Atoms and Support in Compound Distributions

Why Support and Atoms Need Separate Attention

A random sum can combine different kinds of probability structure. A random count may give the sum positive probability of being exactly zero, while the severities may be discrete, continuous, or a mixture of both. To describe the resulting law accurately, we need two related but distinct ideas: support, which records where probability can occur arbitrarily nearby, and atoms, which are individual points assigned positive probability.

Let $N$ be a nonnegative integer-valued random variable, and let $X_1,X_2,\ldots$ be independent and identically distributed severity variables, independent of $N$. Assume $N$ is finite with probability one. Define the compound sum $S$ by $S=0$ when $N=0$, and otherwise $S=X_1+\cdots+X_N$. The empty sum convention makes both cases fit the same definition.

Definition: The support of a real-valued random variable $Y$ is the set of points $y$ such that every open neighborhood of $y$ has positive probability under the law of $Y$. A point $y$ is an atom if $P(Y=y)>0$. A support point need not be an atom.

Write $p_n=P(N=n)$, and let $G$ denote the law of one severity. For each $n\geq 1$, the conditional law of $S$ given $N=n$ is the $n$-fold convolution $G^{*n}$, the law of a sum of $n$ independent severities. For $n=0$, take $G^{*0}$ to be the point mass at zero, denoted $\delta_0$. These conditional laws give a direct route to both support and point probabilities.

Atoms of a Compound Sum

Theorem: For every real number $x$, $$ P(S=x)=\sum_{n=0}^{\infty}p_n\,G^{*n}(\{x\}). $$ In particular, $G^{*0}(\{x\})$ equals one when $x=0$ and zero otherwise.

Proof. The events $\{N=n\}$, for $n=0,1,2,\ldots$, are disjoint and have union of probability one. Partitioning the event $\{S=x\}$ by these events and applying the definition of conditional probability gives

P(S=x) =\sum_{n=0}^{\infty}P(S=x,N=n) =\sum_{n=0}^{\infty}P(N=n)P(S=x\mid N=n) =\sum_{n=0}^{\infty}p_nG^{*n}(\{x\}).

The series is valid by countable additivity; its terms are nonnegative and their sum is at most one. The $n=0$ conditional law is $\delta_0$, which gives the stated convention. $\square$

The formula matters because an atom can come from several different counts. At a given $x$, each count contributes its probability $p_n$ multiplied by the conditional point probability at $x$. Thus, the atom at zero is generally

P(S=0)=p_0+\sum_{n=1}^{\infty}p_nG^{*n}(\{0\}).

If severities are strictly positive with probability one, then a sum with $n\geq1$ is strictly positive, so every term in that sum is zero and $P(S=0)=p_0$. If severities are nonnegative and $q=P(X_1=0)$, then the sum of $n$ severities is zero exactly when all $n$ are zero. Independence gives $G^{*n}(\{0\})=q^n$, so

P(S=0)=\sum_{n=0}^{\infty}p_nq^n=E[q^N],

where the $n=0$ term is $p_0$, including when $q=0$. If severities can be negative, sums of nonzero terms may also cancel to zero, and the general convolution formula is needed.

Worked Example: Discrete Severities and Overlapping Sum Values

Suppose $P(N=0)=1/5$, $P(N=1)=1/2$, and $P(N=2)=3/10$. Each severity independently equals 1 or 3 with probability $1/2$ each. Find the support and all atoms of $S$.

For $N=0$, the sum is zero. For $N=1$, the conditional probabilities are $1/2$ at 1 and $1/2$ at 3. For $N=2$, the four ordered severity pairs $(1,1),(1,3),(3,1),(3,3)$ each have probability $1/4$. Their sums are 2, 4, 4, and 6. Therefore, the conditional probabilities given $N=2$ are $1/4$ at 2, $1/2$ at 4, and $1/4$ at 6.

Weight each conditional probability by its count probability. The resulting nonzero masses are $P(S=1)=(1/2)(1/2)=1/4$, $P(S=2)=(3/10)(1/4)=3/40$, $P(S=3)=(1/2)(1/2)=1/4$, $P(S=4)=(3/10)(1/2)=3/20$, and $P(S=6)=(3/10)(1/4)=3/40$. Also, $P(S=0)=1/5$.

Thus the support is $\{0,1,2,3,4,6\}$, and the complete list of masses is

P(S=0)=\frac15,\quad P(S=1)=\frac14,\quad P(S=2)=\frac{3}{40},\quad P(S=3)=\frac14,\quad P(S=4)=\frac{3}{20},\quad P(S=6)=\frac{3}{40}.

As a check, their sum is $8/40+10/40+3/40+10/40+6/40+3/40=40/40=1$. The count-two outcomes do not produce a value of 3: in particular, $3+1=4$, not 3. The mass at 3 comes only from $N=1$.

Support of a Compound Sum

Theorem: Let $C_n$ be the support of the conditional law of $S$ given $N=n$, and let $I=\{n\geq0:p_n>0\}$. Then the support of $S$ is the closure of $\bigcup_{n\in I}C_n$.

