _{i=1,i\ne j}^n}\frac{{\mu }_j}{{\mu }_j-{\mu }_i},j=1,2,\dots ,n.$$

(3)

We use now the common Laplace transform $\phi \left(t\right):=E\left[{\mathrm{e}}^{-t{X}_i}\right]$. Please note that since ${\mu }_i\ne {\mu }_j$ for $i\ne j$, relation (2) implies that

$$\begin{array}{ccc}\hfill \phi \left({\mu }_{1}t\right)\phi \left({\mu }_{2}t\right)\cdots \phi \left({\mu }_{n}t\right)& =& {\int }_0^{\infty }{\mathrm{e}}^{-tx}{g}_n\left(x\right)dx\hfill \\ & =& {\int }_0^{\infty }{\mathrm{e}}^{-tx}{\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}f\left(\frac{x}{{\mu }_j}\right)dx\hfill \\ & =& {\displaystyle \sum _{j=1}^n}{\ell }_j{\int }_0^{\infty }{\mathrm{e}}^{-tx}\frac{1}{{\mu }_j}f\left(\frac{x}{{\mu }_j}\right)dx={\displaystyle \sum _{j=1}^n}{\ell }_j\phi \left({\mu }_{j}t\right).\hfill \end{array}$$

(4)

The idea now is to start with an arbitrary non-negative random variable X with unknown density f and Laplace transform $\phi$. If the Laplace transform of the linear combination $S_n={\sum }_{i=1}^n{\mu }_i{X}_i$ satisfies (4), we will derive that $\phi \left(t\right)=\lambda {(\lambda +t)}^{-1}$. Thus, the common distribution of $X_j$, $j=1,2,\dots ,n$ is exponential. More precisely, the following characterization result holds.

Theorem 1.

Suppose that$X_1,X_2,\dots ,X_n$,$n\ge 2$, are independent copies of a non-negative random variable X with density f. Assume further that X satisfies Cramér’s condition: there is a number$t_0>0$such that$\mathsf{E}\left[{\mathrm{e}}^{-tX}\right]<\infty$for all$t\in (-t_0,t_0)$. If relation (2) is satisfied for fixed$n\ge 2$and fixed positive mutually different numbers${\mu }_1,{\mu }_2,\dots ,{\mu }_n$, then$X\sim \mathrm{Exp}\left(\lambda \right)$for some$\lambda >0$.

The studies of characterization properties of exponential distributions are abundant. Comprehensive surveys can be found in [6,7,8,9]. More recently, Arnold and Villaseñor [10] obtained a series of exponential characterizations involving sums of two random variables and conjectured possible extensions for sums of more than two variables (see also [11]). Corollary 1 below extends the characterizations in [10,11] to sums of n variables, for any fixed $n\ge 2$.

Consider the special case of (2) when ${\mu }_j=1/j$ for $j=1,2,\dots ,n$. Under this choice of ${\mu }_j$’s, the formula for the weight ${\ell }_j$ simplifies to (see [4], Chapter 1, Problem 13)

$${\ell }_j={\displaystyle \prod _{i=1,i\ne j}^n}\frac{i}{i-j}=\left(\genfrac{}{}{0pt}{}{n}{j}\right){(-1)}^{j-1}.$$

Therefore, Theorem 1 reduces to the following corollary.

Corollary 1.

Suppose that$X_1,X_2,\dots ,X_n$,$n\ge 2$, are independent copies of a non-negative random variable X with density f. Assume further that X satisfies Cramér’s condition: there is a number$t_0>0$such that$\mathsf{E}\left[{\mathrm{e}}^{-tX}\right]<\infty$for all$t\in (-t_0,t_0)$. If for fixed$n\ge 2$,

$$X_1+\frac{1}{2}{X}_2+\dots +\frac{1}{n}{X}_{n}hasdensity{\displaystyle \sum _{j=1}^n}\left(\genfrac{}{}{0pt}{}{n}{j}\right){(-1)}^{j-1}jf\left(jx\right),x\ge 0,$$

(5)

then$X\sim \mathrm{Exp}\left(\lambda \right)$for some$\lambda >0$.

The exponential distribution has the striking property that if $\lambda =1$ (unit exponential), then the density f equals the survival function (the tail of the cumulative distribution function) $\overline{F}=1-F$. Therefore, in case of unit exponential distribution, (2) can be written as follows:

$${\tilde{S}}_n:={\mu }_1{X}_1+{\mu }_2{X}_2+\cdots +{\mu }_n{X}_n\mathrm{has}\mathrm{density}{\tilde{g}}_n\left(x\right):={\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}\overline{F}\left(\frac{x}{{\mu }_j}\right),x\ge 0.$$

(6)

We will show that (6) is a sufficient condition for $X_1,X_2,\dots ,X_n$ to be unit exponential.

