Reciprocal Transformations on One-Sided Supports

Why Reciprocals Need Care

The transformation \(g(x)=1/x\) is simple to write down, but its distribution depends on where \(X\) can lie. It is undefined at zero, and it is strictly decreasing on both \((0,\infty)\) and \((-\infty,0)\). Consequently, reciprocation reverses the order of endpoints on either side of zero. For a continuous random variable, the transformed density also includes a factor that accounts for how the transformation stretches or compresses intervals.

The earlier material on one-to-one transformations provides the general change-of-variables principle. Here we apply it to reciprocation on an interval entirely on one side of zero, keeping the interval’s image distinct from the distribution’s topological support.

Mapping the Interval

Definition: Let \(I\) be an interval contained in either \((0,\infty)\) or \((-\infty,0)\). Its reciprocal image is \(J=\{1/x:x\in I\}\). On either of these half-lines, \(x\mapsto 1/x\) is strictly decreasing, since its derivative is \(-1/x^2<0\). Thus interval endpoints map in reverse order.

For example, if \(0<a<b<\infty\), the reciprocal image of \((a,b)\) is \((1/b,1/a)\). If \(a<b<0\), the image is again \((1/b,1/a)\): the endpoint that was larger maps to the smaller reciprocal. Endpoint inclusion is preserved by taking the image of the actual interval, but endpoints that have probability zero do not affect the distribution of an absolutely continuous random variable.

The reciprocal stays on the same side of zero as its input. It does not, however, preserve endpoint order. The derivative also determines the density adjustment: the inverse transformation is \(x=1/y\), with derivative \(-1/y^2\), so the absolute derivative is \(1/y^2\).

Worked Example: Mapping a Positive Interval

Suppose \(X\) is uniform on \((1,3)\). Since \(1/3<1/x<1\) for \(1<x<3\), the transformed variable \(Y=1/X\) lies in \(J=(1/3,1)\). The left endpoint of the image comes from the right endpoint of the original interval.

The density of \(X\) is \(f_X(x)=1/2\) on \((1,3)\). At \(y\in(1/3,1)\), the inverse is \(x=1/y\), and the absolute derivative of the inverse is \(1/y^2\). Therefore \(f_Y(y)=f_X(1/y)/y^2=1/(2y^2)\) on that interval, and the density is zero outside it.

Check that this density integrates to one:

\int_{1/3}^{1}\frac{1}{2y^2}\,dy =\frac{1}{2}\left[-\frac{1}{y}\right]_{1/3}^{1} =\frac{1}{2}(-1+3)=1.

This normalization check confirms the factor and the mapped endpoints.

The Reciprocal Density

Theorem: Suppose \(X\) is absolutely continuous with density \(f_X\), and \(P(X\in I)=1\) for an interval \(I\) contained in either \((0,\infty)\) or \((-\infty,0)\). Define \(Y=1/X\) and \(J=\{1/x:x\in I\}\). Then \(Y\in J\) almost surely, and a density for \(Y\) is \(f_Y(y)=f_X(1/y)/y^2\) for \(y\in J\), and \(f_Y(y)=0\) for \(y\notin J\).

The statement that \(Y\in J\) almost surely describes a range containing \(Y\) with probability one. It does not assert that every point of \(J\) is in the topological support. The displayed expression is a density on \(J\); as with any density, its values at individual points can be changed without changing the distribution.

Proof. For any Borel set \(A\subseteq\mathbb{R}\), the definition \(Y=1/X\) and the assumption \(P(X\in I)=1\) give

P(Y\in A)=\int_I \mathbf{1}_{A}(1/x)f_X(x)\,dx.

On \(I\), the substitution \(y=1/x\) is one-to-one and continuously differentiable, with inverse \(x=1/y\) and \(\left|dx/dy\right|=1/y^2\). The one-dimensional change-of-variables formula therefore transforms the integral to

P(Y\in A)=\int_J \mathbf{1}_{A}(y)f_X(1/y)\frac{1}{y^2}\,dy.

This is exactly the probability assigned to \(A\) by the stated density, which is nonnegative and integrates to one because the formula gives the distribution of \(Y\). It is zero outside \(J\), since \(Y\in J\) almost surely. \(\square\)

Distribution Functions on Both Sides of Zero

The density formula treats positive and negative inputs in the same way because it uses a squared denominator. A distribution-function calculation makes the order reversal explicit. Let \(F_X\) be the distribution function of \(X\). For \(y\ne 0\), reciprocation reverses inequalities on either half-line, so \(1/X\le y\) is equivalent to \(X\ge 1/y\) when \(X\) and \(y\) are on the same side of zero. Since \(X\) is absolutely continuous, \(P(X=1/y)=0\), and \(P(X\ge 1/y)=1-F_X(1/y)\).

For positive \(X\), this yields the following piecewise expression, including the values outside the possible range:

F_Y(y)= \begin{cases} 0, & y\le 0,\\ 1-F_X(1/y), & y>0. \end{cases}

For negative \(X\), every value of \(Y\) is negative. The corresponding expression is

F_Y(y)= \begin{cases} 1-F_X(1/y), & y<0,\\ 1, & y\ge 0. \end{cases}

Differentiating the applicable expression at points where the density formula applies gives \(f_X(1/y)/y^2\). The positive factor \(1/y^2\), rather than the negative derivative \(-1/y^2\), is required because densities use the absolute value of the inverse derivative.