Proof. Let $U=\bigcup_{n\in I}C_n$. If $x$ lies in the closure of $U$, every open neighborhood $V$ of $x$ contains some point $y\in C_n$ for some $n\in I$. Since $V$ is open and contains $y$, it contains an open neighborhood of $y$. The definition of $C_n$ implies $P(S\in V\mid N=n)>0$. Because $p_n>0$, it follows that $P(S\in V)\geq p_nP(S\in V\mid N=n)>0$. Thus $x$ is in the support of $S$.

Conversely, if $x$ is outside the closure of $U$, some open neighborhood $V$ of $x$ is disjoint from $U$. For each $n\in I$, $V$ is disjoint from $C_n$, so $P(S\in V\mid N=n)=0$. Counts outside $I$ have probability zero. Summing over the count values gives $P(S\in V)=0$, so $x$ is not in the support of $S$. The two inclusions prove the result. $\square$

This theorem makes clear why the count law matters: only counts with positive probability contribute conditional supports. It also explains why support points should not be confused with atoms. For a continuous conditional law, points throughout an interval may lie in the support even though every single point in that interval has probability zero.

Worked Example: A Zero-Count Atom with Continuous Severities

Let $P(N=0)=0.7$ and $P(N=1)=0.3$. Given $N=1$, let the severity be uniform on $[1,2]$. Determine the support and atoms of $S$.

When $N=0$, $S=0$. When $N=1$, $S=X_1$, which has a continuous uniform distribution on $[1,2]$. The mixture therefore assigns probability $0.7$ to zero and probability $0.3$ to the interval component. Its support is $\{0\}\cup[1,2]$: neighborhoods of the endpoints 1 and 2 have positive probability from within the interval, even though the endpoints themselves have no point probability.

The only atom is at zero, with $P(S=0)=0.7$. For every $x\in[1,2]$, the uniform component gives $P(S=x\mid N=1)=0$, and the zero-count component contributes nothing at $x$. Hence $P(S=x)=0$ for every such $x$. This is a concrete case where every point of an interval is in the support, but none is an atom.

Mixed Severities Can Create More Than One Route to an Atom

Suppose a severity law has both discrete and continuous parts. The discrete part can generate atoms in sums, while the continuous part can fill out intervals of support. The atom formula remains valid without separating these components, but careful conditioning is often the clearest way to identify where atoms occur.

Worked Example: A Mixed Severity Law with a Zero Atom

Let $P(N=0)=0.2$, $P(N=1)=0.3$, and $P(N=2)=0.5$. Independently, each severity is zero with probability $0.4$; with probability $0.6$, it has a uniform distribution on $[1,2]$. Find the atom at zero, the other atoms, and the support of $S$.

For $N=0$, the sum is zero. Given $N=1$, zero occurs exactly when the severity takes its zero component, with probability $0.4$. Given $N=2$, zero occurs only if both severities are zero, with probability $0.4^2=0.16$. Therefore

$$ P(S=0)=0.2+(0.3)(0.4)+(0.5)(0.16) =0.2+0.12+0.08=0.4. $$

There are no other atoms. Given $N=1$, the nonzero component is continuous. Given $N=2$, a sum with at least one uniform component has probability zero at any specified point: conditioning on the other severity leaves a continuous uniform variable, which has probability zero of taking the one value needed to make the sum equal that point. The only remaining case is two zero severities, which contributes to zero, already counted.

For $N=1$, the conditional support is $\{0\}\cup[1,2]$. For $N=2$, it includes zero, sums with one uniform severity in $[1,2]$, and sums with two uniform severities in $[2,4]$. Since all three count values have positive probability, the support of $S$ is $\{0\}\cup[1,4]$. There is no support between zero and one, even though the continuous part fills the interval from one to four.

Useful Checks and Common Pitfalls

When the severity law is discrete, sums can coincide even when the ordered severity sequences differ. For example, $(1,3)$ and $(3,1)$ both give 4, so their probabilities must be added. When the severity law has a continuous component, a range of possible values does not imply point masses at those values. And when the count can be zero, the empty sum contributes a point mass at zero even if every severity is strictly positive.

A common mistake is to say that every point in the support has positive probability. That is false: a uniform random variable has support $[1,2]$, but assigns zero probability to each individual point. The correction is to use the neighborhood definition of support and calculate atoms separately with $P(S=x)=\sum_n p_nG^{*n}(\{x\})$. A related mistake is to identify the zero atom with $P(N=0)$ without checking the severity law. That equality holds for strictly positive severities, but a severity that can equal zero may produce $S=0$ even when $N>0$.

Key takeaway: Determine support from the conditional sum laws for count values of positive probability, and determine each atom by adding the corresponding conditional point probabilities. Treat the empty sum explicitly; it creates a zero atom whenever $P(N=0)>0$.

Check Your Understanding

Use the count-conditioned laws to answer each question, distinguishing support from atoms.

If $P(N=0)=0.25$ and severities are strictly positive, what is $P(S=0)$? Which assumption makes the answer immediate?
A severity is uniform on $[2,5]$, and $N$ equals zero or one with positive probability for each. Describe the support of $S$ and identify its atoms.
For two discrete severities, why must probabilities of different ordered pairs be added when those pairs have the same sum?
If severities are nonnegative with $P(X_1=0)=q$, write the formula for $P(S=0)$ in terms of the count probabilities $p_n$.
Can a point belong to the support of a compound sum while having probability zero? Explain using a continuous severity component.