Theorem 2.

Suppose that$X_1,X_2,\dots ,X_n$,$n\ge 2$, are independent copies of a non-negative random variable X with distribution function F. Assume also that X satisfies Cramér’s condition: there is a number$t_0>0$such that$\mathsf{E}\left[{\mathrm{e}}^{-tX}\right]<\infty$for all$t\in (-t_0,t_0)$. If relation (6) is satisfied for fixed$n\ge 2$, then$X\sim \mathrm{Exp}\left(1\right)$.

Setting ${\mu }_j=1/j$ for $j=1,2,\dots ,n$, we obtain the following corollary of Theorem 2.

Corollary 2.

Suppose that$X_1,X_2,\dots ,X_n$,$n\ge 2$, are independent copies of a non-negative random variable X with distribution function F. Assume also that X satisfies Cramér’s condition: there is a number$t_0>0$such that$\mathsf{E}\left[{\mathrm{e}}^{-tX}\right]<\infty$for all$t\in (-t_0,t_0)$. If for fixed$n\ge 2$,

$$X_1+\frac{1}{2}{X}_2+\dots +\frac{1}{n}{X}_{n}hasdensity{\displaystyle \sum _{j=1}^n}\left(\genfrac{}{}{0pt}{}{n}{j}\right){(-1)}^{j-1}j\overline{F}\left(jx\right)x>0,$$

(7)

then$X\sim \mathrm{Exp}\left(1\right)$for some$\lambda >0$.

We organize the rest of the paper as follows. Section 2 contains preliminaries needed in the proofs of the theorems. The proofs themselves are given in Section 3. We discuss the findings in the concluding Section 4.

2. Auxiliaries

We will need the Leibniz rule for differentiating a product of functions. Denote by $v^{\left(k\right)}$ the kth derivative of $v\left(x\right)$ with $v^{\left(0\right)}\left(x\right):=v\left(x\right)$. Let us define a multi-index set $\mathit{\alpha }=({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n)$ as an n-tuple of non-negative integers, and denote $\left|\mathit{\alpha }\right|={\alpha }_1+{\alpha }_2+\dots +{\alpha }_n$. Leibniz considered the problem of determining the kth derivative of the product of n smooth functions $v_1\left(t\right)v_2\left(t\right)\cdots v_n\left(t\right)$ and obtained the formula (e.g., [12])

$$\frac{{\mathrm{d}}^k}{\mathrm{d}{t}^k}\left({\displaystyle \prod _{i=1}^n}{v}_i\left(t\right)\right)=\sum _{\left|\mathit{\alpha }\right|=k}\left(\frac{k!}{{\alpha }_1!{\alpha }_2!\cdots {\alpha }_n!}{\displaystyle \prod _{i=1}^n}{v}_i^{\left({\alpha }_i\right)}\left(t\right)\right).$$

(8)

Here the summation is taken over all multi-index sets $\mathit{\alpha }$ with $\left|\mathit{\alpha }\right|=k$. Formula (8) can easily be proved by induction.

Lemma 1.

Assume that$v\left(t\right)={\sum }_{i=0}^{\infty }{a}_i{t}^i$is a functional series, such that for some${\tilde{t}}_0>0$, the$k^{th}$order derivative$v^{\left(k\right)}\left(t\right)$exists for all$t\in (-{\tilde{t}}_0,{\tilde{t}}_0)$. Then for arbitrary positive real constants${\mu }_1,{\mu }_2,\dots ,{\mu }_n$, we have

$$\frac{{\mathrm{d}}^k}{\mathrm{d}{t}^k}\left({\displaystyle \prod _{i=1}^n}v\left({\mu }_{i}t\right)\right){|}_{t=0}=k!\sum _{\left|\mathit{\alpha }\right|=k}{\displaystyle \prod _{i=1}^n}{\mu }_i^{{\alpha }_i}{a}_{{\alpha }_i}.$$

(9)

Proof. 

Formula (9) is proved by applying Leibniz rule (8) to ${\prod }_{i=1}^{n}v\left({\mu }_{i}t\right)$. □

In addition to (9), we will need some properties of Lagrange basis polynomials ${\ell }_j$ collected below.

Lemma 2

(see [13]). Let ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$ be positive real numbers, such that ${\lambda }_i\ne {\lambda }_j$ for $i\ne j$. Denote

$${\ell }_j={\displaystyle \prod _{i=1,i\ne j}^n}\frac{{\lambda }_i}{{\lambda }_i-{\lambda }_j}j=1,2,\dots ,n.$$

Then, for $n\ge 2$, we have the following:

(i)

${\displaystyle \sum _{j=1}^n{\ell }_j=1}$.

