Divisibility of the exponential distribution

Let \(Z\) be a \(\text{Exp}(1)\) random variable. For \(\alpha_1, \ldots \alpha_N \in \mathbb{R}_{+}^{*}\), we are looking for variables \(X_1, \ldots X_N\) independent and following the same distribution such that:

\[Z = \sum_{j=1}^{N} \alpha_j X_j.\]

For the special case where all \(\alpha_j\) are equal, this corresponds to the problem of infinite divisibility of the exponential distribution, and in this case \(X_1\) follows \(\text{Gamma}(1/N, 1/\alpha_1)\), a gamma variable with shape \(1/N\) and scale \(1/\alpha_1\) (in particular when \(N=1\), we have \(X_1 \sim \text{Exp}(\alpha_1) = \text{Gamma}(1, 1/\alpha_1)\)).

We are willing to extend this divisibility property and give methods to derive the density distribution of \(X_1\) along with a way to generate it.

First, the characteristic function \(\varphi\) of \(X_1\) is given, for \(\vert t \vert \leq \max \alpha_j\), by:

\[\varphi(t) = \exp \left( \sum_{k=1}^{\infty} \frac{i^k t^k}{k \sum_{l = 1}^N \alpha_l^k } \right).\]
Proof.

By writing \(Z = \sum_{j=1}^{N} \alpha_j X_j\) in terms of characteristic function, we are looking for \(\varphi\) verifying for \(t \in \mathbb{R}\):

\[\begin{equation}\label{expo_phi_equality} \frac{1}{1-it} = \prod_{j=1}^{N} \varphi(\alpha_j t). \end{equation}\]

We express the characteristic of the exponential distribution for \(\vert t \vert \leq 1\) as follows:

\[\begin{align*} \frac{1}{1-it} =& \exp \left( - \log \left( 1 - it \right) \right) \\ =& \exp \left( \sum_{k=1}^{\infty} \frac{i^kt^k}{k} \right) \\ =& \exp \left( \sum_{k=1}^{\infty} \frac{i^k t^k \sum_{j = 1}^N \alpha_j^k}{k \sum_{l = 1}^N \alpha_l^k } \right) \\ =& \exp \left( \sum_{j = 1}^N \sum_{k=1}^{\infty} \frac{i^k t^k \alpha_j^k}{k \sum_{l = 1}^N \alpha_l^k } \right) \\ =& \prod_{j = 1}^N \exp \left( \sum_{k=1}^{\infty} \frac{i^k \left(\alpha_j t \right)^k}{k \sum_{l = 1}^N \alpha_l^k } \right), \end{align*}\]

which gives the characteristic function for \(\vert t \vert \leq \max \alpha_j\):

\[\varphi(t) := \exp \left( \sum_{k=1}^{\infty} \frac{i^k t^k}{k \sum_{l = 1}^N \alpha_l^k } \right).\]


From this characteristic function, we deduce:

\[\log \varphi(t) = \sum_{k=1}^{\infty} \frac{(k-1)!}{\sum_{l = 1}^N \alpha_l^k } \frac{\left( it \right)^k}{k!},\]

so the cumulants are given by: \(\kappa_k = (k-1)! / \sum_{l = 1}^N \alpha_l^k.\) For instance the mean and variance of \(X_1\) are \(\mu = \kappa_1 = 1/\sum_{l = 1}^N \alpha_l\) and \(\sigma^2 = \kappa_2 = 1/\sum_{l = 1}^N \alpha_l^2\).

It is also possible to express \(\varphi\) as an analytic function using the (complete exponential) Bell polynomials. For \(k \geq 0\), we define the sequence \(p_k := \left( \sum_{l = 1}^N \alpha_l^k \right)^{-1}\) and we rewrite, for \(\vert t \vert \leq \max \alpha_j\),:

\[\varphi(t) = \sum_{n=0}^{\infty} B_n \left(p_1, p_2, 2! p_3, \ldots, (n-1)! p_{n} \right) \frac{\left(it \right)^n}{n!}.\]
Proof.

