Why Range and Support Are Different
A random variable translates outcomes into numbers, but two different questions can be asked about those numbers. The range asks which values the random variable actually attains on the model’s sample space. The support asks which values are given probability mass nearby by the distribution. Those questions need not have the same answer.
The distinction matters especially when an individual value has probability zero. In a continuous model, an exact point can have probability zero even though values arbitrarily close to it have positive probability. Such a point can belong to the support. Conversely, a value may be attained at an outcome assigned probability zero, putting it in the range without putting it in the support.
The range depends on the particular sample space and on which outcomes are included in it. The support is determined by the distribution of \(X\). A value in the range is attainable in the model, but it need not have positive probability. A value in the support has positive probability in every neighborhood, but it need not itself be attained by any outcome.
It is useful to be precise with the phrase “possible value.” Here, range value means a value attained by at least one outcome, while support value means a value in the support. Some contexts use “possible” informally for either idea, so state which meaning is intended.
Finite Distributions and Positive-Probability Values
For a random variable with a finite range, the support consists exactly of the values that have positive point probability. If \(X\) can take only finitely many values, a value with probability zero is separated from all the positive-probability values by some positive distance. A sufficiently small interval around it therefore has probability zero, so it is not in the support.
This finite case is the setting in which “support” is often used as shorthand for “the values with positive probability.” For an infinite discrete distribution, however, a zero-probability point can still be in the support if positive-probability values accumulate arbitrarily close to it. The definition by neighborhoods handles both cases.
Worked Example: A Finite Model With a Zero-Probability Outcome
A hypothetical model has sample space \(\Omega=\{a,b,c,d\}\), with probabilities \(P(a)=0.20\), \(P(b)=0.30\), \(P(c)=0.50\), and \(P(d)=0\). A random variable records \(X(a)=0\), \(X(b)=1\), \(X(c)=1\), and \(X(d)=2\). Find its range, its support, and the point probabilities.
Method. List the values attained by the outcomes to find the range. Then group outcomes by their \(X\)-values and add their probabilities. The support in this finite model consists of the values with positive point probability.
First check that the outcome probabilities define a probability model: \(0.20+0.30+0.50+0=1.00\). The values attained are \(0\), \(1\), and \(2\), so \(\operatorname{Range}(X)=\{0,1,2\}\).
For \(X=0\), the only relevant outcome is \(a\), so \(P(X=0)=P(a)=0.20\). For \(X=1\), the outcomes are \(b\) and \(c\), so \(P(X=1)=P(b)+P(c)=0.30+0.50=0.80\). For \(X=2\), the only relevant outcome is \(d\), so \(P(X=2)=P(d)=0\).
The point probabilities sum to \(0.20+0.80+0=1.00\), as a check. Thus the support is \(\{0,1\}\), while the range is \(\{0,1,2\}\). The value \(2\) is attained at \(d\), but that outcome has probability zero. Common mistake avoided: treating every range value as a positive-probability value would incorrectly put \(2\) in the support.
Continuous Values and Zero-Probability Points
In a continuous model, single points commonly have probability zero. This does not mean that the point is outside the support. The support concerns probabilities of neighborhoods, not just probabilities at individual points. A point can have zero probability while every interval around it contains a range of values with positive probability.
For instance, if \(U\) is uniform on \([0,1]\), then its range and support are both \([0,1]\). Each exact value, including an endpoint, has probability zero. But any interval around a point in \([0,1]\) intersects \([0,1]\) in an interval of positive length, so it has positive probability. Points outside \([0,1]\) have a sufficiently small neighborhood with probability zero.
Worked Example: An Unattained Endpoint in a Uniform Model
Let \(U\) be uniform on the open interval \((0,1)\), and define \(X=1+2U\). Find the range and support of \(X\). Also find \(P(X=2.2)\) and \(P(1.5\leq X<2.5)\).
Method. Transform the values that \(U\) can attain to find the range. For the support, check whether every neighborhood of each candidate value intersects the transformed interval in a set of positive probability.
Because \(0<U<1\), multiplying by \(2\) gives \(0<2U<2\), and adding \(1\) gives \(1<X<3\). Therefore \(\operatorname{Range}(X)=(1,3)\). The endpoints \(1\) and \(3\) are not attained.
Every neighborhood of an interior point of \((1,3)\) contains an interval of attainable values. Every neighborhood of \(1\) also contains some interval \((1,1+\eta)\) with \(\eta>0\), and every neighborhood of \(3\) contains some interval \((3-\eta,3)\). Each such interval has positive probability. Thus the support is \([1,3]\), including two values outside the range.
The event \(X=2.2\) is equivalent to \(U=(2.2-1)/2=0.6\). A uniform random variable on an interval assigns probability zero to a single point, so \(P(X=2.2)=P(U=0.6)=0\).
For the interval probability, solve the inequalities: \(1.5\leq 1+2U<2.5\) gives \(0.5\leq 2U<1.5\), hence \(0.25\leq U<0.75\). The interval has length \(0.75-0.25=0.50\), so \(P(1.5\leq X<2.5)=0.50\). The calculated probability is between \(0\) and \(1\), as required. Common mistake avoided: excluding \(1\) and \(3\) from the support just because they are unattained, or assigning positive probability to the point \(2.2\).