(ii)

${\displaystyle \sum _{j=1}^n{\ell }_j{\lambda }_j^k=0foranyk,1\le k\le n-1}$.

(iii)

${\displaystyle \sum _{j=1}^n\frac{{\ell }_j}{{\lambda }_j^k}\ge \sum _{j=1}^n\frac{1}{{\lambda }_j^k}foranyk,1\le k\le n-1}$, where the equality holds if and only if$k=1$.

Proof. 

Claim (i) follows by integrating (1) over $z>0$. Claim (ii) is proved in Corollary 1 of [13]. To prove claim (iii) we involve $\mathit{\alpha }$, the multi-index set as in (8). For $k\ge 1$, we have $\mathit{\alpha }={\mathit{\alpha }}^{\prime }\cup {\mathit{\alpha }}^{″}$, where

$$\begin{array}{ccc}\hfill {\mathit{\alpha }}^{\prime }& =& \left\{\right|\mathit{\alpha }|=k:\mathrm{only}\mathrm{one}\mathrm{index}\mathrm{in}\mathit{\alpha }\mathrm{equals}k\mathrm{and}\mathrm{all}\mathrm{others}\mathrm{are}\mathrm{zeros}\}\hfill \\ \hfill {\mathit{\alpha }}^{″}& =& \left\{\right|\mathit{\alpha }|=k:\mathrm{no}\mathrm{single}\mathrm{index}\mathrm{in}\mathit{\alpha }\mathrm{equals}k\}.\hfill \end{array}$$

According to Proposition 5 in [13] we obtain, for $n\ge 2$ and $k\ge 1$, the following chain of relations:

$$\begin{array}{ccc}\hfill {\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\lambda }_j^k}& =& {\displaystyle \sum _{\left|\alpha \right|=k}}\frac{1}{{\lambda }_1^{{\alpha }_1}{\lambda }_2^{{\alpha }_2}\cdots {\lambda }_n^{{\alpha }_n}}\hfill \\ & =& {\displaystyle \sum _{|{\alpha }^{\prime }|}}\frac{1}{{\lambda }_1^{{\alpha }_1}{\lambda }_2^{{\alpha }_2}\cdots {\lambda }_n^{{\alpha }_n}}+{\displaystyle \sum _{|{\alpha }^{\prime \prime }|}}\frac{1}{{\lambda }_1^{{\alpha }_1}{\lambda }_2^{{\alpha }_2}\cdots {\lambda }_n^{{\alpha }_n}}\hfill \\ & =& {\displaystyle \sum _{j=1}^n}\frac{1}{{\lambda }_j^k}+{\displaystyle \sum _{|{\alpha }^{\prime \prime }|}}\frac{1}{{\lambda }_1^{{\alpha }_1}{\lambda }_2^{{\alpha }_2}\cdots {\lambda }_n^{{\alpha }_n}}\hfill \\ & \ge & {\displaystyle \sum _{j=1}^n}\frac{1}{{\lambda }_j^k}.\hfill \end{array}$$

(10)

Clearly, the equality in (10) holds if and only if $k=1$. The proof is complete. □

The properties in Lemma 2 can be easily verified, as an illustration, for $n=2$, $k=1$, and $k=2$. Indeed,

$$\begin{array}{ccc}\hfill {\displaystyle \sum _{j=1}^2}{\ell }_j& =& \frac{{\lambda }_2}{{\lambda }_2-{\lambda }_1}+\frac{{\lambda }_1}{{\lambda }_1-{\lambda }_2}=1,{\displaystyle \sum _{j=1}^2}{\ell }_j{\lambda }_j=\frac{{\lambda }_2{\lambda }_1}{{\lambda }_2-{\lambda }_1}+\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1-{\lambda }_2}=0,\hfill \\ \hfill {\displaystyle \sum _{j=1}^2}\frac{{\ell }_j}{{\lambda }_j}& =& \frac{{\lambda }_2}{({\lambda }_2-{\lambda }_1){\lambda }_1}+\frac{{\lambda }_1}{({\lambda }_1-{\lambda }_2){\lambda }_2}=\frac{{\lambda }_2+{\lambda }_1}{{\lambda }_1{\lambda }_2}={\displaystyle \sum _{i=1}^2}\frac{1}{{\lambda }_i},\hfill \\ \hfill {\displaystyle \sum _{j=1}^2}\frac{{\ell }_j}{{\lambda }_j^2}& =& \frac{{\lambda }_2}{({\lambda }_2-{\lambda }_1){\lambda }_1^2}+\frac{{\lambda }_1}{({\lambda }_1-{\lambda }_2){\lambda }_2^2}=\frac{{\lambda }_2^2+{\lambda }_2{\lambda }_1+{\lambda }_1^2}{{\lambda }_1^2{\lambda }_2^2}={\displaystyle \sum _{i=1}^2}\frac{1}{{\lambda }_i^2}+\frac{1}{{\lambda }_1{\lambda }_2}.\hfill \end{array}$$

3. Proofs of the Characterization Theorems

In the proofs of both theorems we follow the four-step scheme.

