Support from Probability Mass Functions and Densities

Support Is Determined by Nearby Probability

A probability formula can be written on a larger set than the region where the law places probability. A density can also take a positive value at a point that contributes no probability at all. To find support reliably, ask a different question: does every neighborhood of the point have positive probability? This criterion works for discrete laws and for laws with densities, and does not depend on arbitrary choices in how a formula is written.

Definition: For a random variable $X$ taking values in $\mathbb{R}$, the support of its distribution is the set of points $x\in\mathbb{R}$ such that $$ P(x-\varepsilon<X<x+\varepsilon)>0 $$ for every $\varepsilon>0$. In words, every open interval centered at $x$, however small, must have positive probability.

This definition concerns the law of $X$, not a particular formula representing it. The probability in the definition includes any mass at the center as well as probability elsewhere inside the interval. A point can therefore belong to the support even if its own point probability is zero.

Reading Support from a Probability Mass Function

Suppose $X$ takes values in a finite or countable set $S$, with probability mass function $f$, where $f(y)=P(X=y)$. For any interval $I$, its probability is the sum of the masses at the possible values inside it:

P(X\in I)=\sum_{y\in S\cap I} f(y).

Consequently, an interval has positive probability exactly when it contains at least one value with positive mass. The support is the closure of the positive-mass points: points with positive mass are in the support, and so are their limit points. The following result makes the criterion precise.

Theorem: If $X$ takes values in a countable set $S$, then $$ \operatorname{supp}(X)=\overline{\{y\in S:f(y)>0\}}, $$ where the bar denotes closure in $\mathbb{R}$.

Proof. Write $A=\{y\in S:f(y)>0\}$. If $x\in\overline{A}$, every open interval centered at $x$ contains some $y\in A$. That interval therefore has probability at least $P(X=y)=f(y)>0$, so $x$ is in the support.

Conversely, if $x\notin\overline{A}$, the definition of closure gives an open interval centered at $x$ that contains no point of $A$. Since $X$ takes values in $S$, the interval’s probability is the sum of the masses of points in $S$ inside it. Every such mass is zero, so the interval has probability zero. Thus $x$ is not in the support. This proves the equality. $\square$

Worked Example: A Poisson Mass Function

Let $X\sim\operatorname{Poisson}(2)$, so for integers $k\geq 0$, $$ P(X=k)=e^{-2}\frac{2^k}{k!}, $$ and $P(X=k)=0$ for other real values $k$. Find the support.

For every integer $k\geq 0$, $e^{-2}2^k/k!>0$. Thus every nonnegative integer has positive mass. Any open interval containing one of these integers has positive probability.

If $x$ is not a nonnegative integer, the set of nonnegative integers is closed, so there is a sufficiently small interval centered at $x$ containing no nonnegative integer. The random variable cannot fall in that interval, and its probability is zero. Therefore, $$ \operatorname{supp}(X)=\{0,1,2,\ldots\}. $$

For example, the formula $e^{-2}2^x/x!$ is not needed or interpreted at arbitrary real $x$: the pmf is defined on the integer values. The support comes from where the random variable can place probability, not from extending the formula beyond its intended domain.

A less immediate case occurs when the positive-mass points are dense in a whole interval. The support need not be just the set of points that individually have positive mass.

Worked Example: Countable Masses with an Interval as Support

Enumerate the rational numbers in $[0,1]$ without repetition as $q_1,q_2,\ldots$. Define a discrete random variable by $$ P(X=q_k)=2^{-k},\qquad k=1,2,\ldots. $$ These probabilities sum to $1$, since $\sum_{k=1}^{\infty}2^{-k}=1$.

Take any $x\in[0,1]$ and any $\varepsilon>0$. The interval $(x-\varepsilon,x+\varepsilon)$ intersects $[0,1]$ in an interval of positive length, even when $x=0$ or $x=1$. Rational numbers are dense in the real line, so this intersection contains some $q_k$. Hence $$ P(x-\varepsilon<X<x+\varepsilon)\geq P(X=q_k)=2^{-k}>0. $$ Every point of $[0,1]$ is in the support.

If $x<0$, choose $\varepsilon=-x/2$; then $(x-\varepsilon,x+\varepsilon)$ lies below $0$ and has probability zero. If $x>1$, choose $\varepsilon=(x-1)/2$; then the interval lies above $1$ and also has probability zero. Thus $$ \operatorname{supp}(X)=[0,1]. $$

Each rational $q_k$ has positive point probability, but the irrational points in $[0,1]$ also belong to the support: every one of their neighborhoods contains positive mass.

