Power Transformations on Positive Supports

Why the Sign of the Power Matters

If \(X\) is positive and \(r\ne 0\), then \(X^r\) is also positive. But the direction in which values are mapped depends on the sign of \(r\): for \(r>0\), the map \(x\mapsto x^r\) is increasing; for \(r<0\), it is decreasing. That difference changes the event used to calculate a CDF, and it determines the order in which support endpoints are mapped.

The same care applies to \(X^{1/r}\). These transformations are bijections on \((0,\infty)\), so their distributions can be derived directly from the CDF and support of \(X\). When \(X\) has a density, the one-dimensional change-of-variables formula also gives the transformed density. The formulas below distinguish distributional statements, which allow atoms, from density statements, which require absolute continuity.

Definitions and CDF Formulas

For an increasing transformation, the event \(X^r\le y\) becomes an event of the form \(X\le y^{1/r}\). For a decreasing transformation, the inequality reverses and becomes \(X\ge y^{1/r}\). The use of \(F_X(y^{1/r}-)\) in the latter case preserves any probability mass at the threshold.

To justify the formula, when \(r>0\), raising positive values to \(r\) preserves their order, so \(\{X^r\le y\}=\{X\le y^{1/r}\}\). When \(r<0\), the order is reversed, so \(\{X^r\le y\}=\{X\ge y^{1/r}\}\). The complement of \(\{X\ge a\}\) is \(\{X<a\}\), whose probability is \(F_X(a-)\). Finally, \(X^r>0\) almost surely, which gives the formula for \(y\le 0\). This proves the claim. \(\square\)

Now let \(W=X^{1/r}\). Its exponent \(1/r\) has the same sign as \(r\), so the monotonicity cases are unchanged. Solving \(x^{1/r}\le w\) for \(x\) gives \(x\le w^r\) if \(r>0\), and \(x\ge w^r\) if \(r<0\).

How the Support Transforms

We use support relative to the positive real line: a point \(x\in(0,\infty)\) belongs to the support of \(X\) if every open neighborhood of \(x\) relative to \((0,\infty)\) has positive probability under \(X\). In particular, an endpoint such as \(1\) can belong to the support even if \(P(X=1)=0\). For example, a Uniform\((1,3)\) variable has support \([1,3]\) under this definition.

Here is a proof for \(Y=X^r\). The map \(g(x)=x^r\) is a continuous bijection from \((0,\infty)\) to itself, and its inverse \(g^{-1}(y)=y^{1/r}\) is also continuous. Let \(x\) belong to the support of \(X\), and let \(V\) be any relative open neighborhood of \(g(x)\). By continuity of \(g\), \(g^{-1}(V)\) is a relative open neighborhood of \(x\). Thus \(P(Y\in V)=P(X\in g^{-1}(V))>0\), so \(g(x)\) is in the support of \(Y\).

Conversely, if \(x\) is not in the support of \(X\), there is a relative open neighborhood \(U\) of \(x\) with \(P(X\in U)=0\). The image \(g(U)\) is open because \(g^{-1}\) is continuous, and it is a neighborhood of \(g(x)\). Also \(P(Y\in g(U))=P(X\in U)=0\), so \(g(x)\) is not in the support of \(Y\). The two supports are therefore mapped onto each other. Replacing \(r\) by \(1/r\) proves the statement for \(X^{1/r}\). \(\square\)

For a support interval, map both endpoints and then put them in increasing order. An increasing power preserves endpoint order; a decreasing power reverses it. The support endpoints need not carry probability mass. Support describes where probability occurs arbitrarily nearby, not whether a point itself is an atom.

Density Formula

Suppose \(X\) is absolutely continuous on \((0,\infty)\), with density \(f_X\), and set \(Y=X^r\). The inverse transformation is \(x=y^{1/r}\), whose derivative has absolute value \(\frac{1}{|r|}y^{1/r-1}\). Therefore the transformed density is given almost everywhere on \((0,\infty)\) by the following formula.

For a complete derivation, take any Borel set \(A\subseteq(0,\infty)\). Since \(X\) has density \(f_X\),

Make the one-dimensional change of variable \(x=y^{1/r}\). Whether this substitution is increasing or decreasing, the absolute differential is \(dx=\frac{1}{|r|}y^{1/r-1}\,dy\). Hence

This integral identity for every Borel set \(A\) proves that the displayed function is a density version of \(Y\). It also shows why the absolute value \(|r|\) is necessary: a density must be nonnegative for both increasing and decreasing transformations. \(\square\)

A density is defined only up to changes on sets of measure zero. Consequently, the formula need not have a pointwise value of zero at every point outside the transformed support if an arbitrary representative of \(f_X\) is used. The transformed density is zero almost everywhere outside that support. If desired, choose a version that vanishes outside the support and use the corresponding transformed version.

For \(W=X^{1/r}\), apply the same formula with exponent \(1/r\). If \(X\) has density \(f_X\), a density version is

again as an almost-everywhere statement. The factor follows because the inverse map is \(x=w^r\), with derivative \(r w^{r-1}\).

Worked Examples

Common Mistake: Reversing the Map but Not the Support

A common error is to notice that a negative power is decreasing but then report the transformed interval endpoints in their original order. For instance, saying that \(x^{-2}\) maps the support \([1,3]\) to \([1,1/9]\) lists the endpoint images but does not express the support as an interval in increasing order. The correct support is \([1/9,1]\).

A second error is to call the support the interval where the density formula is positive. In the square-root example, the density is positive on \((1,3)\), but the support is \([1,3]\): every relative neighborhood of either endpoint has positive probability. Conversely, a point where an arbitrarily chosen density version is nonzero need not belong to the support. The distribution, not isolated pointwise values of a density, determines support.