• Consider $X_1,X_2,\dots ,X_n$ for $n\ge 2$ to be independent copies of a non-negative random variable X with density f. Suppose ${\mu }_1>{\mu }_2>\dots >{\mu }_n$ are positive real numbers.
• Assume the characterization property

  $$S_n={\mu }_1{X}_1+{\mu }_2{X}_2+\cdots +{\mu }_n{X}_n\mathrm{has}\mathrm{density}{g}_n\left(x\right)={\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}f\left(\frac{x}{{\mu }_j}\right),x\ge 0,$$

  where ${\ell }_j$ is given in (3).
• For the Laplace transform $\phi \left(t\right)=\mathsf{E}\left[{\mathrm{e}}^{-tX}\right]$, $t\ge 0$, obtain the equation

  $$\phi \left({\mu }_{1}t\right)\phi \left({\mu }_{2}t\right)\cdots \phi \left({\mu }_{n}t\right)={\displaystyle \sum _{j=1}^n}{\ell }_j\phi \left({\mu }_{j}t\right).$$

  (11)
• Using Leibniz rule for differentiating product of functions and properties of Lagrange basis polynomials, show that (11) has a unique solution given by $\phi \left(t\right)={(1+{\lambda }^{-1}t)}^{-1}$ for some $\lambda >0$ and conclude that

  $$X_1,X_2,\dots ,X_n\mathrm{are}\mathrm{Exp}\left(\lambda \right)\mathrm{random}\mathrm{variables}.$$

Proof of Theorem 1.

Recall that (see (4))

$$\phi \left({\mu }_{1}t\right)\phi \left({\mu }_{2}t\right)\cdots \phi \left({\mu }_{n}t\right)={\displaystyle \sum _{j=1}^n}{\ell }_j\phi \left({\mu }_{j}t\right).$$

Dividing both sides of this equation by $\phi \left({\mu }_{1}t\right)\phi \left({\mu }_{2}t\right)\cdots \phi \left({\mu }_{n}t\right)$, we obtain

$$1={\displaystyle \sum _{j=1}^n}\left({\ell }_j{\displaystyle \prod _{i=1,i\ne j}^n}\psi \left({\mu }_{i}t\right)\right),$$

(12)

where $\psi :=1/\phi$. Consider the series

$$\psi \left(t\right)={\displaystyle \sum _{k=0}^{\infty }}{a}_k{t}^k,$$

(13)

which, as a consequence of Cramér’s condition for $\phi$, is convergent in a proper neighborhood of $t=0$. To prove the theorem, it is sufficient to show that

$$\psi \left(t\right)=1+{\lambda }^{-1}t,\lambda >0.$$

(14)

We will prove that (12) implies (14) by showing that the coefficients ${\left\{a_k\right\}}_{k=0}^{\infty }$ in (13) satisfy $a_0=1$, $a_1={\lambda }^{-1}>0$, and $a_k=0$ for $k\ge 2$. Notice first that

$$a_0=\frac{1}{\phi \left(0\right)}=1.$$

(15)

Denote

$${\mathsf{\Psi }}_j\left(t\right):={\displaystyle \prod _{i=1,i\ne j}^n}\psi \left({\mu }_{i}t\right)\mathrm{and}H\left(t\right):={\displaystyle \sum _{j=1}^n}{\ell }_j{\mathsf{\Psi }}_j\left(t\right)={\displaystyle \sum _{k=0}^{\infty }}{h}_k{t}^k.$$

By (12) we have $H\left(t\right)\equiv 1$ and therefore $h_0=1$ and $h_k=0$ for all $k\ge 1$. Equating $h_k$’s to the corresponding coefficients of the series in the right-hand side of (12), we will obtain equations for ${\left\{a_k\right\}}_{k=0}^{\infty }$. As a first step, note that

$$h_k=\frac{1}{k!}{H}^{\left(k\right)}{\left(t\right)|}_{t=0}=\frac{1}{k!}{\displaystyle \sum _{j=1}^n}{\ell }_j{\mathsf{\Psi }}_j^{\left(k\right)}\left(t\right){|}_{t=0},k\ge 1.$$

(16)

