Constructing a Law With an Atom and a Density

Combining a Point Mass With a Density

A probability law need not be entirely discrete or entirely continuous. For example, a measurement may be exactly zero with positive probability, while nonzero outcomes vary continuously. To describe such a law, specify how much probability is assigned to a single point and how the remaining probability is spread according to a density.

The key is to keep the two contributions separate. The density describes probabilities for the continuous component after it has been normalized to have total probability one. A weight then assigns only part of the overall probability to that component. The atom receives the rest.

Definition: Let \(a\in\mathbb{R}\), let \(q\in(0,1)\), and let \(g:\mathbb{R}\to[0,\infty)\) be a measurable probability density, so that \(\int_{\mathbb{R}}g(x)\,dx=1\). The law with an atom of mass \(q\) at \(a\) and a continuous component with density \(g\) is defined, for every Borel set \(A\subseteq\mathbb{R}\), by

P(X\in A)=q\,\mathbf{1}_{\{a\in A\}}+(1-q)\int_A g(x)\,dx.

Here \(\mathbf{1}_{\{a\in A\}}\) equals 1 if \(a\in A\) and 0 otherwise. The parameter \(q\) is the atom’s probability; \(1-q\) is the total probability assigned to the density component.

The integral in this definition is with respect to ordinary length on the real line. It gives the probability assigned to \(A\) by the continuous component before that component is given its weight \(1-q\). The indicator term contributes \(q\) precisely when the event contains the atom location. In particular, \(g(a)\) is a density height, not an additional point probability at \(a\).

Why the Formula Defines a Probability Law

The construction must assign nonnegative probabilities, give the whole real line probability one, and be countably additive across disjoint events. These properties follow directly from the corresponding properties of a density and from the point-mass term.

Theorem: Under the conditions in the definition, the formula \(P(X\in A)=q\,\mathbf{1}_{\{a\in A\}}+(1-q)\int_A g(x)\,dx\) defines a probability measure on the Borel subsets of \(\mathbb{R}\).

Proof. Both terms in the formula are nonnegative, since \(q\in(0,1)\), \(g\geq 0\), and an indicator is either zero or one. For the whole real line,

P(X\in\mathbb{R}) =q+(1-q)\int_{\mathbb{R}}g(x)\,dx =q+(1-q)=1.

Now let \(A_1,A_2,\ldots\) be pairwise disjoint Borel sets and let \(A=\bigcup_{j=1}^{\infty}A_j\). Since they are disjoint, the point \(a\) belongs to at most one of them, so \(\mathbf{1}_{\{a\in A\}}=\sum_{j=1}^{\infty}\mathbf{1}_{\{a\in A_j\}}\). Countable additivity of the integral for nonnegative functions gives \(\int_A g(x)\,dx=\sum_{j=1}^{\infty}\int_{A_j}g(x)\,dx\). Substituting these two identities into the defining formula yields

P(X\in A) =\sum_{j=1}^{\infty} \left(q\,\mathbf{1}_{\{a\in A_j\}}+(1-q)\int_{A_j}g(x)\,dx\right) =\sum_{j=1}^{\infty}P(X\in A_j).

Thus the assignment is nonnegative, has total probability one, and is countably additive. It is a probability measure. \(\square\)

Reading Event Probabilities From the Construction

For any event \(A\), first check whether it contains \(a\). That determines whether the atom contributes \(q\). Next, integrate \(g\) over the portion of the real line in \(A\), and multiply that probability by \(1-q\). The two contributions are added, even if the atom location is also in a region where the density is positive.

Result: The probability of the atom location is \(P(X=a)=q\), while the probability of any different single point \(b\ne a\) is \(P(X=b)=0\). More generally, for any Borel set \(A\), its probability is the atom contribution, if \(a\in A\), plus the weighted density integral over \(A\).

To verify the point-probability statements, apply the event formula to a singleton. A singleton has integral zero under a density, so

P(X=a)=q+(1-q)\int_{\{a\}}g(x)\,dx=q, \qquad P(X=b)=0+(1-q)\int_{\{b\}}g(x)\,dx=0 \quad (b\ne a).

A useful consequence is that changing whether an interval includes an endpoint does not change its density integral: a single point has zero probability under a density. But if that endpoint is \(a\), including it changes the atom contribution by \(q\). This is why brackets and parentheses can matter for an event under a mixed law even though they do not change probabilities under a purely continuous density.

Worked Example: An Atom Alongside a Uniform Density

Put probability \(q=0.25\) at \(a=-1\), and let the continuous component have the uniform density on \([0,4]\):

g(x)= \begin{cases} \frac14, & 0\leq x\leq 4,\\ 0, & \text{otherwise}. \end{cases}

Find the probability of \(A=[-1,1]\). The event contains the atom at \(-1\), so the atom contributes \(0.25\). Its intersection with the density region is \([0,1]\), whose length is \(1\). Therefore

P(X\in[-1,1]) =0.25+(1-0.25)\int_0^1\frac14\,dx =0.25+0.75\left(\frac14\right) =0.4375.