Reading Support from a Density

If $X$ has density $f$, then for any Borel set $B$, $$ P(X\in B)=\int_B f(t)\,dt. $$ In particular, support is determined by integrals over neighborhoods. A pointwise check such as $f(x)>0$ is not enough: the integral over the neighborhood, rather than the value at its center, decides whether that neighborhood has positive probability.

Theorem: If $X$ has a density $f$, then $$ x\in\operatorname{supp}(X) \quad\Longleftrightarrow\quad \int_{x-\varepsilon}^{x+\varepsilon} f(t)\,dt>0 \ \text{for every }\varepsilon>0. $$

Proof. By the density formula for probabilities, for every $\varepsilon>0$, $$ P(x-\varepsilon<X<x+\varepsilon)=\int_{x-\varepsilon}^{x+\varepsilon} f(t)\,dt. $$ The defining condition for support says that the probability on the left is positive for every $\varepsilon>0$. Substituting the integral expression gives exactly the condition on the right. $\square$

Worked Example: A Density on Two Intervals and an Isolated Positive Value

Define $f$ on $\mathbb{R}$ by $$ f(x)= \begin{cases} \frac12, & x\in(0,1)\cup(2,3),\\ 17, & x=4,\\ 0, & \text{otherwise}. \end{cases} $$ This is a valid density: it is nonnegative and measurable, and its integral is $$ \int_{\mathbb{R}}f(x)\,dx =\frac12(1-0)+\frac12(3-2)+\int_{\{4\}}17\,dx =\frac12+\frac12+0=1. $$

For $x\in[0,1]$, every neighborhood of $x$ intersects $(0,1)$ in an interval of positive length. Since the density there is $1/2$, the neighborhood has positive probability. For $x\in[2,3]$, the same argument applies to $(2,3)$.

If $1<x<2$, choose $\varepsilon<\min(x-1,2-x)$. The interval centered at $x$ then misses both regions with positive integral, so its probability is zero. Points outside $[0,1]\cup[2,3]$ and away from $4$ also have a sufficiently small neighborhood with zero probability.

Although $f(4)=17>0$, the point $4$ does not belong to the support. For example, the neighborhood $(3.5,4.5)$ has probability $$ \int_{3.5}^{4.5} f(t)\,dt=0, $$ because a single point has Lebesgue measure zero and the density is zero elsewhere in this interval. The support is therefore $$ \operatorname{supp}(X)=[0,1]\cup[2,3]. $$

Formula Values, Formula Domains, and the Law

The examples show two common reasons a formula can mislead. First, its stated domain can include points where the probability is zero. For a pmf, only its values as probabilities at possible outcomes matter; for a density, the fact that a formula is written at a point does not place probability there. Second, pointwise density values do not uniquely determine a distribution. Changing a density at one point leaves all its integrals unchanged, so it leaves the law and support unchanged.

In particular, a density may be zero at a point that is in the support. In the preceding example, $f(0)=0$, but every neighborhood of $0$ includes a positive-length portion of $(0,1)$, and therefore has positive probability. Likewise, a density that is positive at an isolated point does not make that point part of the support.

Key takeaway: For a pmf, check whether each neighborhood contains a positive-mass value. For a density, integrate over each neighborhood. In both cases, support is a property of the probability law, not merely of the formula’s domain or pointwise values.

Common Pitfall

Wrong reasoning: “The density is positive at $x$, so $x$ is in the support,” or “the pmf formula is defined at $x$, so $x$ is in the support.”

Correction: Support requires positive probability in every neighborhood. For a density, a positive value at one point can contribute zero integral; for a pmf, a point with zero mass may still be in the support if positive-mass values occur arbitrarily close to it. Test the neighborhood probabilities directly.

Check Your Understanding

Use the neighborhood definition in each question.

A pmf assigns positive mass to every positive integer and zero mass elsewhere. What is its support?
Why can a point with zero point probability still belong to the support of a discrete law?
A density is positive only at one isolated point and zero everywhere else. Can it integrate to one? Does that point belong to the support?
For the density in the worked example, explain why $0$ is in the support although $f(0)=0$.
What probability calculation should you use to decide whether a point is in the support of a law with density $f$?