Validating a Proposed Mixed Distribution

What Must Be Checked?

A formula involving point masses and a density can look plausible without defining a probability distribution. A negative coefficient cannot represent probability, a density that is negative on a set of positive length cannot represent probability, and even nonnegative pieces fail if their total mass is not one. If a CDF is also supplied, it must agree with the proposed pieces—not merely look like a familiar step-plus-curve.

We will check a candidate representation consisting of countably many atoms and an absolutely continuous part. This is a specific claim about the whole law: it says that all probability is accounted for by those atoms and that density. A singular continuous component, for example, is not included in this form.

Definition: Let $a_1,a_2,\ldots$ be real numbers, let $p_j$ be the proposed mass at $a_j$, and let $f:\mathbb{R}\to\mathbb{R}$ be a measurable proposed density. The associated atom-and-density representation is $$ P(X\in A)=\sum_{j=1}^{\infty}p_j\,\mathbf{1}\{a_j\in A\}+\int_A f(x)\,dx $$ for every Borel set $A\subseteq\mathbb{R}$. A finite list of atoms is included by taking only finitely many nonzero $p_j$. Repeated locations are allowed; their listed masses add.

The indicator $\mathbf{1}\{a_j\in A\}$ is one when $a_j\in A$ and zero otherwise. Thus an atom contributes probability only to sets containing its location. The density contributes probability through integration with respect to length; a single point has length zero, so the density itself contributes no point mass.

The Basic Validity Conditions

Theorem: The proposed representation defines a probability distribution if

$p_j\geq 0$ for every $j$;
$f(x)\geq 0$ for almost every $x$, and $f$ is Lebesgue measurable;
$\sum_{j=1}^{\infty}p_j+\int_{-\infty}^{\infty}f(x)\,dx=1$.

Under these conditions, its CDF is $$ F(x)=\sum_{j=1}^{\infty}p_j\,\mathbf{1}\{a_j\leq x\}+\int_{-\infty}^{x}f(t)\,dt. $$

The use of “almost every” matters: changing a density at isolated points, or on any set of length zero, does not change its integrals. By contrast, changing a listed atom weight changes the probability assigned to its location. The total-mass equation counts every listed atom, including repeated locations, and the entire integral of the density.

Check signs.
Every proposed point mass must be nonnegative, and the density must be nonnegative almost everywhere.

Check the total.
Add the weights and integrate the density over the whole real line. The result must be exactly one.

Check the locations and CDF.
Combine weights at repeated locations, then compare the claimed CDF with the step contributions and density integral at every $x$.

Here is why the conditions suffice. For any Borel set $A$, each term $p_j\mathbf{1}\{a_j\in A\}$ is nonnegative, and the integral over $A$ is nonnegative. The sum and integral are countably additive as functions of $A$; their sum is therefore a measure. Its total mass, found by taking $A=\mathbb{R}$, is the stated sum plus integral, which equals one. Hence it is a probability measure.

Taking $A=(-\infty,x]$ in the representation gives the displayed CDF formula. The infinite sum is well-defined: its terms are nonnegative, and it is bounded above by $\sum_j p_j\leq 1$. Each atom produces a jump at its location, while the integral term is continuous in $x$ because $f$ is integrable. This proves the theorem. $\square$

Worked Checks

Worked Example: One Atom and a Uniform Density Piece

Consider a proposed atom of mass $0.3$ at $0$ and the density $f(x)=0.35$ for $0<x<2$, with $f(x)=0$ elsewhere. Determine whether this is a probability distribution and derive its CDF.

Signs. The weight $0.3$ is nonnegative and $f(x)\geq 0$ everywhere, so the sign checks pass.

Total mass. The density contributes $$ \int_{-\infty}^{\infty} f(x)\,dx=\int_0^2 0.35\,dx=0.35(2-0)=0.7. $$ Therefore the total is $0.3+0.7=1$, so the representation is valid.

CDF. For $x<0$, neither component has contributed mass. For $0\leq x<2$, the atom contributes $0.3$ and the density contributes $\int_0^x0.35\,dt=0.35x$. For $x\geq2$, all mass has accumulated. Thus $$ F(x)= \begin{cases} 0, & x<0,\\ 0.3+0.35x, & 0\leq x<2,\\ 1, & x\geq2. \end{cases} $$ At $x=0$, the CDF jumps from $0$ to $0.3$, exactly the listed atom mass. At $x=2$, it reaches $0.3+0.35(2)=1$; there is no atom at $2$, so the CDF does not jump there.

Worked Example: Correct Total Mass Does Not Fix a Negative Density

A proposal assigns mass $0.2$ to $0$, uses $f(x)=-0.1$ on $0<x<1$, $f(x)=0.9$ on $1<x<2$, and sets $f(x)=0$ elsewhere. Is it valid?

The proposed integral is $$ \int_{-\infty}^{\infty}f(x)\,dx=(-0.1)(1)+(0.9)(1)=0.8, $$ so the total listed mass is $0.2+0.8=1$. But $f(x)=-0.1$ throughout an interval of length one, so the density is negative on a set of positive length. The representation is invalid despite having total one.