For comparison, the event \([0,1]\) excludes the atom, so its probability is only \(0.75\int_0^1\frac14\,dx=0.1875\). The two events differ by the atom’s mass, even though the density integral over their continuous portions is the same.

Worked Example: An Atom at the Endpoint of an Exponential Component

Suppose \(q=0.20\) is assigned to \(a=0\), and the density component is exponential with rate 2: \(g(x)=2e^{-2x}\) for \(x\geq0\), and \(g(x)=0\) otherwise. Find the probability of \(A=[0,\frac{\log 2}{2}]\).

The endpoint \(0\) is the atom, so its contribution is \(0.20\). For the density component, set \(t=\frac{\log 2}{2}\). Then

\int_0^t 2e^{-2x}\,dx =\left[-e^{-2x}\right]_0^t =1-e^{-2t} =1-e^{-\log 2} =\frac12.

The density component has overall weight \(1-q=0.80\), so

P\left(0\leq X\leq\frac{\log 2}{2}\right) =0.20+0.80\left(\frac12\right) =0.60.

If the event were instead \((0,\frac{\log 2}{2}]\), it would exclude the atom. Its probability would be \(0.80(1/2)=0.40\). The density integral is unchanged because the single endpoint has zero probability under the density.

Worked Example: A Normal Density With a Point Mass at Its Center

Let \(q=0.30\), put the atom at \(a=0\), and use the standard normal density as the continuous component. Find \(P(X\geq0)\) and \(P(X>0)\).

By symmetry of the standard normal density, its integral over \([0,\infty)\) is \(1/2\). The event \(X\geq0\) includes the atom, so

P(X\geq0) =0.30+0.70\int_0^\infty\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,dx =0.30+0.70\left(\frac12\right) =0.65.

The event \(X>0\) excludes the atom. Removing the single point \(0\) does not change the normal integral, hence

P(X>0) =0.70\int_0^\infty\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,dx =0.70\left(\frac12\right) =0.35.

The difference, \(0.65-0.35=0.30\), is exactly the probability assigned to the atom.

A Reliable Calculation Procedure

For an event described by one or more intervals, the following steps keep the point mass and the density contribution distinct.

Locate the atom.
Write down its position \(a\) and mass \(q\). Decide whether the event includes \(a\), paying attention to open and closed endpoints.

Find the density-region intersection.
Identify the part of the event where \(g\) can be nonzero. Integrate \(g\) over that set.

Apply the component weight.
Multiply the density integral by \(1-q\), not by \(q\).

Add and check.
Add \(q\) if the atom is included. Confirm the result is between 0 and 1 and that no contribution has been counted twice.

The check is particularly useful for unions of separated intervals. If the event contains the atom, count its mass once, regardless of how many interval descriptions touch that location. The density integral is over the event as a set, so overlapping pieces must likewise not be integrated twice.

Common Mistake: Treating Density Height as Point Probability

A frequent error is to calculate the event probability by adding \(g(a)\) to the atom weight. This is incorrect: \(g(a)\) is a height, with units inverse to those of \(X\), whereas a probability is dimensionless. The density’s probability contribution is an integral over a set, multiplied by \(1-q\). Since the density component assigns zero probability to every singleton, the mass at \(a\) is exactly \(q\), not \(q+(1-q)g(a)\).

A related mistake is to multiply the atom mass by the interval probability from the density, or to use \(q\) as the weight on the density term. The defining formula makes the allocation explicit: \(q\) weights the point contribution, and \(1-q\) weights the entire density component. As a final check, taking \(A=\mathbb{R}\) must give \(q+(1-q)=1\).

Key takeaway: A law with one atom and one density component assigns an event the atom’s mass if the event contains the atom, plus the remaining weight times the density integral over the event. Endpoint inclusion can change the first contribution, but not the density integral at a single point.

Check Your Understanding

For each question, keep the atom term and density integral separate.

Let \(q=0.40\), place the atom at \(3\), and let the density component be uniform on \([0,2]\). Find \(P(X\in[1,3])\).
For a law with atom mass \(q\) at \(a\) and density component \(g\), explain why \(P(X=a)=q\), even if \(g(a)>0\).
An event contains the atom and has density-component probability \(0.2\). Write its probability under the mixed law in terms of \(q\).
In the normal example, why do \(P(X\geq0)\) and \(P(X>0)\) differ although the normal density assigns the same integral to both intervals?
What condition on \(g\) ensures that the density component contributes total probability one before it is weighted by \(1-q\)?