Support of Mixed Laws

Why Mixed Supports Need Care

A mixed law can put positive probability at individual points and also spread probability continuously over a region. Its support must describe both features. It may also contain limit points of atoms, or boundaries where a discrete component and a continuous component meet. Such points can belong to the support even when the law assigns them probability zero.

The key idea is to combine the supports of the components that actually receive positive mixture weight, then take the closure. The closure matters because support is defined through probabilities of neighborhoods, not only by asking which individual points have positive probability. In particular, a density component can contribute endpoints, and a sequence of atoms can contribute its accumulation points.

Support of a Mixture

Definition: For a probability law $P$ on the real line, its support is the set $$ S(P)=\{x\in\mathbb{R}: P((x-\varepsilon,x+\varepsilon))>0\text{ for every }\varepsilon>0\}. $$ Thus $x$ belongs to the support when every open interval around $x$, no matter how small, has positive probability. A point in the support need not itself have positive probability.

A mixture of component laws $P_1,P_2,\ldots$ with weights $w_j$ is the law $P$ satisfying $P(B)=\sum_j w_jP_j(B)$ for every measurable set $B$. Here the weights are nonnegative and sum to one; only indices with $w_j>0$ contribute. The formula includes finite mixtures as a special case.

Theorem: For a finite or countable mixture of probability laws on the real line with weights $w_j\geq 0$ summing to one, $$ S(P)=\overline{\bigcup_{j:w_j>0}S(P_j)}. $$ The bar denotes closure: the union together with all its limit points.

This result does not say that every point in the mixture support must be an atom, nor that the support is just the set of points where a density is positive. It says to identify each positive-weight component's support and then close their union. We prove the result directly from the neighborhood definition.

Take a point in the component-support union.
If $x\in S(P_j)$ for some $j$ with $w_j>0$, every neighborhood $U$ of $x$ has $P_j(U)>0$. Hence $P(U)\geq w_jP_j(U)>0$, so $x\in S(P)$.

Include limit points of that union.
Every support $S(P)$, as defined by positive probability in every neighborhood, is closed. Therefore it contains the closure of the component-support union.

Exclude points outside the closure.
If $x$ is outside the closure, some open neighborhood $U$ of $x$ misses every $S(P_j)$ with positive weight. Since $U$ is open, each point in $U$ has a smaller neighborhood disjoint from $S(P_j)$; equivalently, the complement of $S(P_j)$ is open and $P_j(U)=0$. Thus $P(U)=\sum_jw_jP_j(U)=0$, so $x\notin S(P)$.

For completeness, the fact used in the last step follows from the support definition: if $y\notin S(P_j)$, some open interval around $y$ has $P_j$-probability zero. The complement of $S(P_j)$ is therefore a union of open intervals of zero probability. Any open set contained in that complement has probability zero: it can be covered by a countable collection of such intervals, using the countable base of intervals with rational endpoints. This also verifies the required measure-zero conclusion for $U$. The two inclusions prove the theorem. $\square$

Atoms and Density Components

For a discrete component with positive point masses at locations in a set $A$, its support is $\overline{A}$, not necessarily just $A$. Every neighborhood of a limit point of $A$ contains at least one positive-mass location. For a component with density $g$, its support consists of the points every neighborhood of which has positive integral of $g$. One must use neighborhood integrals rather than the value $g(x)$ at a single point, since changing a density at an individual point does not change its law.

Corollary: Suppose $P$ is a mixture of a purely atomic law with positive-mass locations $A$ and a law with density $g$, and both mixture weights are positive. Define $$ D_g=\left\{x:\int_{x-\varepsilon}^{x+\varepsilon}g(t)\,dt>0\text{ for every }\varepsilon>0\right\}. $$ Then $$ S(P)=\overline{A\cup D_g}. $$

Indeed, the atomic component has support $\overline{A}$: each point of $A$ has positive probability in every neighborhood, and each point outside $\overline{A}$ has a neighborhood containing no mass locations. The density component has support $D_g$ by the definition of its probabilities through density integrals. Applying the mixture theorem gives $\overline{\overline{A}\cup D_g}=\overline{A\cup D_g}$, as claimed. $\square$

The positive-weight condition is essential. A component assigned weight zero contributes nothing, even if its own support is large. For positive weights, their numerical sizes do not change the support: any positive contribution to every neighborhood suffices.

Worked Example: A Point Mass at the Edge of a Uniform Component

Let $P$ put probability $1/3$ at $0$, and with probability $2/3$ draw from the uniform distribution on $[0,1]$. Find the support and identify which component supports contain $0$.