Next, we apply Leibniz rule for differentiation. To fix the notation, let us define a multi-index set ${\mathit{\alpha }}_{-j}=({\alpha }_1,\dots ,{\alpha }_{j-1},{\alpha }_{j+1},\dots ,{\alpha }_n)$, $1\le j\le n$ as a set of $(n-1)$-tuples of non-negative integer numbers, with $|{\mathit{\alpha }}_{-j}|={\alpha }_1+\dots +{\alpha }_{j-1}+{\alpha }_{j+1}+\dots +{\alpha }_n$. Applying Lemma 1 for fixed $k\ge 1$ and fixed $1\le j\le n$, we obtain

$${\mathsf{\Psi }}_j^{\left(k\right)}\left(t\right){|}_{t=0}=k!{\displaystyle \sum _{\left\{\right|{\alpha }_{-j}|=k\}}}{\displaystyle \prod _{i=1,i\ne j}^n}{\mu }_i^{{\alpha }_i}{a}_{{\alpha }_i}.$$

(17)

Introduce the set ${\mathsf{\Lambda }}_{k,j}:=\{{\mathit{\alpha }}_{-j}:|{\mathit{\alpha }}_{-j}|=k\}$ and partition it into three disjoint subsets as follows:

$${\mathsf{\Lambda }}_{k,j}={\mathsf{\Lambda }}_{k,j}^{\prime }\cup {\mathsf{\Lambda }}_{k,j}^{″}\cup {\mathsf{\Lambda }}_{k,j}^{\prime \prime \prime },$$

where for $k\ge 1$

$$\begin{array}{ccc}\hfill {\mathsf{\Lambda }}_{k,j}^{\prime }& =& \left\{\right|{\mathit{\alpha }}_{-j}|=k:\mathrm{only}\mathrm{one}\mathrm{index}\mathrm{in}{\mathit{\alpha }}_{-j}\mathrm{equals}k,\mathrm{all}\mathrm{others}\mathrm{are}\mathrm{zeros}\}\hfill \\ \hfill {\mathsf{\Lambda }}_{k,j}^{″}& =& \left\{\right|{\mathit{\alpha }}_{-j}|=k:k\ge 2\mathrm{and}\mathrm{exactly}k\mathrm{of}\mathrm{the}\mathrm{indices}\mathrm{in}{\mathit{\alpha }}_{-j}\mathrm{equal}1,\mathrm{all}\mathrm{others}\mathrm{are}\mathrm{zeros}\}\hfill \\ \hfill {\mathsf{\Lambda }}_{k,j}^{\prime \prime \prime }& =& \left\{\right|{\mathit{\alpha }}_{-j}|=k:k\ge 3\mathrm{and}\mathrm{there}\mathrm{is}\mathrm{an}\mathrm{index}{\alpha }_i\mathrm{with}2\le {\alpha }_i<k\}.\hfill \end{array}$$

For example, if $n=5$, $k=3$, and $j=5$, then ${\mathsf{\Lambda }}_{3,5}^{\prime }=\{(3,0,0,0),(0,3,0,0),(0,0,3,0),(0,0,0,3)\}$, ${\mathsf{\Lambda }}_{3,5}^{″}=\{(1,1,1,0),(1,1,0,1),(1,0,1,1),(0,1,1,1)\}$, and ${\mathsf{\Lambda }}_{3,5}^{\prime \prime \prime }=\{(1,2,0,0),(1,0,2,0),\dots ,(0,0,2,1)\}$. Referring to (16) and (17), we have for $k\ge 1$

$$\begin{array}{ccc}\hfill h_k& =& {\displaystyle \sum _{j=1}^n}\left({\ell }_j{\displaystyle \sum _{{\mathsf{\Lambda }}_{k,j}}}{\displaystyle \prod _{i=1,i\ne j}^n}{\mu }_i^{{\alpha }_i}{a}_{{\alpha }_i}\right)\hfill \\ & =& {\displaystyle \sum _{j=1}^n}{\ell }_j\left({\displaystyle \sum _{{\mathsf{\Lambda }}_{k,j}^{\prime }}}(·)+{\displaystyle \sum _{{\mathsf{\Lambda }}_{k,j}^{\prime \prime }}}(·)+{\displaystyle \sum _{{\mathsf{\Lambda }}_{k,j}^{\prime \prime \prime }}}(·)\right)\hfill \\ & =:& {\displaystyle \sum _{j=1}^n}{\ell }_j\left(S_{1,j}+S_{2,j}+S_{3,j}\right),\mathrm{say}.\hfill \end{array}$$

(18)