We further define \(q_k := (k-1)! p_k\) and we express \(\varphi\) following Wikipedia:

\[\begin{align*} \varphi(t) =& \exp \left( \sum_{k=1}^{\infty} \frac{(k-1)!}{\sum_{l = 1}^N \alpha_l^k} \frac{\left(it \right)^k}{k! } \right) \\ =& \exp \left( \sum_{k=1}^{\infty} q_k \frac{\left(it \right)^k}{k! } \right) \\ =& \sum_{n=0}^{\infty} B_n(q_1, q_2, q_3, \ldots, q_n) \frac{\left(it \right)^n}{n!} \\ =& \sum_{n=0}^{\infty} B_n \left(p_1, p_2, 2! p_3, \ldots, (n-1)! p_{n} \right) \frac{\left(it \right)^n}{n!}. \end{align*}\]


We would like to extend the characteristic function on \(t \in \mathbb{R}\). We order the indexes to get: \(\alpha_1 \leq \ldots \leq \alpha_N\). In the case where all coefficients are equal, we already expressed the solution in the introduction. Otherwise, there exists \(p \in [1,N-1]\) such that \(\alpha_1 \leq \ldots \leq \alpha_{N-p} < \alpha_{N-p+1} = \ldots = \alpha_{N}\). The integer \(p\) is the number of indexes equal to \(\alpha_N\); in most of the cases we have \(p = 1\) hence \(\alpha_{N-1} < \alpha_{N}\).

For \(\vert t \vert > \alpha_N\), we let \(D := \lceil \frac{\log \vert t \vert}{\log \alpha_N - \log \alpha_{N-p}} \rceil\) and deduce \(\varphi(t)\) from:

\[\begin{align*} \log \varphi(t) = -& \frac{1}{p} \log \left( 1- \frac{1}{\alpha_N}it \right) \\ +& \sum_{d=2}^{D} \frac{\left( -1 \right)^d}{p^d} \left[\sum_{j_{1}=1}^{N-p} \ldots \sum_{j_{d-1}=1}^{N-p} \log \left( 1-\frac{\prod_{k=1}^{d-1} \alpha_{j_k}}{\alpha_N^d} it \right) \right] \\ +& \frac{\left(-1 \right)^D}{p^D} \sum_{j_{1}=1}^{N-p} \ldots \sum_{j_{D}=1}^{N-p} \log \varphi \left( \frac{\prod_{k=1}^{D} \alpha_{j_k}}{\alpha_N^D} t \right). \end{align*},\]

where the terms \(\log \varphi \left( \frac{\prod_{k=1}^{D} \alpha_{j_k}}{\alpha_N^D} t \right)\) are computed using the formula valid for \(\vert t \vert \leq \alpha_N\).

Proof.

We rewrite the functional relation of the characteristic function as follows, for \(t \in \mathbb{R}\):

\[\frac{1}{1-it} = \prod_{j=1}^{N-p} \varphi(\alpha_j t) \prod_{j=N-p+1}^{N} \varphi(\alpha_j t) = \prod_{j=1}^{N-p} \varphi(\alpha_j t) \varphi(\alpha_N t)^p,\]

and by taking the logarithm:

\[\log \varphi(\alpha_N t) = - \frac{1}{p} \log \left( 1-it \right) - \frac{1}{p} \sum_{j=1}^{N-p} \log \varphi(\alpha_j t)\]

which gives for \(t \in \mathbb{R}\):

\[\log \varphi(t) = - \frac{1}{p} \log \left( 1-\frac{1}{\alpha_N}it \right) - \frac{1}{p} \sum_{j=1}^{N-p} \log \varphi \left(\frac{\alpha_j}{\alpha_N} t \right).\]

We use the functionality of the equation: We first evaluate the previous equation in \(\frac{\alpha_j}{\alpha_N} t\), and we then use the result to replace the terms in the same equation.

