When a Derivative or Jacobian Vanishes

Why a Vanishing Derivative Needs Care

For a smooth one-to-one transformation, the density formula divides by the absolute derivative, or by the absolute Jacobian determinant. If that quantity is zero at an isolated point, the formula cannot be evaluated there. Nearby, its reciprocal may also become large. The right conclusion is not that the transformation has failed: the ordinary formula can still describe the transformed distribution almost everywhere, while the density may be unbounded near the exceptional value.

The distinction between an exceptional point and a whole region matters. A continuous random variable has probability zero of taking any particular value, so an isolated point where the derivative vanishes contributes no probability by itself. But a small neighborhood of that point can be stretched or compressed enough to produce a large density nearby. The resulting density must still integrate to one.

One-Dimensional Formula Away From Critical Points

Definition: Let $I$ be an open interval and let $g:I\to g(I)$ be strictly monotone and continuously differentiable. A point $x_0\in I$ is a critical point here if $g'(x_0)=0$; its image $g(x_0)$ is a critical value. We consider transformations for which the critical points form an isolated set.

Write $Y=g(X)$, where $X$ has density $f_X$ and $P(X\in I)=1$. At points $y$ whose inverse $x=g^{-1}(y)$ satisfies $g'(x)\ne 0$, the usual one-to-one transformation formula is

f_Y(y)=\frac{f_X(g^{-1}(y))}{|g'(g^{-1}(y))|}.

This formula applies on the image of the regular points, not necessarily at every point of $g(I)$. At a critical value the expression may be undefined, and a density is only determined up to changes on sets of Lebesgue measure zero. It is therefore appropriate to specify the formula almost everywhere and to treat the critical values separately.

Theorem: Suppose $g:I\to g(I)$ is strictly monotone and continuously differentiable, its zero-derivative set $C=\{x\in I:g'(x)=0\}$ consists of isolated points, and $X$ has density $f_X$ with $P(X\in I)=1$. Then $Y=g(X)$ has a density, given for almost every $y\in g(I)\setminus g(C)$ by $$ f_Y(y)=\frac{f_X(g^{-1}(y))}{|g'(g^{-1}(y))|}. $$ The density can be assigned arbitrary values at the critical values $g(C)$. In particular, $Y$ has no atoms.

Proof. An isolated subset of an interval is countable, so $C$ and $g(C)$ are countable and have Lebesgue measure zero. Also, $P(X\in C)=0$, since $X$ has a density. The open set $I\setminus C$ is a countable union of disjoint open intervals. On each such interval $J$, the derivative is continuous and nonzero, so it has constant sign; the inverse function theorem makes $g$ a differentiable one-to-one change of variable there. For any Borel set $A\subseteq g(I)$, the one-dimensional change-of-variables formula on each $J$ gives

P(Y\in A) =\int_{g^{-1}(A)} f_X(x)\,dx =\sum_J \int_{A\cap g(J)} \frac{f_X(g^{-1}(y))}{|g'(g^{-1}(y))|}\,dy.

The omitted set $C$ has probability zero. Since $g$ is injective, the intervals $g(J)$ do not overlap, and the countable sum is the integral of the displayed density over $A$, apart from the null set $g(C)$. Taking $A=g(I)$ shows that the density integrates to one. Finally, for any $y$, injectivity makes $g^{-1}(\{y\})$ either empty or a singleton; $X$ assigns probability zero to either possibility. Thus $P(Y=y)=0$. $\square$

Worked Example: A Cubic Map With an Interior Critical Value

Let $X\sim\operatorname{Uniform}(-1,1)$, so $f_X(x)=1/2$ on $(-1,1)$, and set $Y=X^3$. The map $g(x)=x^3$ is strictly increasing, with inverse $g^{-1}(y)=y^{1/3}$. Its derivative $g'(x)=3x^2$ vanishes only at $x=0$, and the critical value is $y=0$.

For $y\in(-1,0)\cup(0,1)$, substitute the inverse and its derivative into the formula:

f_Y(y) =\frac{1/2}{3(y^{1/3})^2} =\frac{1}{6|y|^{2/3}}.

The density tends to infinity as $y$ approaches zero. This does not mean there is positive probability at zero: $Y=0$ exactly when $X=0$, an event of probability zero. The singularity is integrable, as a direct check confirms:

\int_{-1}^{1}\frac{1}{6|y|^{2/3}}\,dy =2\cdot\frac{1}{6}\int_0^1 y^{-2/3}\,dy =\frac{1}{3}\cdot 3 =1.

The density may be assigned any finite value at $y=0$ without changing the distribution. It is the formula away from zero, rather than its value at that single point, that determines the law.

How the Input Density Affects the Singularity

A zero derivative does not force every transformed density to diverge. The numerator $f_X(g^{-1}(y))$ matters too: if the input density is small near the critical point, it can offset the small derivative. The next example keeps the same map but changes the input distribution.

Worked Example: A Vanishing Input Density Cancels the Jacobian

Let $X$ have density $f_X(x)=\frac{3}{2}x^2$ on $(-1,1)$. This is a probability density because

\int_{-1}^{1}\frac{3}{2}x^2\,dx =\frac{3}{2}\left(\frac{2}{3}\right)=1.

Again take $Y=X^3$. For $y\ne0$, the inverse is $x=y^{1/3}$, and

f_Y(y) =\frac{(3/2)(y^{1/3})^2}{3(y^{1/3})^2} =\frac{1}{2},\qquad -1<y<1,\quad y\ne0.

Thus $Y$ is uniform on $(-1,1)$. Although the change-of-variables expression at zero initially has the indeterminate form $0/0$, the density is $1/2$ on either side. Assigning that value at zero gives a convenient version of the density. The transformation itself still has zero derivative there; it is the input density that cancels the potential blow-up.

