Combining Countably Many Atoms With a Density
A probability law can place probability at individual points and also spread probability continuously across a range of values. The atomic part can contain infinitely many points: each point may carry positive probability even when the total of all those probabilities is finite. To construct a valid law, the atomic weights and the density must together account for exactly one unit of probability.
The basic calculation is additive. For an event \(A\), add the weights of all atoms that lie in \(A\), then add the integral of the density over \(A\). This rule applies whether \(A\) is an interval or, more generally, a Borel set. It also clarifies why a density value at a point is not an additional point probability.
The distinctness condition makes each listed point correspond to one weight. Zero weights are permitted in the formula, although a point is an atom in the usual sense when its probability is positive. The density is also allowed to be zero outside a specified region. The total atomic mass is the infinite series \(\sum_j p_j\); it need not be found by adding terms one at a time if a standard series formula applies.
Why the Construction Defines a Probability Law
The normalization equation makes the probability of the whole real line equal to one. Nonnegativity of the weights and of \(g\) makes every assigned probability nonnegative. The remaining requirement is countable additivity, which ensures that probabilities behave consistently when an event is split into disjoint pieces.
Proof. Write \(\mu(A)\) for the stated formula. Both terms are nonnegative, so \(\mu(A)\geq 0\), and normalization gives $$ \mu(\mathbb{R})=\sum_{j=1}^{\infty}p_j+\int_{\mathbb{R}}g(x)\,dx=1. $$ Now let \(A_1,A_2,\ldots\) be pairwise disjoint Borel sets. For each atom \(x_j\), it can belong to at most one of these sets, so $$ \sum_{n=1}^{\infty}\mathbf{1}_{\{x_j\in A_n\}} =\mathbf{1}_{\{x_j\in \cup_n A_n\}}. $$ All terms in the atomic series are nonnegative. Therefore, interchanging the two nonnegative sums is valid, and the atomic contribution satisfies $$ \sum_{n=1}^{\infty}\sum_{j=1}^{\infty}p_j\mathbf{1}_{\{x_j\in A_n\}} =\sum_{j=1}^{\infty}p_j\mathbf{1}_{\{x_j\in \cup_n A_n\}}. $$ Countable additivity of the integral for a nonnegative measurable function on disjoint sets gives $$ \sum_{n=1}^{\infty}\int_{A_n}g(x)\,dx =\int_{\cup_n A_n}g(x)\,dx. $$ Adding these equalities yields \(\sum_n\mu(A_n)=\mu(\cup_n A_n)\). Thus \(\mu\) is nonnegative, countably additive, and has total mass one, as required. \(\square\)
A useful consequence identifies the probability at a point. A density integrated over a singleton contributes zero, since a singleton has Lebesgue measure zero. Because the \(x_j\) are distinct, the formula therefore gives $$ P(X=x_j)=p_j,\qquad P(X=x)=0\ \text{if }x\notin\{x_1,x_2,\ldots\}. $$ The second statement concerns point probabilities only; a point outside the list may still lie where \(g\) is positive and thus be surrounded by intervals with positive probability.
Worked Probability Calculations
Worked Example: Atoms at Reciprocal Integers
For \(k=1,2,\ldots\), put an atom at \(x_k=1/k\) with weight \(p_k=2^{-k}/4\). Let \(g(x)=3/8\) on \([0,2]\) and \(g(x)=0\) elsewhere. Verify the total probability, then calculate \(P(0\leq X\leq 1/2)\) and \(P(X=1/2)\).
The atomic weights sum to $$ \sum_{k=1}^{\infty}\frac{2^{-k}}{4} =\frac{1}{4}\left(\frac{1/2}{1-1/2}\right) =\frac{1}{4}. $$ The density contributes $$ \int_{\mathbb{R}}g(x)\,dx=\int_0^2\frac{3}{8}\,dx =\frac{3}{8}(2)=\frac{3}{4}. $$ Thus the total is \(1/4+3/4=1\), so the construction is normalized.
The atoms in \([0,1/2]\) are those with \(1/k\leq 1/2\), or \(k\geq 2\). Their total weight is $$ \sum_{k=2}^{\infty}\frac{2^{-k}}{4} =\frac{1}{4}\left(\frac{1/4}{1-1/2}\right)=\frac{1}{8}. $$ The density contribution is \(\int_0^{1/2}(3/8)\,dx=3/16\). Hence $$ P(0\leq X\leq 1/2)=\frac{1}{8}+\frac{3}{16}=\frac{5}{16}. $$ At \(1/2\), only the atom with \(k=2\) contributes point probability; the density assigns zero probability to the singleton. Therefore \(P(X=1/2)=p_2=2^{-2}/4=1/16\).