The point-mass component has support $\{0\}$. The uniform component has support $[0,1]$: every neighborhood of any point in that interval intersects it in a set of positive length, while points outside the interval have a sufficiently small neighborhood with zero probability. Both component weights are positive, so $$ S(P)=\overline{\{0\}\cup[0,1]}=[0,1]. $$

The point $0$ belongs to both component supports. The mixture assigns it probability $1/3$, since the uniform component assigns zero probability to a single point. Points in $(0,1]$ are supported by the uniform component, though none of them is an atom. The left endpoint is included even though it is a boundary of the density region.

Worked Example: Atoms Accumulating at a Boundary

Suppose a mixture assigns weight $1/2$ to a discrete law that puts mass $2^{-n}$ at $1/n$ for each integer $n\geq 1$, and weight $1/2$ to a uniform law on $[1,2]$. The masses of the discrete law sum to one because $\sum_{n=1}^{\infty}2^{-n}=1$. Find the full support.

The locations with positive point mass are $A=\{1,1/2,1/3,\ldots\}$. Their closure is $$ \overline{A}=\{1/n:n\geq1\}\cup\{0\}, $$ because $1/n\to0$, and there are no other limit points. The uniform component has support $[1,2]$. Consequently, $$ S(P)=\overline{A\cup[1,2]}=\{0\}\cup\{1/n:n\geq1\}\cup[1,2]. $$

The point $0$ belongs to the mixture support although neither component assigns it positive probability: every neighborhood of zero contains some $1/n$, which has positive mixture mass. The point $1$ is a shared support point: it is an atom of the discrete component and also the left endpoint of the uniform component's support. The interval $(0,1)$ is not filled in by taking closure; apart from the listed atom locations, its points are not in the support.

Worked Example: Separate Regions and a Shared Endpoint

Let $P$ be a mixture with weights $1/4,1/2,1/4$ on, respectively, a point mass at $0$, a uniform law on $[-2,-1]$, and a uniform law on $[0,2]$. Determine the support and describe the role of $0$.

The three component supports are $\{0\}$, $[-2,-1]$, and $[0,2]$. Each weight is positive. Their union is already closed, so $$ S(P)=[-2,-1]\cup[0,2]. $$

The point $0$ belongs to the support of both the point-mass component and the uniform component on $[0,2]$. Its point probability is exactly $1/4$: the first component contributes $1/4$, while the uniform component gives probability zero to the singleton $\{0\}$. The interval $(-1,0)$ is outside the support; no component gives positive probability to a sufficiently small neighborhood of any point strictly between $-1$ and $0$. This example shows why components should be combined by union and closure, not by filling the interval between their regions.

A Reliable Support Calculation

For a finite or countable mixture, use the following procedure. First discard any zero-weight components. Next determine the support of each remaining component using neighborhoods: for an atom, its location is supported; for a density, test local integrals; for a discrete component with infinitely many locations, include accumulation points. Finally take the union and its closure. Record separately which component supports contain a boundary point; that information can clarify whether the point is atomic, continuous, or shared.

A shared boundary does not mean that probability should be counted twice. The mixture formula adds the component probabilities of a set, and a continuous component assigns probability zero to a singleton boundary point. Support membership is about positive probability in every neighborhood, whereas an atom is about positive probability at the singleton itself.

Common Mistakes

Using only atom locations. This misses limit points such as $0$ for atoms at $1/n$. Include the closure of the component-support union.
Including a zero-weight component. Its support has no effect on the mixture. Use only components with positive weights.
Demanding positive point probability. A density-supported point or an accumulation point can be in the support without being an atom.
Filling gaps between components. Closure adds limit points, not every point between separated support regions.
Assuming a shared support point receives mass from both components. A density component can contain a boundary in its support while assigning that boundary zero point probability.

Key takeaway: The support of a mixed law is the closure of the union of the supports of its positive-weight components. Check atoms, density neighborhoods, accumulation points, and shared boundaries separately; support membership alone does not imply positive point mass.

Check Your Understanding

For each question, distinguish support membership from positive probability at a single point.

A law is a positive-weight mixture of a point mass at $2$ and a uniform law on $[2,5]$. What is its support, and which components support $2$?
A discrete component has positive masses at $3+1/n$ for every positive integer $n$. Is $3$ in the support of that component? Explain using neighborhoods.
A mixture has positive weights on the uniform laws on $[-3,-2]$ and $[1,4]$. Is $0$ in its support? Why or why not?
What changes in the support formula if one component has weight zero?
Can a point belong to the support of a density component while having probability zero as a singleton? Give an example.

What You'll Learn