Infinite Divisibility and Convolution Roots

What It Means to Be Infinitely Divisible

A distribution is infinitely divisible when it can be split into any prescribed number of identically distributed independent contributions. The number of contributions can be chosen arbitrarily, and the distribution of each contribution may change with that number. This idea connects the behavior of sums to the structure of a single distribution.

For example, if a Poisson count has mean 12, it can be represented in distribution as the sum of four independent Poisson counts, each with mean 3. Infinite divisibility requires such a representation not just for four pieces, but for every positive integer number of pieces.

Definition: A distribution $F$ on the real line is infinitely divisible if, for every integer $m\geq 1$, there exist independent and identically distributed random variables $Y_{m,1},\ldots,Y_{m,m}$ such that $$ Y_{m,1}+\cdots+Y_{m,m}\ \text{has distribution }F. $$ The common distribution of the $Y_{m,j}$ is called an $m$-th convolution root of $F$, and is denoted $G_m$. Equivalently, $F=G_m^{*m}$, where $G_m^{*m}$ is the $m$-fold convolution of $G_m$ with itself.

The root is allowed to depend on $m$. Infinite divisibility does not require one fixed distribution whose repeated sums produce $F$ for every number of terms. The case $m=1$ is automatic, since $F$ itself is a first convolution root.

Characteristic Functions and Convolution Roots

The characteristic function of a random variable $X$ is $\phi_X(t)=E[e^{itX}]$, for real $t$. If $X$ and $Y$ are independent, the characteristic function of their sum is the product of their characteristic functions. This turns convolution into multiplication and gives a convenient way to check proposed roots.

Theorem: A distribution $F$, with characteristic function $\phi_F$, has an $m$-th convolution root $G_m$ if and only if there is a probability distribution $G_m$ with characteristic function $\phi_{G_m}$ such that $$ \phi_F(t)=\bigl(\phi_{G_m}(t)\bigr)^m \quad\text{for every real }t. $$ Consequently, $F$ is infinitely divisible if and only if such a probability distribution exists for every integer $m\geq 1$.

Proof. First suppose $F$ has an $m$-th convolution root. Let $Y_{m,1},\ldots,Y_{m,m}$ be independent with common distribution $G_m$, and let $S=\sum_{j=1}^m Y_{m,j}$. Independence gives

\phi_S(t) =E\!\left[e^{it\sum_{j=1}^m Y_{m,j}}\right] =E\!\left[\prod_{j=1}^m e^{itY_{m,j}}\right] =\prod_{j=1}^m E[e^{itY_{m,j}}] =\bigl(\phi_{G_m}(t)\bigr)^m.

Since $S$ has distribution $F$, $\phi_S=\phi_F$, proving the required identity. Conversely, suppose $G_m$ is a probability distribution whose characteristic function satisfies the identity. Take $m$ independent random variables with distribution $G_m$. Their sum has characteristic function $(\phi_{G_m})^m=\phi_F$. By the uniqueness theorem for characteristic functions, that sum has distribution $F$. Thus $G_m$ is an $m$-th convolution root. Applying this argument for every $m$ proves the final assertion. $\square$

The condition that $\phi_{G_m}$ be the characteristic function of a probability distribution matters. An algebraic $m$-th root of $\phi_F(t)$, considered merely as a complex-valued function, need not be a characteristic function. A valid root must itself satisfy the conditions required of a characteristic function; transform identification alone cannot turn an arbitrary algebraic expression into a distribution.

Worked Examples: Poisson, Normal, and Gamma Roots

Worked Example: Splitting a Poisson Count

Let $X\sim\operatorname{Poisson}(12)$. Show that $X$ has a fourth convolution root and identify it.

The characteristic function of a Poisson random variable with rate $\lambda$ follows from its probability mass function:

\phi_X(t) =\sum_{k=0}^{\infty}e^{itk}e^{-\lambda}\frac{\lambda^k}{k!} =e^{-\lambda}\exp(\lambda e^{it}) =\exp\!\left(\lambda(e^{it}-1)\right).

For $\lambda=12$, this is $\phi_X(t)=\exp(12(e^{it}-1))$. If $Y\sim\operatorname{Poisson}(3)$, then

\bigl(\phi_Y(t)\bigr)^4 =\left[\exp\!\left(3(e^{it}-1)\right)\right]^4 =\exp\!\left(12(e^{it}-1)\right) =\phi_X(t).

Thus four independent $\operatorname{Poisson}(3)$ variables sum to a $\operatorname{Poisson}(12)$ variable in distribution. More generally, for every $m\geq1$, a $\operatorname{Poisson}(12/m)$ variable is an $m$-th convolution root, since raising its characteristic function to the $m$-th power gives $\exp(12(e^{it}-1))$. The Poisson distribution is therefore infinitely divisible.

Worked Example: Splitting a Normal Distribution

Let $X\sim N(5,9)$, where the second parameter is the variance. Find a fourth convolution root and verify the sum.

