Why Support Boundaries Need Care
When a random variable is transformed, its possible values are determined by the values the original variable can take and by the function applied to them. The endpoint details matter: a function may approach a finite value without ever attaining it, and a random variable may have support at an endpoint even when the probability of equaling that endpoint is zero. These are different questions—what values are attained, and what values belong to the support.
The distinction is useful whenever a transformed variable is described by an interval. It prevents a common error: treating the image of the values actually attained as the support, without checking whether the image has limiting boundary points.
For a continuous random variable with density \(f_X\), a useful interval check is this: if \(f_X=0\) almost everywhere outside an interval and \(f_X\) is positive almost everywhere in the interval’s interior, then the support is the closure of that interval. The condition outside the interval is essential; positivity within an interval alone does not rule out density elsewhere. Density values at isolated points do not determine support, because changing a density at a single point does not change any probabilities.
Map the Support, Then Close the Image
Suppose \(Y=g(X)\), where \(g\) is continuous on the support of \(X\). The general rule is to apply \(g\) to the support of \(X\), then take the closure of the resulting set. Taking the closure includes finite boundary values that can be approached even if they are never attained.
If \(K\) is an interval and \(g\) is continuous and strictly monotone there, its image is an interval. For an increasing function, the lower endpoint comes from the lower end of \(K\), and the upper endpoint from the upper end. For a decreasing function, the order is reversed. At an infinite endpoint, use a limit. The resulting range may omit a limiting endpoint, but the support includes that endpoint when it is finite.
| Original support endpoint | What to calculate | How it affects transformed support |
|---|---|---|
| Finite endpoint included in \(K\) | Evaluate \(g\) at that endpoint | The endpoint value is attained and is in the image |
| Finite endpoint approached but not included in the values of \(X\) | Take the one-sided limit of \(g\) | A finite limiting value belongs to the support |
| \(+\infty\) or \(-\infty\) | Take the corresponding limit of \(g(x)\) | A finite limit is a support boundary; an infinite limit gives an unbounded boundary |
For interval supports, the endpoint procedure can be organized into four steps.
Use the distribution or a density condition, not just a list of values that happen to be attained.
Determine whether the function preserves or reverses the order of points on that support.
Use function values at included finite endpoints and limits at excluded or infinite endpoints.
The support is the closure of the range on the original support.
Worked Examples
Worked Example: A Decreasing Transformation on a Finite Interval
Let \(X\) have density \(f_X(x)=1/4\) for \(1<x<5\) and \(f_X(x)=0\) otherwise. Define \(Y=10-\frac{3}{2}X\). Find the range of \(Y\) on the values attained by \(X\), and then find the support of \(Y\).
Method. First identify the support of \(X\). The density is zero outside \((1,5)\) and positive throughout its interior, so the support is its closure: $$ \operatorname{supp}(X)=[1,5]. $$
The transformation \(g(x)=10-\frac{3}{2}x\) has slope \(-3/2<0\), so it is strictly decreasing. On the actual values \(1<X<5\), its limiting endpoint values are $$ g(1)=10-\frac{3}{2}=\frac{17}{2}=8.5, \qquad g(5)=10-\frac{15}{2}=\frac{5}{2}=2.5. $$ Because \(X\) does not equal 1 or 5, these two values are not attained. Thus the range is \((2.5,8.5)\).
The support is the closure of this range: $$ \operatorname{supp}(Y)=[2.5,8.5]. $$ The order of the endpoints is reversed because \(g\) is decreasing. The answer is checked by noting that \(g(x)\) lies between 2.5 and 8.5 for every \(x\) between 1 and 5, approaching both boundaries from within. The common mistake is to report \((2.5,8.5)\) as the support because neither boundary is attained; the correction is that support includes limit points.
Worked Example: An Exponential Variable With a Finite Lower Limit
Suppose \(X\sim\operatorname{Exponential}(\lambda=2)\), with density \(f_X(x)=2e^{-2x}\) for \(x\geq 0\). Let \(Y=e^{-X}\). Find the range of \(Y\) and its support, explaining the status of both boundaries.
The exponential density is positive for \(x\geq0\) and zero for \(x<0\), so \(\operatorname{supp}(X)=[0,\infty)\). The function \(g(x)=e^{-x}\) is strictly decreasing because \(g'(x)=-e^{-x}<0\). At the included endpoint, \(g(0)=1\). At the unbounded endpoint, $$ \lim_{x\to\infty}e^{-x}=0. $$ No finite \(x\) maps to zero, so the range is \((0,1]\). Its closure, and therefore the support of \(Y\), is \([0,1]\).