Worked Example: Geometric Atoms and an Exponential Density
For \(n=0,1,\ldots\), put an atom at \(n\) with weight \(p_n=(4/15)(1/3)^n\). Let \(g(x)=(3/5)e^{-x}\) for \(x\geq 0\) and zero for \(x<0\). Verify the total mass and calculate \(P(X>1)\).
The atomic weights form a geometric series: $$ \sum_{n=0}^{\infty}\frac{4}{15}\left(\frac{1}{3}\right)^n =\frac{4}{15}\frac{1}{1-1/3} =\frac{4}{15}\frac{3}{2}=\frac{2}{5}. $$ The integral of the continuous part is $$ \int_0^\infty \frac{3}{5}e^{-x}\,dx =\frac{3}{5}\left[-e^{-x}\right]_0^\infty=\frac{3}{5}. $$ These contributions sum to one.
The atoms strictly greater than 1 are at \(n=2,3,\ldots\), so their total weight is $$ \sum_{n=2}^{\infty}\frac{4}{15}\left(\frac{1}{3}\right)^n =\frac{4}{15}\frac{(1/3)^2}{1-1/3} =\frac{4}{15}\frac{1}{6}=\frac{2}{45}. $$ The continuous contribution is $$ \int_1^\infty\frac{3}{5}e^{-x}\,dx=\frac{3}{5}e^{-1}. $$ Consequently \(P(X>1)=2/45+3/(5e)\). The atom at \(1\) is excluded because the event is a strict inequality.
Worked Example: Negative Atoms and a Density on a Bounded Interval
For \(k=1,2,\ldots\), put an atom at \(-1/k\) with weight \(2^{-k-1}\), and set \(g(x)=x\) on \(0\leq x\leq 1\), with \(g(x)=0\) elsewhere. Verify normalization and find \(P(-1/2\leq X\leq 1/2)\).
The sum of atomic weights is $$ \sum_{k=1}^{\infty}2^{-k-1} =\frac{1}{2}\sum_{k=1}^{\infty}2^{-k} =\frac{1}{2}. $$ The density mass is $$ \int_{\mathbb{R}}g(x)\,dx=\int_0^1 x\,dx =\left[\frac{x^2}{2}\right]_0^1=\frac{1}{2}. $$ Thus the combined mass is one.
The interval contains atoms at \(-1/k\) exactly when \(k\geq 2\). Their total weight is $$ \sum_{k=2}^{\infty}2^{-k-1} =\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\cdots =\frac{1/8}{1-1/2}=\frac{1}{4}. $$ Its continuous contribution comes from \([0,1/2]\), since the density is zero for negative \(x\): $$ \int_0^{1/2}x\,dx=\left[\frac{x^2}{2}\right]_0^{1/2}=\frac{1}{8}. $$ Therefore \(P(-1/2\leq X\leq 1/2)=1/4+1/8=3/8\). In particular, the left endpoint \(-1/2\) contributes the atom with \(k=2\); it is included.
Interpreting the Two Contributions
For any Borel event, the atomic sum includes exactly the listed points belonging to that event. The integral measures the continuous contribution over the event. In the interval examples, open or closed endpoints matter for the atomic sum whenever an endpoint is an atom. Changing an endpoint does not change the integral, because adding or removing a singleton does not alter an integral with respect to Lebesgue measure.
An atom can lie within a region where the density is positive. Its probability is still its assigned weight \(p_j\); the integral over a singleton is zero. For a larger event containing that point, both contributions are included: the atomic weight and the integral over the event. These represent different parts of the law, not duplicate calculations of the same point mass.
A reliable calculation can be organized in three steps: first check that the atomic series converges and evaluate it; next integrate the nonnegative density over the whole line; finally verify that these two totals add to one. For a requested event, separately identify the included atom indices and the region of integration. Keeping the two contributions separate makes endpoint decisions and normalization checks transparent.
Check Your Understanding
For each question, keep the atomic sum separate from the density integral.
- Suppose atoms at positive integers \(k\) have weights \(2^{-k-2}\), and a density contributes total mass \(3/4\). What is the total atomic mass, and is the combined law normalized?
- A law has an atom of weight \(1/5\) at \(x=2\) and a density that is positive near 2. What is \(P(X=2)\), and why is the density not added to that point probability?
- For atoms at \(x_k=1/k\) with weights \(p_k=3\cdot 4^{-k}\), which indices contribute to \(P(X\leq 1/3)\)?
- In the geometric-atom example, why is the atom at 1 not included in \(P(X>1)\)?
- When an interval endpoint is an atom, what changes if that endpoint is included rather than excluded?