A Short Justification of the Support Test
The neighborhood definition explains why support is not simply a list of values with positive point probability. Suppose \(X\) has a finite distribution. If \(P(X=x)>0\), then for every \(\varepsilon>0\), the event \(X=x\) is contained in \(x-\varepsilon<X<x+\varepsilon\). Consequently,
So \(x\) is in the support. If \(P(X=x)=0\), and the distribution has only finitely many positive-probability values, choose a small enough interval around \(x\) that contains none of those other values. That interval has probability zero, so \(x\) is not in the support. This proves that for a finite distribution the support is exactly the set of positive-probability values.
The finite assumption matters. In an infinite distribution, positive-probability values may approach a point without ever equalling it. Every neighborhood of that point can then have positive probability, even though the point itself has probability zero.
Worked Example: A Mixed Distribution With Endpoint Support
A hypothetical random variable \(X\) is generated as follows: with probability \(0.40\), set \(X=0\); with probability \(0.60\), draw \(U\) uniformly from \((0,1)\) and set \(X=2+U\). Find the range, support, \(P(X=2)\), and \(P(2.25<X<2.75)\).
Method. Treat the point outcome and the continuous outcome separately. Combine their attainable values for the range, then use neighborhood probabilities to identify the support.
The point component attains \(0\). In the continuous component, \(0<U<1\), so \(2<2+U<3\). Therefore the range is \(\{0\}\cup(2,3)\). The probabilities of choosing the two components add to \(0.40+0.60=1.00\).
The value \(0\) has point probability \(0.40\), so it is in the support. Every neighborhood of \(2\) contains values just greater than \(2\) from the continuous component, and every neighborhood of \(3\) contains values just less than \(3\). Such neighborhoods have positive probability. The interior of \((2,3)\) is also in the support. Hence the support is \(\{0\}\cup[2,3]\).
The continuous component never attains \(2\), and the point component only produces \(0\), so \(P(X=2)=0\). For the interval probability, \(2.25<X<2.75\) is equivalent in the continuous component to \(0.25<U<0.75\). This interval has length \(0.75-0.25=0.50\), so its conditional probability within that component is \(0.50\). Multiplying by the probability \(0.60\) of selecting the component gives \(P(2.25<X<2.75)=0.60(0.50)=0.30\).
As a check, the event is impossible in the point component, and \(0.30\leq 0.60\), the total probability of the continuous component. Common mistake avoided: treating the entire support as continuous or forgetting to multiply the interval probability by the probability of selecting the continuous component.
Exam Traps and a Practical Checklist
A frequent error is to equate “probability zero” with “impossible.” That conclusion is not valid for continuous distributions: an exact point can have probability zero while remaining in the support. Another error is to equate the range with the support. The finite example showed a range value excluded from the support; the uniform example showed support points not in the range.
A third trap appears with infinite discrete distributions. For \(P(X=1/n)=2^{-n}\), where \(n=1,2,\ldots\), the point \(0\) has probability zero but belongs to the support: every interval around \(0\) contains some \(1/n\) with positive probability. The support is \(\{1,1/2,1/3,\ldots\}\cup\{0\}\), even though the range does not contain \(0\).
List or calculate the values attained as the sample-space outcomes vary. Use the stated sample space, including any outcomes assigned probability zero.
For a discrete finite model, add the probabilities of outcomes that give each value. Do not infer positive probability merely from attainability.
Ask whether every open interval around a candidate value has positive probability. Include limit points when positive-probability values accumulate there.
When stating a result, report the range and support separately, and identify any important zero-probability values. This makes clear whether a value is attainable, supported by nearby probability, or both.
Check Your Understanding
For each problem, distinguish the range from the support and use point probabilities or neighborhood probabilities as appropriate.
- A model has outcomes \(a,b,c\) with probabilities \(0.25,0.75,0\), respectively, and \(X(a)=4\), \(X(b)=4\), \(X(c)=9\). Give the range, support, and \(P(X=4)\). (Answer: Range \(\{4,9\}\); support \(\{4\}\); \(P(X=4)=1\).)
- Let \(U\) be uniform on \([0,1]\) and \(Y=5U\). Give the range and support, and state \(P(Y=0)\) and \(P(Y=5)\). (Answer: Range and support are both \([0,5]\); both point probabilities are \(0\).)
- Let \(U\) be uniform on \((0,1)\) and \(W=3-2U\). Give the range and support, and state whether the endpoints have positive point probability. (Answer: Range \((1,3)\); support \([1,3]\); \(P(W=1)=P(W=3)=0\).)
- With probability \(0.30\), set \(X=-1\); otherwise, let \(X\) be uniform on \([2,4]\). Give the range and support, and calculate \(P(X=3)\) and \(P(2.5<X<3.5)\). (Answer: Range and support are both \(\{-1\}\cup[2,4]\); \(P(X=3)=0\); \(P(2.5<X<3.5)=0.70(1/2)=0.35\).)
- Let \(P(X=1/n)=2^{-n}\) for \(n=1,2,\ldots\). Give the range and support, and find \(P(X=0)\). (Answer: Range \(\{1,1/2,1/3,\ldots\}\); support \(\{1,1/2,1/3,\ldots\}\cup\{0\}\); \(P(X=0)=0\).)