Sums With Atoms and Mixed Components

Why a Sum Can Have Both Atoms and a Density

When random variables have continuous distributions, their sum is described by a density. But if a summand can equal a particular value with positive probability, it has an atom, and the sum may have point masses as well as a density. To find the full law, we must distinguish pairs of outcomes that give discrete sums from combinations that spread probability continuously.

Throughout, assume $X$ and $Y$ are independent and each has a law consisting only of countably many atoms and an absolutely continuous component. The continuous component may have total probability less than one, so its density is a subdensity: its integral is the probability assigned to that component. This setup includes purely discrete and purely continuous distributions as special cases.

Definition: A distribution has an atom at $a$ if $P(X=a)>0$. For the class of mixed distributions considered here, its law can be written $P(X\in A)=\sum_i p_i\mathbf{1}_{\{a_i\in A\}}+\int_A f_X(x)\,dx$ for every Borel set $A$, where the $a_i$ are distinct, $p_i=P(X=a_i)>0$, and $f_X$ is a nonnegative integrable subdensity. The atom probabilities and continuous mass satisfy $\sum_i p_i+\int_{\mathbb R} f_X(x)\,dx=1$. The same notation applies to $Y$, with atoms $b_j$, probabilities $q_j$, and subdensity $f_Y$.

If there are no atoms, the corresponding sum over atoms is empty. If there is no continuous component, the subdensity is zero. This decomposition excludes a singular-continuous component; the formula below is for laws made up of atoms and an absolutely continuous part.

The Law of the Sum

Let $S=X+Y$. When both variables land on atoms, their sum is an atom candidate. When one lands on an atom and the other comes from a continuous component, that continuous component is shifted. When both come from continuous components, their contribution is the usual convolution density. The resulting formula keeps these sources separate.

Theorem: If $X$ and $Y$ are independent and their laws have the atomic and absolutely continuous decompositions above, then the mass of $S$ at $s$ is $$ P(S=s)=\sum_{i,j:\,a_i+b_j=s}p_iq_j. $$ The remaining part of the law is absolutely continuous, with subdensity $$ g(s)=\sum_i p_i f_Y(s-a_i)+\sum_j q_j f_X(s-b_j) +\int_{\mathbb R} f_X(x)f_Y(s-x)\,dx $$ for almost every $s$. The sums and integral are nonnegative; the expression defines an integrable density up to changes on a set of Lebesgue measure zero.

The restriction $a_i+b_j=s$ in the mass formula matters: if several pairs of atoms add to the same value, all their probabilities contribute to that one atom. The three terms in $g$ correspond, respectively, to an atom of $X$ with the continuous component of $Y$, an atom of $Y$ with the continuous component of $X$, and two continuous components.

Why the Formula Holds

Independence means the joint law of $(X,Y)$ is the product of their marginal laws. For any Borel set $A$, divide the product law into its four combinations: atom–atom, atom–continuous, continuous–atom, and continuous–continuous. This gives

\begin{aligned} P(S\in A) ={}&\sum_{i,j}p_iq_j\mathbf{1}_{\{a_i+b_j\in A\}}\\ &+\sum_i p_i\int_{\mathbb R}\mathbf{1}_{\{a_i+y\in A\}}f_Y(y)\,dy\\ &+\sum_j q_j\int_{\mathbb R}\mathbf{1}_{\{x+b_j\in A\}}f_X(x)\,dx\\ &+\int_{\mathbb R}\int_{\mathbb R}\mathbf{1}_{\{x+y\in A\}}f_X(x)f_Y(y)\,dy\,dx. \end{aligned}

The first line assigns mass $p_iq_j$ to $a_i+b_j$; grouping terms with the same sum gives the stated point-mass formula. In the second line, substitute $s=a_i+y$, so $y=s-a_i$ and $dy=ds$. Its contribution becomes $\int_A\sum_i p_i f_Y(s-a_i)\,ds$. In the third line, substitute $s=x+b_j$, obtaining $\int_A\sum_j q_j f_X(s-b_j)\,ds$.

For the last line, fix $x$ and substitute $s=x+y$. Then $y=s-x$ and $dy=ds$. The contribution is $\int_A\int_{\mathbb R} f_X(x)f_Y(s-x)\,dx\,ds$. These substitutions are translations, so their Jacobian is one. Because all integrands are nonnegative, Tonelli’s theorem permits the rearrangement of the sums and integrals, including when there are countably many atoms. Combining the three integrals gives $\int_A g(s)\,ds$.

Each integral part is absolutely continuous with respect to Lebesgue measure, so it assigns probability zero to every singleton. Thus the only point masses are those from the atom–atom line. Finally, taking $A=\mathbb R$ shows that the total mass in the formula is one: the four combinations have total mass $(\sum_i p_i+\int f_X)(\sum_j q_j+\int f_Y)=1$. This proves both the decomposition and its normalization. $\square$

Worked Examples

Worked Example: Two Mixed Variables With the Same Continuous Support

Let $X$ equal 0 with probability $1/3$, and otherwise have a Uniform$(0,1)$ component. Let $Y$ equal 0 with probability $1/2$, and otherwise have a Uniform$(0,1)$ component. Assume independence. Find the atom and density of $S=X+Y$.