When $f_X(x_0)>0$, a derivative that approaches zero often does produce an unbounded transformed density near $g(x_0)$. The cubic example illustrates the mechanism: a change in $y$ corresponds to a comparatively larger change in $x$ near zero, so probability from a wider input interval is concentrated in a narrow output interval. The precise rate depends on how $g'$ approaches zero and on how $f_X$ behaves; it is not determined by the fact of vanishing alone.

Multivariate Jacobians With Isolated Zeros

The same issue occurs for a one-to-one transformation of random vectors. Let $\Omega\subset\mathbb{R}^d$ be open and $T:\Omega\to T(\Omega)$ be injective and continuously differentiable. Suppose $\det DT(x)\ne0$ except at isolated points of $\Omega$, and let $X$ have density $f_X$ with $P(X\in\Omega)=1$. Away from the critical points, the inverse function theorem and the multivariate change-of-variables formula give

f_Y(y)= \frac{f_X(T^{-1}(y))} {\left|\det DT(T^{-1}(y))\right|}, \qquad Y=T(X).

This is an almost-everywhere formula. To see why the exceptional points do not invalidate it, the set $C=\{x:\det DT(x)=0\}$ is countable: every isolated point can be assigned a distinct element of a countable base of open balls in $\mathbb{R}^d$. Hence $P(X\in C)=0$, and injectivity makes $T(C)$ countable as well. On $\Omega\setminus C$, the inverse function theorem gives a local differentiable inverse at every point. The restriction of $T$ to this open set is injective, so these local inverses agree wherever their image neighborhoods overlap and together define a differentiable inverse on $T(\Omega\setminus C)$. Applying the multivariate change-of-variables formula there gives the density above. The omitted critical set has probability zero, and its image has Lebesgue measure zero. As in one dimension, the density’s behavior near a critical value can nevertheless be singular.

Worked Example: A Radial Map With an Isolated Jacobian Zero

Let $X$ be uniform on the open unit disk in $\mathbb{R}^2$, so $f_X(x)=1/\pi$ inside the disk. Define $T(x)=\|x\|^2x$. In polar coordinates, this map preserves direction and sends radius $r$ to radius $s=r^3$. It is one-to-one from the unit disk onto itself.

Writing $x$ as a column vector, its derivative matrix is $DT(x)=\|x\|^2 I+2xx^{\mathsf T}$. For $x\ne0$, vectors perpendicular to $x$ have eigenvalue $r^2$, while the direction of $x$ has eigenvalue $3r^2$. Therefore $\det DT(x)=3r^4$, which vanishes only at the origin. For $y=T(x)\ne0$, $\|y\|=r^3$, so $r=\|y\|^{1/3}$. The transformed density is

f_Y(y) =\frac{1/\pi}{3r^4} =\frac{1}{3\pi\|y\|^{4/3}}, \qquad 0<\|y\|<1.

This density is unbounded near the origin, but it integrates to one. Using polar coordinates in the output disk,

\int_{\|y\|<1} f_Y(y)\,dy =\int_0^{2\pi}\int_0^1 \frac{1}{3\pi s^{4/3}}s\,ds\,d\theta =\frac{2}{3}\int_0^1 s^{-1/3}\,ds =1.

The origin is not an atom: $T(X)=0$ only when $X=0$, which has probability zero. The density’s value at that one point is immaterial.

A Boundary Critical Point

Worked Example: A Critical Point at the Edge of the Support

Let $X$ be uniform on $(0,1)$ and set $Y=X^3$. The map is strictly increasing, and its derivative vanishes at zero, the boundary of the input support. For $0<y<1$, the inverse formula gives

f_Y(y)=\frac{1}{3y^{2/3}}.

The density becomes unbounded as $y$ approaches the boundary value zero, yet

\int_0^1 \frac{1}{3y^{2/3}}\,dy=1.

There is no point mass at the boundary, since $P(X=0)=0$. A boundary singularity in a density is therefore compatible with a continuous distribution supported on $(0,1)$.

Common Mistake and Practical Check

A common mistake is to substitute a critical value directly into the reciprocal-derivative formula, obtain division by zero, and conclude that the transformed distribution has an atom or that no density exists. Neither conclusion follows. The formula is established off the exceptional set; a density’s value on a null set is arbitrary. To decide whether the transformed law has a density and whether it is integrable, use the change-of-variables argument on the regular region and account for the probability of the critical set.

For an absolutely continuous input and an injective transformation, an isolated critical point has input probability zero and cannot generate an atom. A reciprocal Jacobian may still create an unbounded density nearby. Its integrability is guaranteed by the change-of-variables identity when the stated hypotheses hold; the explicit integral is a useful check in examples. Do not infer the size or rate of a singularity solely from the statement that a derivative or Jacobian vanishes.

Key takeaway: If an injective transformation is regular except at isolated points, apply the usual density formula wherever the derivative or Jacobian is nonzero. Interpret it almost everywhere: the density may diverge near a critical value, but an isolated critical point does not itself create an atom for an absolutely continuous input.

Check Your Understanding

Use the definitions and change-of-variables arguments above to answer these questions.

For $Y=X^3$ with $X\sim\operatorname{Uniform}(-1,1)$, identify the critical point and value, and explain why the density is unbounded at the value but $P(Y=0)=0$.
Why is it legitimate to assign a finite value to a density at an isolated critical value?
For $g(x)=x^3$ and $f_X(x)=\frac{3}{2}x^2$ on $(-1,1)$, show how the numerator and derivative cancel in the density formula.
In the radial map $T(x)=\|x\|^2x$ on the unit disk, where does the Jacobian vanish, and why is the resulting density still integrable?
What additional information, beyond knowing that a derivative vanishes, is needed to determine the rate at which a transformed density grows near a critical value?