The problem also appears in the implied CDF formula. For $0\leq x<1$, it would give $F(x)=0.2-0.1x$, which decreases as $x$ increases. A CDF must be nondecreasing: for $x\leq y$, the event $\{X\leq x\}$ is contained in $\{X\leq y\}$, so $F(x)\leq F(y)$. The negative density violates this necessary property.

Worked Example: Repeated Atom Locations and an Endpoint

A proposal lists mass $0.2$ at $0$, mass $0.1$ at $1$, and another mass $0.15$ at $1$. It also gives $f(x)=0.55$ on $0<x<1$ and $f(x)=0$ elsewhere. Validate the representation and find its CDF.

All weights and density values are nonnegative. The density integral is $$ \int_{-\infty}^{\infty}f(x)\,dx=0.55(1-0)=0.55. $$ The total mass is $0.2+0.1+0.15+0.55=1$, so the representation is valid. The two listed masses at $1$ combine to $0.25$.

For $x<0$, the CDF is zero. For $0\leq x<1$, it includes the atom at $0$ and the density accumulated between $0$ and $x$, giving $0.2+0.55x$. At $x=1$, the atom masses at $1$ are included, as is the entire density integral. Hence $$ F(x)= \begin{cases} 0, & x<0,\\ 0.2+0.55x, & 0\leq x<1,\\ 1, & x\geq1. \end{cases} $$ The left limit at $1$ is $0.2+0.55=0.75$, while $F(1)=1$. The jump is $1-0.75=0.25$, matching the combined atom mass there. The density contributes no additional mass at the endpoint $1$.

Consistency With a Claimed CDF

Suppose a proposal supplies both components and a claimed CDF $G$. Passing the sign and total checks does not excuse a mismatch between $G$ and those components. The precise check is to compare $G(x)$, for every real $x$, with $$ \sum_j p_j\,\mathbf{1}\{a_j\leq x\}+\int_{-\infty}^{x}f(t)\,dt. $$ If they agree for all $x$, the CDF is exactly the one induced by the representation. Conversely, if the listed components describe the law, evaluating their probability on $(-\infty,x]$ gives this formula, so the claimed CDF must agree with it at every $x$.

This comparison checks more than the jump sizes. At any location $a$, the jump of the CDF is $F(a)-F(a-)$, where $F(a-)=\lim_{x\uparrow a}F(x)$. For the proposed representation, this jump is the sum of the weights listed at $a$. The density integral is continuous and therefore adds no jump. Also, where the CDF is differentiable, its derivative agrees with $f$ almost everywhere, as follows from the fundamental theorem for Lebesgue integrals. These are useful diagnostics, but they are not enough by themselves to verify the entire representation.

Theorem: For a nonnegative atom-and-density representation of total mass one, its CDF is determined at every $x$ by the component formula above. A claimed CDF represents precisely those components if and only if it equals that formula for every real $x$.

For the forward direction, if the components give the law, then the event $\{X\leq x\}$ contains exactly the listed atoms at locations $a_j\leq x$, and the density contributes $\int_{-\infty}^{x}f(t)\,dt$. Adding these probabilities gives the formula. For the reverse direction, the formula is the CDF of the probability measure defined by the components. If the claimed CDF equals it everywhere, the two probability distributions have the same CDF and therefore assign the same probabilities to all intervals $(a,b]$, by subtracting their CDF values. These intervals determine a probability measure on the Borel sets, so the claimed distribution is the one defined by the components. $\square$

Why Local Checks Can Mislead

A jump check only examines atoms, and a derivative check only examines the rate of change where that derivative exists. Neither rules out a leftover singular continuous component. The Cantor distribution, for instance, has no atoms and has derivative zero almost everywhere, yet it is not the constant-zero CDF. Thus matching all listed jumps and an almost-everywhere density does not establish that a proposed representation accounts for the whole law. The full CDF identity—or an equivalent check of the probability measure—is needed when the claim is specifically “atoms plus density only.”

A common mistake is to integrate a density over an interval and then add an atom at an endpoint without checking what the CDF expression includes. For the interval $(a,b]$, the probability is $F(b)-F(a)$: an atom at $b$ is included, and an atom at $a$ is excluded. For $[a,b]$, an atom at $a$ is included too. Keeping the interval convention explicit prevents endpoint errors.

In practice, validate in order: signs, total mass, atom locations and combined weights, then the claimed CDF. A representation that fails any one of these checks cannot be the probability law claimed, even if its other features look plausible.

Check Your Understanding

For each proposed representation, identify which checks are needed before accepting it.

A proposed density is nonnegative and integrates to $0.8$, while the listed atom weights sum to $0.2$. What additional issue, if any, must be checked?
Two listed atoms are both located at $3$, with weights $0.1$ and $0.25$. What jump should the CDF have at $3$, assuming the density is integrable?
Why is a density that is negative only at one isolated point not, by itself, evidence that its induced integral measure assigns negative probability?
A claimed CDF has the same jumps and almost-everywhere derivative as the listed components. Explain why that need not verify the proposed atom-and-density representation.
For a proposed atom at $a$, which endpoint convention includes its mass: $(a,b]$, $[a,b]$, or both?