The atom probabilities are $p_0=1/3$ and $q_0=1/2$. The subdensities are $f_X(x)=(2/3)\mathbf{1}_{(0,1)}(x)$ and $f_Y(y)=(1/2)\mathbf{1}_{(0,1)}(y)$. The only atom–atom pair gives $S=0$, with mass $p_0q_0=1/6$.

For $0<s<1$, the atom–continuous contributions are $p_0f_Y(s)=1/6$ and $q_0f_X(s)=1/3$. For $1<s<2$, both shifted subdensities are zero. The continuous–continuous contribution is

\int_{\mathbb R} f_X(x)f_Y(s-x)\,dx =\frac{1}{3}\,\operatorname{length}\big((0,1)\cap(s-1,s)\big) = \begin{cases} s/3,&0<s<1,\\ (2-s)/3,&1\le s<2,\\ 0,&\text{otherwise}. \end{cases}

Therefore, a density for the continuous part is $g(s)=1/2+s/3$ for $0<s<1$, $g(s)=(2-s)/3$ for $1<s<2$, and zero elsewhere. Its integral is $\int_0^1(1/2+s/3)\,ds+\int_1^2(2-s)/3\,ds=2/3+1/6=5/6$. Adding the atom mass $1/6$ gives total probability one.

Worked Example: Different Atom Pairs Coincide

Let $X$ have mass $0.2$ at 0, mass $0.3$ at 1, and a continuous component of mass $0.5$ that is Uniform$(0,1)$. Let $Y$ be discrete, with masses $0.4$ at 0 and $0.6$ at 1. Assume independence. Find the law of $S=X+Y$.

The atom–atom pairs give $S=0$ from $(0,0)$, $S=1$ from both $(0,1)$ and $(1,0)$, and $S=2$ from $(1,1)$. Thus $P(S=0)=0.2(0.4)=0.08$, $P(S=1)=0.2(0.6)+0.3(0.4)=0.12+0.12=0.24$, and $P(S=2)=0.3(0.6)=0.18$. The two distinct pairs producing 1 must be added.

There is no continuous component in $Y$, so only the term that shifts $X$'s continuous component by atoms of $Y$ contributes to the density. The continuous subdensity of $X$ is $0.5$ on $(0,1)$. The atom of $Y$ at 0 contributes density $0.4(0.5)=0.2$ on $(0,1)$, and its atom at 1 contributes density $0.6(0.5)=0.3$ on $(1,2)$. The density integrates to $0.2+0.3=0.5$; the atom masses sum to $0.08+0.24+0.18=0.5$. Together they account for all probability.

Worked Example: A Continuous Summand Removes Point Masses

Let $X$ be 0 with probability $1/2$, and otherwise have an Exponential$(1)$ component. Let $Y$ be Exponential$(2)$, independently of $X$. Here the rate parameter is used, so the exponential densities are $e^{-x}\mathbf{1}_{(0,\infty)}(x)$ and $2e^{-2y}\mathbf{1}_{(0,\infty)}(y)$. Find the density of $S=X+Y$.

Since $Y$ has no atoms, there are no atom–atom pairs and $S$ has no point masses. The atom of $X$ contributes $(1/2)2e^{-2s}=e^{-2s}$ for $s>0$. The continuous component of $X$ has subdensity $(1/2)e^{-x}$ for $x>0$, so the continuous–continuous contribution is

\begin{aligned} \int_0^s \frac12 e^{-x}\,2e^{-2(s-x)}\,dx &=\int_0^s e^{-x}e^{-2(s-x)}\,dx\\ &=e^{-2s}\int_0^s e^x\,dx\\ &=e^{-2s}(e^s-1)=e^{-s}-e^{-2s},\qquad s>0. \end{aligned}

Adding the two contributions gives $g(s)=e^{-2s}+e^{-s}-e^{-2s}=e^{-s}$ for $s>0$, and zero for $s<0$. This integrates to one. Although $X$ has an atom, adding the independent continuous $Y$ spreads that part of the probability into a density rather than preserving an atom.

Common Mistake and a Practical Check

A common mistake is to assume that any atom in either summand becomes an atom in the sum. For an atom at $a$ in $X$, the event $X=a$ contributes to $S$, but $S$ is then $a+Y$. If $Y$ is continuous, $a+Y$ is continuous and has no point mass. Only an atom of $X$ paired with an atom of $Y$ can create an atom of the sum under the assumptions here.

Another error is to count just one pair when distinct atom pairs yield the same value. The point mass at $s$ is the sum over all pairs satisfying $a_i+b_j=s$. As a check on any calculation, add the total atom mass to the integral of the continuous subdensity; the result must be one.

Key takeaway: For independent mixed summands, atom–atom pairs determine the point masses, while atom–continuous and continuous–continuous pairs determine the density. Keep these contributions separate, combine all pairs that yield the same atom, and verify that their total probability is one.

Check Your Understanding

Use the atom–atom and density contributions separately when answering.

If $X$ has atoms at 0 and 1 and $Y$ has atoms at 0 and 1, which pairs contribute to $P(X+Y=1)$?
Suppose $X$ has an atom at 2 and $Y$ has a density but no atoms. Can $X+Y$ have an atom arising from $X=2$? Explain.
In the density formula, what does the term $\sum_i p_i f_Y(s-a_i)$ represent?
Why must probabilities from multiple atom pairs be added when the pairs have the same sum?
If the atomic masses of $X+Y$ sum to $0.35$, what must the integral of its continuous subdensity be?