For the term $S_{2,j}$ in the middle, since $a_0=1$, we have $S_{2,j}=0$ when $k=1$ and for any $k\ge 2$

$$\begin{array}{ccc}\hfill S_{2,j}& =& {\displaystyle \sum _{{\mathsf{\Lambda }}_{k,j}^{\prime \prime }}}{\displaystyle \prod _{i=1,i\ne j}^n}{\mu }_i^{{\alpha }_i}{a}_{{\alpha }_i}\hfill \\ & =& a_0^{n-1-k}{a}_1^k\sum {}^{\prime }({\mu }_{i_1}{\mu }_{i_2}\cdots {\mu }_{i_k})\hfill \\ & =& a_1^k{\mu }_j^{-1}\sum {}^{\prime }({\mu }_j{\mu }_{i_1}{\mu }_{i_2}\cdots {\mu }_{i_k})\hfill \end{array}$$

where the summation in $\sum {}^{\prime }$ is over all k-tuples (with $i_j$th component dropped) $i_1,\dots ,i_{j-1},i_{j+1}\dots ,i_k$, such that $i_m\in \{1,2,\dots ,n\}$ and $i_1<i_2<\dots <i_k$. Using that ${\sum }_{j=1}^n{\ell }_j{\mu }_j^{-1}=0$ by Lemma 2(ii) with ${\lambda }_i={\mu }_i^{-1}$, we obtain for any $k\ge 2$

$${\displaystyle \sum _{j=1}^n}{\ell }_j{S}_{2,j}=a_1^k\left({\displaystyle \sum _{j=1}^n}{\ell }_j{\mu }_j^{-1}\right)\sum {}^{″}({\mu }_{i_1}{\mu }_{i_2}\cdots {\mu }_{i_k})=0.$$

(19)

Here the summation in $\sum {}^{″}$ is over all k-tuples $i_1,i_2,\dots ,i_k$, such that $i_m\in \{1,2,\dots ,n\}$ and $i_1<i_2<\dots <i_k$. For the first term $S_{1,j}$ in the last expression of (18), we have for any $k\ge 1$

$$\begin{array}{ccc}\hfill S_{1,j}& =& {\displaystyle \sum _{{\mathsf{\Lambda }}_{k,j}^{\prime }}}{\displaystyle \prod _{i=1,i\ne j}^n}{\mu }_i^{{\alpha }_i}{a}_{{\alpha }_i}=a_0^{n-2}{a}_k{\displaystyle \sum _{i=1,i\ne j}^n}{\mu }_i^k\hfill \\ & =& a_k\left({\displaystyle \sum _{i=1}^n}{\mu }_i^k-{\mu }_j^k\right).\hfill \end{array}$$

Furthermore, since ${\sum }_{j=1}^n{\ell }_j=1$ by Lemma 2(i) with ${\lambda }_i={\mu }_i^{-1}$, we have for any $k\ge 1$

$$\begin{array}{ccc}\hfill {\displaystyle \sum _{j=1}^n}{\ell }_j{S}_{1,j}& =& a_k{\displaystyle \sum _{j=1}^n}{\ell }_j\left({\displaystyle \sum _{i=1}^n}{\mu }_i^k-{\mu }_j^k\right)\hfill \\ & =& a_k{\displaystyle \sum _{i=1}^n}{\mu }_i^k{\displaystyle \sum _{j=1}^n}{\ell }_j-a_k{\displaystyle \sum _{j=1}^n}{\ell }_j{\mu }_j^k\hfill \\ & =& a_k\left({\displaystyle \sum _{i=1}^n}{\mu }_i^k-{\displaystyle \sum _{j=1}^n}{\ell }_j{\mu }_j^k\right)\hfill \\ & =:& a_k{c}_k.\hfill \end{array}$$

(20)

Lemma 2(iii) with ${\lambda }_i={\mu }_i^{-1}$ implies that $c_1=0$ and $c_k<0$ for any $k\ge 2$. It follows from (18)–(20) that

$$h_k=c_k{a}_k+{\displaystyle \sum _{j=1}^n}{\ell }_j{S}_{3,j},$$

(21)

where $c_1=0$ and $c_k<0$ for $k\ge 2$.

