Locating Support, Atoms, and Density Regions

Reading a Mixed Probability Formula

A probability formula can assign probability in more than one way. Some probability may be concentrated at individual points, while another part is spread across intervals and calculated by integrating a density. To interpret such a formula, keep these contributions separate: a point mass determines the probability of its specific point, and a density determines probabilities through integrals over sets.

Here we consider laws represented by a countable collection of point masses together with a density. The goal is to locate the places that contribute probability in that representation. This does not require treating every point where a density is positive as a point mass, or deciding that a point has positive probability merely because it lies in an interval where a density contributes.

Definition: Suppose that, for every Borel set $A\subseteq\mathbb{R}$, a random variable $X$ has probabilities of the form $$ P(X\in A)=\sum_{j}p_j\,\mathbf{1}_{\{x_j\in A\}}+\int_A f(x)\,dx, $$ where the $x_j$ are distinct real numbers, $p_j>0$, $f$ is a nonnegative Lebesgue-integrable function, and $\sum_j p_j+\int_{\mathbb R}f(x)\,dx=1$. The $x_j$ are the listed atom locations, with point masses $p_j$. The density region for a chosen version of $f$ is the set where $f(x)>0$; the density contributes to an event through the integral of $f$ over that event.

The indicator $\mathbf{1}_{\{x_j\in A\}}$ is 1 if $x_j\in A$ and 0 otherwise. If a point mass and the density region include the same location, the formula still applies: the event containing that point receives its point mass, while the density contributes according to an integral. A density is defined only up to changes on sets of Lebesgue measure zero, so its pointwise positive region can change under such a change. The probabilities it determines do not change.

Locating Atoms and Density Contributions

A point $a$ is an atom when $P(X=a)>0$. In a formula of the displayed form, the atom locations are precisely the listed $x_j$, and each atom has probability $p_j$. The density contributes no probability to a singleton, even when $f(a)>0$, because a singleton has Lebesgue measure zero.

Result: For a law with the representation above, $$ P(X=a)= \begin{cases} p_j,& a=x_j\text{ for some }j,\\ 0,& a\notin\{x_j:j\}. \end{cases} $$ Thus, a location in a density region need not be an atom, and a listed atom remains an atom whether or not it lies in a density region.

Proof. Apply the probability formula to the singleton $A=\{a\}$. Its atomic sum is $p_j$ if $a=x_j$, and is zero otherwise, since the atom locations are distinct. The density term is $\int_{\{a\}}f(x)\,dx=0$, because $\{a\}$ has Lebesgue measure zero. Adding the two contributions gives the result. $\square$

The density region identifies where a density can contribute, but a positive density value at one point does not by itself ensure positive probability for a particular event. For an event $A$, the density contribution is $\int_A f(x)\,dx$. It is positive exactly when $A$ overlaps the positive-density region in a set of positive Lebesgue measure (up to changes on null sets).

Result: Let $D=\{x:f(x)>0\}$ for a chosen nonnegative version of the density. For every Borel set $A$, $$ \int_A f(x)\,dx>0 \quad\Longleftrightarrow\quad \lambda(A\cap D)>0, $$ where $\lambda$ denotes Lebesgue measure. This criterion concerns the density contribution only; the event may also receive probability from atoms in $A$.

Proof. If $\lambda(A\cap D)=0$, then $f=0$ almost everywhere on $A$, so $\int_A f(x)\,dx=0$. Conversely, suppose $\lambda(A\cap D)>0$. Since $$ A\cap D=\bigcup_{m=1}^{\infty}\{x\in A:f(x)\geq 1/m\}, $$ at least one set in this union has positive Lebesgue measure; otherwise their countable union would have measure zero. For that $m$, $$ \int_A f(x)\,dx \geq \frac{1}{m}\lambda\bigl(\{x\in A:f(x)\geq 1/m\}\bigr)>0. $$ This proves both directions. $\square$

Worked Examples

Worked Example: A Point Mass Inside a Density Region

Let $X$ have a point mass of $0.3$ at $1$, and let the remaining $0.7$ be uniform on $[0,2]$. Identify its atom and density region, then calculate $P(X=1)$ and $P(0.5<X<1.5)$.

The uniform component has density $1/2$ on $[0,2]$, multiplied by its mixture weight $0.7$. Thus a density for the continuous contribution is $$ f(x)= \begin{cases} 0.35,&0\leq x\leq 2,\\ 0,&\text{otherwise}. \end{cases} $$ Its integral is $0.35(2)=0.7$, as required. The formula is $P(X\in A)=0.3\mathbf{1}_{\{1\in A\}}+\int_A f(x)\,dx$. The atom is at $1$; the density contributes on $[0,2]$. These locations overlap.

For the singleton $\{1\}$, the density integral is zero, so $$ P(X=1)=0.3+\int_{\{1\}}f(x)\,dx=0.3+0=0.3. $$ The interval $(0.5,1.5)$ contains the atom and has length $1$. Therefore $$ P(0.5<X<1.5)=0.3+\int_{0.5}^{1.5}0.35\,dx =0.3+0.35(1)=0.65. $$ The interval probability includes both types of contribution.

