Why Endpoint Mapping Matters
A monotone transformation changes not only the values of a random variable but also the interval on which its distribution can lie. The transformed interval’s endpoints may switch places under a decreasing function, may be excluded from the image, or may appear in the support as limiting values even when the transformation never attains them. These distinctions determine where a density formula applies and whether the resulting density accounts for all probability.
The key is to keep three objects distinct: the interval on which the transformation is applied, its image under the function, and the topological support of the transformed random variable. In particular, the support of the transformed variable need not equal the image of the original support.
Images of Intervals and Their Endpoints
For an increasing function, the lower endpoint of the image comes from the lower endpoint of the input interval, and the upper endpoint comes from the upper endpoint. For a decreasing function, those roles reverse. At an infinite input endpoint, a finite limiting value is not attained at that endpoint, so it is excluded from the image. A limit of \(+\infty\) or \(-\infty\) indicates an unbounded image endpoint, not a real endpoint to include.
| Transformation | Input boundary | Corresponding image boundary | When a finite boundary is included |
|---|---|---|---|
| Increasing | Lower endpoint | Lower endpoint | When the input lower endpoint belongs to \(I\) |
| Increasing | Upper endpoint | Upper endpoint | When the input upper endpoint belongs to \(I\) |
| Decreasing | Lower endpoint | Upper endpoint | When the input lower endpoint belongs to \(I\) |
| Decreasing | Upper endpoint | Lower endpoint | When the input upper endpoint belongs to \(I\) |
This endpoint rule follows from continuity, strict monotonicity, and the intermediate value theorem. Every value strictly between the endpoint limits is attained: choose input points on either side of that value’s preimage range, and continuity supplies an input point between them. A finite boundary value is attained only if the relevant input endpoint is itself included.
Worked Example: An Increasing Map With Open Endpoints
Suppose \(X\sim \operatorname{Uniform}(2,5)\) and \(Y=3X-1\). The values of \(X\) lie in \(I=(2,5)\), and \(g(x)=3x-1\) is increasing. Its endpoint limits are \(g(2)=5\) and \(g(5)=14\). Since neither input endpoint belongs to \(I\), the image of \(I\) is \((5,14)\).
The density of \(X\) is \(1/3\) on \((2,5)\). The inverse is \(g^{-1}(y)=(y+1)/3\), and the absolute derivative of the inverse is \(1/3\). Thus, for \(5<y<14\),
Set the density to zero outside \((5,14)\); its values at \(5\) and \(14\) can be chosen arbitrarily without changing the distribution. The integral is \(\int_5^{14}(1/9)\,dy=1\). Although the density is supported on this open interval in the sense that \(Y\) lies there with probability one, the topological support of \(Y\) is \([5,14]\): every neighborhood of either endpoint contains values with positive probability.
Support of a Transformed Random Variable
For a real-valued random variable \(X\), its topological support, written \(\operatorname{supp}(X)\), is the set of points \(x\) such that every open neighborhood of \(x\) has positive probability under the distribution of \(X\). It is a closed set, and \(P(X\in\operatorname{supp}(X))=1\). The support describes not just values attained with positive probability at individual points, but also values approached by sets having positive probability nearby.
The closure is essential in general. A continuous one-to-one function need not send a closed, noncompact set to a closed set. A homeomorphism between its domain and image does not change that fact: the image may be nonclosed in the ambient real line.
Proof. Write \(S=\operatorname{supp}(X)\) and \(T=\overline{g(S)}\). Since \(P(X\in S)=1\), we have \(P(g(X)\in g(S))=1\), and therefore \(P(g(X)\in T)=1\). If \(y\notin T\), some open neighborhood of \(y\) is disjoint from \(T\); hence \(g(X)\) cannot lie in that neighborhood except on an event of probability zero. Thus \(y\notin\operatorname{supp}(g(X))\).
Now take \(y\in T\). Every open neighborhood \(V\) of \(y\) intersects \(g(S)\), so there is an \(x\in S\) with \(g(x)\in V\). By continuity, \(g^{-1}(V)\) is an open set containing \(x\). Since \(x\in S\), \(P(X\in g^{-1}(V))>0\); consequently \(P(g(X)\in V)>0\). Every neighborhood of \(y\) has positive probability under the distribution of \(g(X)\), so \(y\in\operatorname{supp}(g(X))\). The two inclusions prove the result. \(\square\)
If \(S\) is compact, then its continuous image \(g(S)\) is compact and hence closed in \(\mathbb{R}\). In that case the theorem gives \(\operatorname{supp}(g(X))=g(S)\). Compactness is a sufficient condition for exact equality, not a requirement for the closure formula.