Let $k=1$. Since $h_1=0$ and the sets ${\mathsf{\Lambda }}_1^{\prime \prime }$ and ${\mathsf{\Lambda }}_2^{\prime \prime \prime }$ are empty, we obtain $c_1{a}_1=0$, where $c_1=0$. Hence, there are no restrictions on the coefficient $a_1$, other than $a_1>0$, since X has positive mean. Therefore, there is a number ${\lambda }^{-1}>0$ such that

$$a_1={\lambda }^{-1}>0.$$

(22)

Let $k=2$. Since the set ${\mathsf{\Lambda }}_2^{\prime \prime \prime }$ is empty, Equation (21) yields $h_2=c_2{a}_2=0,$ where recall that $c_2<0$. Thus, $a_2=0$. Next, applying (21) and taking into account that $h_k=0$ for $k\ge 2$, we will show by induction that $a_k=0$ for any $k\ge 2$. Assuming $a_k=0$ for $k=2,3,\dots ,r$, we will show that $a_{r+1}=0$. Indeed, by (21) we have

$$h_{r+1}=c_{r+1}{a}_{r+1}+{\displaystyle \sum _{j=1}^n}\left({\ell }_j{\displaystyle \sum _{{\mathsf{\Lambda }}_{r+1,j}^{\prime \prime \prime }}}{\displaystyle \prod _{i=1,i\ne j}^n}{\mu }_i^{{\alpha }_i}{a}_{{\alpha }_i}\right)=c_{r+1}{a}_{r+1},$$

because at least one index ${\alpha }_i$, satisfies $2\le {\alpha }_i\le r$ and hence $a_{{\alpha }_i}=0$, by assumption. Therefore, $h_{r+1}=c_{r+1}{a}_{r+1}=0$ and, since $c_{r+1}<0$, we have $a_{r+1}=0$, which completes the induction. Hence,

$$a_k=0\mathrm{for}\mathrm{any}k\ge 2.$$

(23)

The Equations (15) and (22)–(23) imply (14), which completes the proof of the theorem. □

Proof of Theorem 2.

Taking into account (6), similarly to (4) and using integration-by-parts, we obtain

$$\begin{array}{ccc}\hfill \phi \left({\mu }_{1}t\right)\phi \left({\mu }_{2}t\right)\cdots \phi \left({\mu }_{n}t\right)& =& {\int }_0^{\infty }{\mathrm{e}}^{-tx}{g}_n\left(x\right)dx={\int }_0^{\infty }{\mathrm{e}}^{-tx}{\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}\overline{F}\left(\frac{x}{{\mu }_j}\right)dx\hfill \\ & =& {\displaystyle \sum _{j=1}^n}{\ell }_j{\int }_0^{\infty }{\mathrm{e}}^{-tx}\frac{1}{{\mu }_j}\overline{F}\left(\frac{x}{{\mu }_j}\right)dx\hfill \\ & =& \frac{1}{t}{\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}\left(1-\phi \left({\mu }_{j}t\right)\right).\hfill \end{array}$$

Using the fact that ${\sum }_{j=1}^n{\ell }_j/{\mu }_j=0$ (see Lemma 2(ii)), this simplifies to

$$\phi \left({\mu }_{1}t\right)\phi \left({\mu }_{2}t\right)\cdots \phi \left({\mu }_{n}t\right)=-\frac{1}{t}{\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}\phi \left({\mu }_{j}t\right).$$

(24)

Dividing both sides of (24) by $-\phi \left({\mu }_{1}t\right)\phi \left({\mu }_{2}t\right)\cdots \phi \left({\mu }_{n}t\right)/t$, for $t>0$, we obtain

$$-t={\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}{\displaystyle \prod _{i=1,i\ne j}^n}\psi \left({\mu }_{i}t\right),$$

(25)

where, as before, $\psi =1/\phi$. Consider the series $\psi \left(t\right)={\sum }_{k=0}^{\infty }{a}_k{t}^k$, which is convergent by assumption. To prove the theorem, it is sufficient to show that $\psi \left(t\right)=1+t$, $t\ge 0$, or, equivalently, that the coefficients ${\left\{a_k\right\}}_{k=0}^{\infty }$ of the above series satisfy $a_0=1$, $a_1=1$, and $a_k=0$ for $k\ge 2$. Clearly, $a_0=1/\phi \left(0\right)=1$. Recall that

$${\mathsf{\Psi }}_j\left(t\right):={\displaystyle \prod _{i=1,i\ne j}^n}\psi \left({\mu }_{i}t\right)\mathrm{and}\mathrm{denote}-Q\left(t\right):={\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}{\mathsf{\Psi }}_j\left(t\right)=-{\displaystyle \sum _{k=0}^{\infty }}{q}_k{t}^k.$$

By (25) we have $Q\left(t\right)\equiv t$ and therefore $q_1=1$ and $q_k=0$ for all $k\ne 1$. We will express $q_k$ in terms of $a_j$’s. Proceeding as in the proof of Theorem 1, applying Leibniz rule for differentiating a product of functions, and using the same notation, we obtain for $k\ge 1$ that

$$-q_k={\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}\left(S_{1,j}+S_{2,j}+S_{3,j}\right).$$