Worked Example: A Reciprocal on a Negative Interval

Let \(X\) be uniform on \((-4,-1)\). Its density is \(1/3\) there. The reciprocal image is \((-1,-1/4)\): specifically, \(1/(-1)=-1\) is the smaller image endpoint and \(1/(-4)=-1/4\) is the larger one. Applying the density formula gives

f_Y(y)= \begin{cases} \dfrac{1}{3y^2}, & -1<y<-1/4,\\ 0, & \text{otherwise}. \end{cases}

To check normalization, integrate with the stated endpoint order:

\int_{-1}^{-1/4}\frac{1}{3y^2}\,dy =\frac{1}{3}\left[-\frac{1}{y}\right]_{-1}^{-1/4} =\frac{1}{3}\bigl(4-1\bigr)=1.

For an additional check, take \(-1<y<-1/4\). The uniform distribution function of \(X\) on its interval is \(F_X(x)=(x+4)/3\), so

F_Y(y)=1-F_X(1/y) =1-\frac{1/y+4}{3} =\frac{-1/y-1}{3}.

At \(y=-1/2\), this gives \(F_Y(-1/2)=1/3\). Indeed, \(1/X\le-1/2\) is equivalent here to \(X\ge-2\), and the interval \([-2,-1)\) occupies one third of the length of \((-4,-1)\).

Support Is Not Always the Reciprocal Image of an Assumed Interval

The theorem assumes that \(X\) lies in \(I\) almost surely. That does not imply that \(X\) has positive probability arbitrarily close to every point of \(I\). Accordingly, the reciprocal image \(J\) is a valid almost-sure range and the density formula applies there, but \(J\) need not be the topological support.

The topological support records which points have positive probability in every open neighborhood. When the support of \(X\) is considered on a positive or negative half-line, reciprocation maps that support by the continuous one-to-one map \(x\mapsto1/x\). If the support reaches zero as a boundary point, its reciprocal image is taken over nonzero support points and then closed in \(\mathbb{R}\). In particular, the support may include a boundary point not attained by \(Y\).

Worked Example: An Exponential Input

Let \(X\) have an exponential distribution with rate \(2\), so \(f_X(x)=2e^{-2x}\) for \(x>0\), and let \(Y=1/X\). Choose \(I=(0,\infty)\), for which \(J=(0,\infty)\). The transformed density is

f_Y(y)= \begin{cases} \dfrac{2}{y^2}e^{-2/y}, & y>0,\\ 0, & y\le 0. \end{cases}

To check its integral, use \(x=1/y\), for which \(dy=-dx/x^2\) and \(y^{-2}=x^2\). As \(y\) increases from \(0\) to \(\infty\), \(x\) decreases from \(\infty\) to \(0\). Thus

\int_0^\infty \frac{2}{y^2}e^{-2/y}\,dy =\int_\infty^0 2x^2 e^{-2x}\left(-\frac{1}{x^2}\right)\,dx =\int_0^\infty 2e^{-2x}\,dx =1.

The distribution function is \(F_Y(y)=e^{-2/y}\) for \(y>0\), since \(1-F_X(1/y)=e^{-2/y}\). The random variable \(Y\) is positive almost surely and has density zero at and below zero. Nevertheless, its topological support is \([0,\infty)\): every neighborhood of zero contains positive values of \(Y\) with positive probability, because \(X\) can be arbitrarily large. Thus \(J=(0,\infty)\) is not the topological support.

Common Mistakes and Practical Checks

Reciprocal transformations are especially easy to check by tracking the input interval before manipulating a density. Two errors account for many incorrect answers:

Keeping the endpoints in their original order. For \(1<X<3\), it is incorrect to claim \(1<1/X<3\). Because reciprocation is decreasing, the correct range is \(1/3<1/X<1\). The same reversal applies to negative intervals.
Using a negative density factor. The derivative of \(1/y\) is \(-1/y^2\), but a density uses the absolute derivative, \(1/y^2\). A negative candidate density cannot represent probabilities.
Calling the image interval the support automatically. A full-probability interval need not be the set where the distribution has mass arbitrarily close to every point. State the almost-sure range and the topological support separately when they differ.

A reliable check is to identify the side of zero, map the interval by testing its endpoints, apply the absolute inverse derivative, and verify that the resulting density integrates to one. For a distribution function, also check its boundary limits: it must increase from zero to one as \(y\) ranges over the real line.

Key takeaway: If \(X\) has density \(f_X\) and lies almost surely in an interval \(I\) that excludes zero on one side, then \(Y=1/X\) lies almost surely in \(J=\{1/x:x\in I\}\) and has density \(f_X(1/y)/y^2\) on \(J\). The reciprocal reverses interval order; the image range and topological support must not be conflated.

Check Your Understanding

For each question, track the interval and the inverse derivative before writing a density.

If \(X\) is uniform on \((2,5)\), what is the range of \(1/X\), and what is its density on that range?
If \(X\) is uniform on \((-6,-2)\), which endpoint of its interval maps to the smaller reciprocal? Give the reciprocal image.
Why does the reciprocal density contain \(1/y^2\) rather than \(-1/y^2\)?
For a positive exponential random variable, explain why the reciprocal can have topological support including zero even though it is always strictly positive.
What does the assumption \(P(X\in I)=1\) establish about \(Y=1/X\), and what does it not establish about the topological support?