The characteristic function of $N(\mu,\sigma^2)$ is $\exp(it\mu-\sigma^2t^2/2)$. Choose $Y\sim N(5/4,9/4)$. Its fourth power is

\bigl(\phi_Y(t)\bigr)^4 =\left[\exp\!\left(it\frac54-\frac{(9/4)t^2}{2}\right)\right]^4 =\exp\!\left(5it-\frac{9t^2}{2}\right) =\phi_X(t).

Therefore, four independent variables with distribution $N(5/4,9/4)$ sum to a variable with distribution $N(5,9)$. For any $m\geq1$, the choice $N(5/m,9/m)$ is an $m$-th root: multiplying the mean and variance contributions by $m$ recovers 5 and 9, respectively. Thus every nondegenerate normal distribution is infinitely divisible.

The variance must be divided by $m$, not by $m^2$. Variances of independent summands add, so $m$ identical summands with variance $9/m$ have total variance $9$.

Worked Example: Splitting a Gamma Distribution

Use the shape-rate parameterization and let $X\sim\operatorname{Gamma}(6,2)$, with shape 6 and rate 2. Find a third convolution root.

For shape $\alpha>0$ and rate $\beta>0$, the gamma density is $f(x)=\beta^\alpha x^{\alpha-1}e^{-\beta x}/\Gamma(\alpha)$ for $x>0$. Its characteristic function is

\phi_X(t) =\int_0^\infty e^{itx}\frac{\beta^\alpha x^{\alpha-1}e^{-\beta x}}{\Gamma(\alpha)}\,dx =\left(\frac{\beta}{\beta-it}\right)^\alpha.

The last equality follows from the gamma integral with complex argument $\beta-it$, whose real part is $\beta>0$. Take $Y\sim\operatorname{Gamma}(2,2)$, retaining rate 2 and dividing shape 6 by 3. Then

\bigl(\phi_Y(t)\bigr)^3 =\left(\frac{2}{2-it}\right)^6 =\phi_X(t).

Hence three independent $\operatorname{Gamma}(2,2)$ variables sum to a $\operatorname{Gamma}(6,2)$ variable. For every integer $m\geq1$, $\operatorname{Gamma}(6/m,2)$ is a valid $m$-th root, since its characteristic function to the $m$-th power is $(2/(2-it))^6$. This proves infinite divisibility for every gamma distribution with positive shape and rate.

A Closure Property

Infinite divisibility is preserved when independent infinitely divisible variables are added. The reason is that roots of their distributions can be added as well.

Theorem: If $X$ and $Z$ are independent and each has an infinitely divisible distribution, then the distribution of $X+Z$ is infinitely divisible.

Proof. Fix an integer $m\geq1$. Let $G_m$ and $H_m$ be $m$-th convolution roots of the distributions of $X$ and $Z$, respectively. Construct independent variables $Y_1,\ldots,Y_m$ with distribution $G_m$, and independent variables $W_1,\ldots,W_m$ with distribution $H_m$, with the two collections also independent. The paired sums $V_j=Y_j+W_j$ are independent and identically distributed. Their total is

\sum_{j=1}^m V_j =\sum_{j=1}^m Y_j+\sum_{j=1}^m W_j.

The first sum has the distribution of $X$, the second has the distribution of $Z$, and the two sums are independent. Their combined distribution is therefore that of $X+Z$. The common distribution of the $V_j$ is an $m$-th convolution root of the distribution of $X+Z$. Since this works for every $m$, the distribution of $X+Z$ is infinitely divisible. $\square$

Why the Root Must Be a Distribution

The characteristic-function method is powerful because it reduces a sum to a product, but the method has a logical checkpoint: each proposed root must correspond to an actual probability distribution. For the Poisson, normal, and gamma examples, this was explicit. We selected a known distribution with an allowed parameter and then checked that its characteristic function raised to the desired power matched the target.

Common mistake: “The target characteristic function has an algebraic $m$-th root, so the target distribution is infinitely divisible.” This does not follow. Pointwise algebraic roots can fail to be characteristic functions, and even a formal expression involving a fractional power may require a consistent definition as a function of $t$. The correction is to exhibit, for each $m$, a probability distribution $G_m$ and verify $\phi_F=(\phi_{G_m})^m$. The transform identity then gives the sum representation by uniqueness.

Infinite divisibility is useful because it identifies distributions that remain in the same family under arbitrary equal-piece decompositions. The Poisson rate, normal mean and variance, and gamma shape provide simple examples of parameters that can be apportioned across the summands. The general concept is broader than these three families, but the convolution-root definition and transform criterion are the essential starting points.

Check Your Understanding

Use the convolution-root definition and characteristic-function criterion to answer the following.

State precisely what it means for a distribution to be infinitely divisible. Does the same root distribution have to work for every $m$?
What is an $m$-th convolution root of $\operatorname{Poisson}(\lambda)$, and how does its characteristic function verify the claim?
For $X\sim N(8,16)$, give a fifth convolution root, stating both its mean and variance.
For $X\sim\operatorname{Gamma}(10,4)$ in the shape-rate parameterization, give a second convolution root.
Why is an algebraic $m$-th root of a characteristic function not, by itself, enough to establish infinite divisibility?