To check the limiting boundary directly, take \(0<\varepsilon<1\). Then $$ P(Y<\varepsilon) =P(e^{-X}<\varepsilon) =P(X>-\ln\varepsilon) =e^{-2(-\ln\varepsilon)} =\varepsilon^2>0. $$ Thus every neighborhood of 0 has positive probability, although \(P(Y=0)=0\). At the other boundary, $$ P(Y>1-\varepsilon) =P(X<-\ln(1-\varepsilon)) =1-(1-\varepsilon)^2>0. $$ So every neighborhood of 1 also has positive probability. These checks confirm the support \([0,1]\), not merely the attained range \((0,1]\). The rate parameter is \(2\); changing it to another positive rate would alter these probabilities but not this support.
The common mistake is to exclude 0 from the support because no finite \(X\) gives \(e^{-X}=0\). That reasoning identifies the range, not the support. The positive-probability neighborhood calculation shows why 0 belongs to the support.
Worked Example: A Normal Input and Two Limiting Boundaries
Let \(X\sim N(0,1)\), so its mean is \(0\) and its variance is \(1\). Define $$ Y=2+\frac{5}{1+e^{-X}}. $$ Find the range of \(Y\) and its support.
The standard normal density is positive on the entire real line, so \(\operatorname{supp}(X)=\mathbb{R}\). Set \(g(x)=2+5/(1+e^{-x})\). Its derivative is $$ g'(x)=\frac{5e^{-x}}{(1+e^{-x})^2}>0 $$ for every real \(x\), so \(g\) is strictly increasing. Its endpoint limits are $$ \lim_{x\to-\infty}g(x)=2, \qquad \lim_{x\to\infty}g(x)=7. $$ Neither limit is attained at a finite \(x\). Therefore the range is \((2,7)\), and the support is its closure: $$ \operatorname{supp}(Y)=[2,7]. $$
The boundaries are genuine support points: for any neighborhood of 2, sufficiently negative but finite values of \(X\) map into that neighborhood, and the standard normal assigns positive probability to such intervals. The same reasoning applies to sufficiently large finite \(X\) near 7. The answer is consistent with the formula: \(1/(1+e^{-x})\) lies strictly between 0 and 1 for finite \(x\), so \(Y\) lies strictly between 2 and 7 while approaching both limits. The common mistake is to confuse that open range with the closed support.
Why the Closure Rule Holds
The support rule follows from the definition of support and continuity. Write \(K=\operatorname{supp}(X)\). A real random variable lies in its support with probability 1. Consider first a point \(y\) in the closure of \(g(K)\). Every open neighborhood \(V\) of \(y\) contains some value \(g(x)\) with \(x\in K\). Because \(g\) is continuous, points sufficiently close to \(x\) have images in \(V\). Since \(x\) is in the support of \(X\), every such neighborhood of \(x\) has positive probability. Consequently, \(P(g(X)\in V)>0\). This holds for every neighborhood \(V\), so \(y\) is in the support of \(g(X)\).
Now let \(y\) be outside the closure of \(g(K)\). There is an open neighborhood \(V\) of \(y\) that does not intersect \(g(K)\). Since \(X\in K\) with probability 1, \(g(X)\in g(K)\) with probability 1, and therefore \(P(g(X)\in V)=0\). Thus \(y\) is not in the support of \(g(X)\). The two directions prove $$ \operatorname{supp}(g(X))=\overline{g(K)}. $$
Common Mistakes and Checks
- Stopping at the attained range. A boundary that is approached but never attained can still belong to the support. Take the closure of the image.
- Reversing endpoints incorrectly. A decreasing function reverses their order. Evaluate or take limits at both ends, then write the smaller transformed value first.
- Using only an interval where a density is positive. Positivity in that interval does not prove there is no probability outside it. For the interval-support conclusion, also establish that the density is zero almost everywhere outside.
- Using endpoint values when the endpoint is infinite. Evaluate the appropriate limit instead. A finite limit is a potential support boundary even if the function never attains it.
- Applying the two-endpoint shortcut without monotonicity. For a function that is not monotone, endpoint values alone may miss values reached inside the interval. The general rule is still to map the full support and take the closure.
Check Your Understanding
For each problem, find the support of the transformed variable. Distinguish the attained range from the support when they differ.
- Let \(U\) be uniform on \((-2,3)\), and set \(Y=4-U\). Find the support of \(Y\). (Answer: \([1,6]\))
- Let \(X\sim\operatorname{Exponential}(\lambda=3)\), with support \([0,\infty)\), and set \(W=1/(1+X)\). Give the range and support of \(W\). (Answer: Range \((0,1]\); support \([0,1]\))
- Suppose \(X\) has support \([2,\infty)\) and \(T=5+1/(X-1)\). Find the range and support of \(T\). (Answer: Range \((5,6]\); support \([5,6]\))
- Let \(X\sim N(-1,9)\), where the second parameter is the variance, and define \(Z=\arctan(3X+2)\). Find the support of \(Z\). (Answer: \([-\pi/2,\pi/2]\))
- Let \(X\) be uniform on \([0,6]\), and define \(R=(X-2)/(X+1)\). Find the support of \(R\), checking monotonicity before mapping the endpoints. (Answer: \([-2,4/7]\))