Worked Example: A Decreasing Transformation Reverses the Endpoints
Let \(X\sim\operatorname{Uniform}(1,4)\) and \(Y=1/X\). The function \(g(x)=1/x\) is strictly decreasing on \([1,4]\), so its image reverses the endpoints: \(g([1,4])=[1/4,1]\). The density of \(X\) is \(1/3\) on this interval. The inverse is \(x=1/y\), with absolute derivative \(1/y^2\). Therefore,
Take \(f_Y(y)=0\) outside \((1/4,1)\); values at the two endpoints do not affect probabilities. Normalization verifies the result:
The image of the closed input support is the closed interval \([1/4,1]\), and it is also the topological support of \(Y\). If instead the input interval were \((1,4)\), its image would be \((1/4,1)\), while the topological support would still include the limiting endpoints.
From the Image to a Valid Density
The one-dimensional density change-of-variables formula uses the actual image interval as its domain. A useful version is the following. Suppose \(X\) has density \(f_X\), \(P(X\in I)=1\) for an interval \(I\), and \(g\) is strictly monotone and continuously differentiable on the interior of \(I\), with \(g'(x)\ne 0\) there. Then \(Y=g(X)\) has density, for \(y\in g(I)\) away from any endpoints,
The density is zero off the image, up to choices on boundary points. To verify that this gives the transformed law, take any Borel set \(A\). Since \(Y=g(X)\) and \(X\) lies in \(I\) with probability one,
The second equality is the ordinary one-dimensional substitution formula, with the absolute derivative handling either direction of monotonicity. This identity verifies the probabilities of all Borel sets, not just intervals. In particular, taking \(A=\mathbb{R}\) shows that the proposed density integrates to one. Endpoints have Lebesgue measure zero, so assigning a different density value at an endpoint does not change the distribution.
The formula describes a density on \(g(I)\), but the topological support may be its closure. For a continuous transformation of a random variable with topological support \(S\), the precise support statement remains \(\overline{g(S)}\). Do not infer from an open endpoint of the image that the transformed distribution has no support at the corresponding limiting point.
Worked Example: A Finite Limiting Endpoint Not in the Image
Let \(X\) have the exponential density \(f_X(x)=e^{-x}\) for \(x\geq 0\), and set \(Y=e^{-X}\). The support of \(X\) is \(S=[0,\infty)\). The function \(g(x)=e^{-x}\) is continuous and strictly decreasing there. It maps \(0\) to \(1\), while \(g(x)\) tends to \(0\) as \(x\) tends to infinity without ever reaching \(0\). Thus \(g(S)=(0,1]\).
The inverse is \(g^{-1}(y)=-\ln y\), and the absolute derivative is \(1/y\). For \(0<y\leq1\), the transformed density is
Set the density to zero outside \((0,1)\), with any value at \(y=1\). Its integral is \(\int_0^1 1\,dy=1\). The image \(g(S)=(0,1]\) is not closed, so it cannot equal the topological support. The support theorem gives \(\operatorname{supp}(Y)=\overline{(0,1]}=[0,1]\). Every neighborhood of zero has positive probability, even though \(Y\) is never equal to zero.
Common Mistakes and a Reliable Check
A common error is to say, “A continuous one-to-one transformation maps the support exactly to the transformed support.” That reasoning overlooks the possibility that the image of a noncompact closed set is not closed. For example, \(e^{-x}\) maps \([0,\infty)\) onto \((0,1]\), while the support of \(e^{-X}\) for an exponential \(X\) is \([0,1]\). The correction is to use the closure formula for supports and to claim exact equality only when the image is closed, as it is when the original support is compact.
A second error is to reverse the endpoint values under a decreasing transformation without checking which input endpoints are included. The values reverse, but inclusion follows the input endpoint: an included lower input endpoint contributes an included upper image endpoint. Finally, a density need not assign a special value to an endpoint. What matters for the distribution is its integral over sets, and changing a density at finitely many points has no effect.
When checking a transformed density, first find the image interval and its endpoint status. Then find the inverse and its absolute derivative, write the formula only on the image, and verify that its integral is one. Separately determine the topological support by taking the closure of the image of the original support.
Check Your Understanding
Use endpoint limits and the support distinction to answer these questions.
- If \(g\) is increasing on \((a,b]\), which endpoint of \(g((a,b])\) is included, and why?
- If \(g\) is decreasing on \([a,b)\), which endpoint of the image comes from \(a\), and is that image endpoint included?
- Why can \(\operatorname{supp}(g(X))\) differ from \(g(\operatorname{supp}(X))\) even when \(g\) is a continuous one-to-one function?
- For an exponential \(X\) and \(Y=e^{-X}\), state the image of \(\operatorname{supp}(X)\) and the support of \(Y\). Explain the difference.
- Why can a density formula be specified only on the interior of the image interval without assigning a consequential value at its endpoints?