Worked Example: Gaps in the Density Region and an Endpoint Atom

Suppose $X$ has mass $0.2$ at $1$, and its remaining probability has density $$ f(x)= \begin{cases} 0.4,&x\in[0,1]\cup[2,3],\\ 0,&\text{otherwise}. \end{cases} $$ Find the atom locations, the density region, $P(X=1)$, and $P(0.5\leq X\leq 2.5)$.

The two density intervals each have length $1$, so the density integrates to $0.4(1)+0.4(1)=0.8$. Adding the mass $0.2$ gives total probability $1$. The only atom is at $1$. The density contributes on the two indicated intervals; it contributes nothing in the gap $(1,2)$.

At $1$, the atom contributes $0.2$, while the integral over the singleton is zero: $$ P(X=1)=0.2+\int_{\{1\}}f(x)\,dx=0.2. $$ For $[0.5,2.5]$, the density overlaps $[0.5,1]$ and $[2,2.5]$, each of length $0.5$. The atom at $1$ is also included. Hence $$ P(0.5\leq X\leq 2.5) =0.2+0.4(0.5)+0.4(0.5) =0.2+0.2+0.2=0.6. $$ Including or excluding the endpoints of either density interval would not change these density integrals. Including or excluding the atom at $1$, however, would change the point-mass contribution.

Worked Example: Countably Many Atoms and an Accumulation Point

Consider a law with mass $0.25$ at $-1$, density $f(x)=0.25$ on $[0,2]$ and zero elsewhere, and atoms at $x_k=3+1/k$, for $k=1,2,\ldots$, with masses $p_k=2^{-(k+2)}$. Locate the probability contributions and find the probability that $X$ is one of the $x_k$ with even $k$.

The density integrates to $0.25(2)=0.5$. The countable atom masses sum to $$ \sum_{k=1}^{\infty}2^{-(k+2)} =\frac18\sum_{k=1}^{\infty}2^{-(k-1)} =\frac18\cdot 2=\frac14. $$ Including the mass $0.25$ at $-1$, the total is $0.25+0.5+0.25=1$.

The atoms are $-1$ and $3+1/k$ for every positive integer $k$. The density contributes on $[0,2]$. These regions are disjoint. The locations $3+1/k$ approach $3$, but $3$ is not one of the listed atom locations and has no density contribution. In particular, $P(X=3)=0$.

For even indices, write $k=2j$, where $j=1,2,\ldots$. The required probability is the sum of the corresponding point masses: $$ P\bigl(X\in\{x_k:k\text{ even}\}\bigr) =\sum_{j=1}^{\infty}2^{-(2j+2)} =\frac{1}{16}\sum_{j=0}^{\infty}\left(\frac14\right)^j =\frac{1}{16}\cdot\frac{1}{1-1/4} =\frac{1}{12}. $$ The density contributes nothing to this countable event because a countable set has Lebesgue measure zero.

A Reliable Reading Procedure

When a formula combines masses and a density, read it component by component. The following procedure keeps point probabilities distinct from probabilities accumulated across regions.

List the point masses.
Record every location $x_j$ and its weight $p_j$. These locations are the atoms in the given representation.

Locate the density contribution.
Read where $f$ is positive, allowing for changes on sets of Lebesgue measure zero. For an event, calculate its density contribution by integrating over the event.

Check which components the event includes.
Add the weights of included atoms to the density integral over the event. Do not replace one contribution with the other when their locations overlap.

A recurring mistake is to treat a positive density value as the probability at that point. For example, if $f(1)=0.35$, it is incorrect to conclude that $P(X=1)=0.35$. The density value is not a point mass: $\int_{\{1\}}f(x)\,dx=0$. In the first worked example, $P(X=1)=0.3$ because there is a separate atom of size $0.3$; without that atom, the point probability would be zero.

Another mistake is to assume that every point approached by a sequence of atoms must itself be an atom. In the third example, the atom locations approach $3$, but the formula assigns no mass at $3$. An accumulation of locations is not itself a point-mass contribution. To determine whether a particular point is an atom, inspect the mass at that point.

Key takeaway: In a discrete-plus-density representation, atoms are identified by positive point weights, while the density contributes probabilities through integrals. For any event, its probability is the sum of the included atom weights and the integral of the density over that event.

Check Your Understanding

For each question, separate any point-mass contribution from the density contribution.

A law has mass $0.4$ at $0$ and density $0.3$ on $[1,3]$, zero elsewhere. What is $P(X=0)$, and what is $P(1<X<2)$?
If a density satisfies $f(2)>0$ but there is no point mass at $2$, what is $P(X=2)$? Explain using the integral over a singleton.
A mixed law has an atom at $1$ and a density that is positive on $[0,2]$. Which contribution determines $P(X=1)$, and which contributions enter $P(0.5<X<1.5)$?
For a density region $D=\{x:f(x)>0\}$, what condition on a Borel set $A$ makes its density contribution positive?
Atoms occur at $2+1/k$ for all positive integers $k$, with no listed mass at $2$ and no density near $2$. Is $2$ an atom? Why or why not?

Locating Support, Atoms, and Density Regions

What You'll Learn