First, for all \(\alpha_j\), we have (taking care of using another index name for the sum in the right term):

\[\log \varphi \left( \frac{\alpha_j}{\alpha_N} t \right) = - \frac{1}{p} \log \left( 1-\frac{\alpha_j}{\alpha_N^2} it \right) - \frac{1}{p} \sum_{k=1}^{N-p} \log \varphi \left( \frac{\alpha_j\alpha_k}{\alpha_N^2} t \right)\]

Then, we put it in the previous equation to get:

\[\log \varphi(t) = - \frac{1}{p} \log \left( 1-\frac{1}{\alpha_N}it \right) + \frac{1}{p^2} \sum_{j=1}^{N-p} \log \left( 1-\frac{\alpha_j}{\alpha_N^2} it \right) + \frac{1}{p^2} \sum_{j=1}^{N-p} \sum_{k=1}^{N-p} \log \varphi \left(\frac{\alpha_j\alpha_k}{\alpha_N^2} t \right).\]

We use again the functionality of the equation. We first write:

\[\log \varphi \left(\frac{\alpha_j\alpha_k}{\alpha_N^2} t \right) = - \frac{1}{p} \log \left( 1- \frac{\alpha_j\alpha_k}{\alpha_N^3} it \right) - \frac{1}{p} \sum_{l=1}^{N-p} \log \varphi \left(\frac{\alpha_j\alpha_k\alpha_l}{\alpha_N^3} t \right)\]

to then get:

\[\begin{align*} \log \varphi(t) = -& \frac{1}{p} \log \left( 1-\frac{1}{\alpha_N}it \right) \\ +& \frac{1}{p^2} \sum_{j=1}^{N-p} \log \left( 1-\frac{\alpha_j}{\alpha_N^2} it \right) - \frac{1}{p^3} \sum_{j=1}^{N-p} \sum_{k=1}^{N-p} \log \left( 1- \frac{\alpha_j\alpha_k}{\alpha_N^3} it \right) \\ -& \frac{1}{p^3} \sum_{j=1}^{N-p} \sum_{k=1}^{N-p} \sum_{l=1}^{N-p} \log \varphi \left(\frac{\alpha_j\alpha_k\alpha_l}{\alpha_N^3} t \right). \end{align*}\]

We continue by induction to obtain, for all \(D \geq 1\):

\[\begin{align*} \log \varphi(t) = -& \frac{1}{p} \log \left( 1- \frac{1}{\alpha_N}it \right) \\ +& \sum_{d=2}^{D} \frac{\left( -1 \right)^d}{p^d} \left[\sum_{j_{1}=1}^{N-p} \ldots \sum_{j_{d-1}=1}^{N-p} \log \left( 1-\frac{\prod_{k=1}^{d-1} \alpha_{j_k}}{\alpha_N^d} it \right) \right] \\ +& \frac{\left(-1 \right)^D}{p^D} \sum_{j_{1}=1}^{N-p} \ldots \sum_{j_{D}=1}^{N-p} \log \varphi \left( \frac{\prod_{k=1}^{D} \alpha_{j_k}}{\alpha_N^D} t \right). \end{align*}\]

Given \(t \in \mathbb{R}\), there exists \(D\) such that \(\left( \frac{\alpha_{N-p}}{\alpha_N} \right)^D \vert t \vert \leq 1\) and from this \(D\), the previous formula has all the terms evaluated in the interval \([-1, 1]\), hence \(\varphi(t)\) is well defined. An explicit value of \(D\) is given by: \(D := \lceil \frac{\log \vert t \vert}{\log \alpha_N - \log \alpha_{N-p}} \rceil\).


The naive number of terms to compute for a given \(t\) linked with a certain \(D\) is \(\frac{(N-p)^{D+1} - 1}{N-p-1}\) for \(p < N-1\) and \(D\) for \(p = N-1\).

Ideas: case with negative \(\alpha\) (or more complex), e.g. \(Z = \vert X - Y \vert\)? It’s not possible to get directly \(Z = X - Y\) with same distributed \(X\) and \(Y\), because \(Z\) is nonnegative.

Written on May 11, 2021