As with (19), applying Lemma 2(ii), we obtain

$${\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}{S}_{2,j}=a_1^k\left({\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j^2}\right)\sum {}^{\prime \prime }({\mu }_{i_1}{\mu }_{i_2}\cdots {\mu }_{i_k})=0,$$

(26)

where the summation in $\sum {}^{\prime \prime }$ is over all k-tuples $i_1,\dots ,i_k$, such that $i_m\in \{1,\dots ,n\}$ and $i_1<\dots <i_k$. Furthermore, since ${\sum }_{j=1}^n{\ell }_j=1$ and ${\sum }_{j=1}^n{\ell }_j/{\mu }_j^2=0$ by Lemma 2, we have for any $k\ge 1$

$$\begin{array}{ccc}\hfill {\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}{S}_{1,j}& =& a_k{\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}\left({\displaystyle \sum _{i=1}^n}{\mu }_i^k-{\mu }_j^k\right)\hfill \\ & =& a_k{\displaystyle \sum _{i=1}^n}{\mu }_i^k{\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}-a_k{\displaystyle \sum _{j=1}^n}{\ell }_j{\mu }_j^{k-1}\hfill \\ & =& -a_k{\displaystyle \sum _{j=1}^n}{\ell }_j{\mu }_j^{k-1}\hfill \\ & =:& -a_k{d}_k.\hfill \end{array}$$

(27)

It follows from (26) and (27) that for $k\ge 1$,

$$-q_k=-a_k{d}_k+{\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}{S}_{3,j}.$$

(28)

Let $k=1$. Since $q_1=1$ and the set ${\mathsf{\Lambda }}_1^{\prime \prime \prime }$ is empty, we obtain $a_1{d}_1=1$, where $d_1=1$ by Lemma 2(iii). Therefore, $a_1=1$. Let $k=2$. Since ${\mathsf{\Lambda }}_2^{\prime \prime \prime }$ is empty, Equation (28) yields $q_2=d_2{a}_2=0,$ where $d_2>0$ by Lemma 2(iii). Thus, $a_2=0$. Assuming $a_k=0$ for $2\le k\le r$, we will show that $a_{r+1}=0$. Indeed,

$$q_{r+1}=d_{r+1}{a}_{r+1}+{\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}{S}_{3,j}=d_{r+1}{a}_{r+1},$$

because at least one index ${\alpha }_i$, satisfies $2\le {\alpha }_i\le r$, in which case $a_{{\alpha }_i}=0$, by assumption. Therefore, $q_{r+1}=d_{r+1}{a}_{r+1}=0$ and, since $d_{r+1}<0$, we have $a_{r+1}=0$, which completes the induction proof. Hence, $a_k=0$ for any $k\ge 2$. Since $a_0=a_1=1$ and $a_k=0$ for $k\ge 2$, we obtain $\psi \left(t\right)=1+t$, which clearly completes the proof of the theorem. □

4. Concluding Remarks

Arnold and Villaseñor [10] proved that if $X_1$ and $X_2$ are two independent and non-negative random variables with common density f and $\mathsf{E}\left[X_1\right]<\infty$, then

$$X_1+\frac{1}{2}{X}_2\mathrm{has}\mathrm{density}2f\left(x\right)-2f\left(2x\right),x>0,$$

if and only if $X_1\sim \mathrm{Exp}\left(\lambda \right)$ for some $\lambda >0$. Motivated by this result, we extended it in two directions considering: (i) arbitrary number $n\ge 2$ of independent identically distributed non-negative random variables and (ii) linear combination of independent variables with arbitrary positive and distinct coefficients ${\mu }_1,{\mu }_2,\dots ,{\mu }_n$. Namely, our main result is that

$$S_n={\mu }_1{X}_1+{\mu }_2{X}_2+\dots +{\mu }_n{X}_n\mathrm{has}\mathrm{density}{g}_n\left(x\right)={\displaystyle \sum _{j=1}^n}\frac{{\ell }_j}{{\mu }_j}f\left(\frac{x}{{\mu }_j}\right)x\ge 0,$$

where ${\ell }_j={\prod }_{i=1,i\ne j}^n{\mu }_j{({\mu }_j-{\mu }_i)}^{-1}$, if and only if $X_i\sim \mathrm{Exp}\left(\lambda \right)$ for some $\lambda >0$.

In this paper, we dealt with the situation where the rate parameters ${\lambda }_i$ are all distinct from each other. The other extreme case of equal ${\lambda }_i$’s is trivial. The obtained characterization seems of interest on its own, but it can also serve as a basis for further investigations of intermediate cases of mixed type with some ties and at least two distinct parameters (see [2]). Of certain interest is also the case where not all weights ${\mu }_i$’s are positive (see [1]).