Common Jacobian and Support Errors

Why Transformation Errors Happen

A change-of-variables calculation can look algebraically simple and still give an invalid density. Three checks catch many of the most common mistakes: use the determinant associated with the inverse map in the density formula, include every valid inverse branch, and restrict the density to the image of the original support. These checks are especially important when a transformation is many-to-one or when its image has a boundary that is not obvious from the formula alone.

The earlier material on Jacobian determinants and many-to-one transformations provides the general change-of-variables framework. Here the emphasis is on diagnosing errors in short calculations. A density is not just an algebraic expression: it must be evaluated on the right set, with the right contributions, and must integrate to one.

The Three Checks

Definition: Suppose \(X\) has density \(f_X\) on a support \(S_X\), and \(Y=T(X)\). The transformed support is the image \(T(S_X)=\{T(x):x\in S_X\}\), up to boundary sets of probability zero. For a one-to-one differentiable transformation with differentiable inverse \(x=T^{-1}(y)\), the transformed density is \(f_Y(y)=f_X(T^{-1}(y))\left|\det D T^{-1}(y)\right|\) on \(T(S_X)\), and is zero outside it. The determinant is that of the inverse map.

The inverse determinant is not an arbitrary convention. A small volume around \(y\) corresponds, under the inverse map, to a volume around \(x\) scaled by \(\left|\det D T^{-1}(y)\right|\). Equivalently, if \(y=T(x)\), then \(\left|\det D T^{-1}(y)\right|=1/\left|\det D T(x)\right|\), provided the determinant is nonzero. Taking an absolute value makes the density nonnegative even when the transformation reverses orientation.

For a many-to-one map, each possible preimage contributes. In one dimension, a convenient form is to divide the original support into intervals on which \(T\) is one-to-one. For a value \(y\), the formula sums contributions from the inverse roots that fall in those intervals and in the original support.

Definition: If \(T\) is differentiable and one-to-one on each of a finite collection of disjoint intervals covering the original support apart from probability-zero points, and \(T'(x)\ne 0\) on those intervals, then for almost every \(y\), \(f_Y(y)=\sum_{x:T(x)=y,\ x\in S_X} f_X(x)/|T'(x)|\). A root outside \(S_X\) contributes nothing; a value outside \(T(S_X)\) has no valid root and density zero.

Worked Examples

Worked Example: An Affine Map and the Inverse Determinant

Let \(X=(X_1,X_2)^\mathsf{T}\) be uniform on the rectangle \(0<x_1<1,\ 0<x_2<2\). Its density is \(1/2\) on that rectangle, whose area is \(2\). Set \(Y=AX\), where \(A=\begin{pmatrix}1&2\\3&4\end{pmatrix}\). Find the density of \(Y\) and identify what goes wrong if the forward determinant is used as a multiplier.

The determinant is \(\det A=1(4)-2(3)=-2\), so \(|\det A|=2\). The inverse matrix and inverse map are

A^{-1}=\begin{pmatrix}-2&1\\[2pt]3/2&-1/2\end{pmatrix}, \qquad x_1=-2y_1+y_2,\quad x_2=\frac32y_1-\frac12y_2.

The inverse determinant has absolute value \(1/2\). Therefore the density equals \((1/2)(1/2)=1/4\) wherever the inverse coordinates satisfy the original support restrictions:

-2y_1+y_2\in(0,1), \qquad \frac32y_1-\frac12y_2\in(0,2).

These inequalities describe the parallelogram that is the image of the original rectangle. Its area is \(|\det A|\) times the original area, namely \(2\cdot2=4\), so the density integrates to \((1/4)\cdot4=1\). If we mistakenly multiplied by \(|\det A|=2\), the proposed density would be \(1\) on this same parallelogram and would integrate to \(4\), not \(1\). The forward determinant is not the density multiplier for the inverse substitution.

Worked Example: A Squaring Map with a Missing Inverse Branch

Let \(X\) be uniform on \((-1,2)\), so \(f_X(x)=1/3\) there, and define \(Y=X^2\). Find the density of \(Y\), taking care to include only roots in the original support.

For \(y>0\), the equation \(x^2=y\) has roots \(x=\sqrt y\) and \(x=-\sqrt y\), and \(|T'(x)|=|2x|=2\sqrt y\) at either root. The positive root lies in \((-1,2)\) precisely when \(0<y<4\). The negative root lies in that support only when \(0<y<1\). Thus both roots contribute below \(1\), but only the positive root contributes between \(1\) and \(4\):

f_Y(y)= \begin{cases} \dfrac{1}{3\sqrt y},&0<y<1,\\[6pt] \dfrac{1}{6\sqrt y},&1<y<4,\\[6pt] 0,&\text{otherwise}. \end{cases}

The total probability confirms the branch accounting:

\int_0^1\frac{1}{3\sqrt y}\,dy+ \int_1^4\frac{1}{6\sqrt y}\,dy =\frac23+\frac{2-1}{3}=1.

A common incorrect answer uses both roots for every \(0<y<4\). That assigns a contribution from \(-\sqrt y\) even when \(-\sqrt y\le -1\), outside the support of \(X\). It produces too much density on \(1<y<4\). The point \(x=0\), where the derivative vanishes, has probability zero under this continuous distribution; the density formula is understood almost everywhere, so this single point does not create an additional probability mass.

Worked Example: A Curved Image with a Restrictive Support

Let \(X\) and \(Z\) be independent uniform random variables on \((0,1)\), and define \(U=X,\ V=XZ\). Find the joint density of \((U,V)\). The letter \(Z\) is used here for the second original variable so that it is not confused with the transformed coordinate \(V\).

The inverse transformation is \(x=u,\ z=v/u\), which is defined for \(u>0\). Its Jacobian matrix and absolute determinant are

\frac{\partial(x,z)}{\partial(u,v)} = \begin{pmatrix} 1&0\\[2pt] -v/u^2&1/u \end{pmatrix}, \qquad \left|\det\frac{\partial(x,z)}{\partial(u,v)}\right|=\frac1u.

The original support imposes both \(0<x<1\) and \(0<z<1\). Substituting the inverse gives \(0<u<1\) and \(0<v/u<1\); since \(u>0\), the latter is \(0<v<u\). The original joint density is \(1\) on the unit square, so

f_{U,V}(u,v)= \begin{cases} 1/u,&0<v<u<1,\\ 0,&\text{otherwise}. \end{cases}

The support restriction is essential. Direct integration gives \(\int_0^1\int_0^u (1/u)\,dv\,du=\int_0^1 1\,du=1\). If one incorrectly allowed \(0<v<1\) for every \(0<u<1\), the integral would be \(\int_0^1\int_0^1 (1/u)\,dv\,du=\int_0^1 du/u\), which diverges. The divergent integral is a clear signal that the proposed support is wrong.

Why the Many-to-One Formula Works

The root-sum rule can be justified by testing it against probabilities of sets. The assumptions matter: the original density is supported on a set that can be partitioned into finitely many intervals where the transformation is continuously differentiable, one-to-one, and has nonzero derivative. Endpoints and any excluded critical points must have probability zero. Under these conditions, ordinary one-dimensional change of variables applies on each interval.

Theorem: Under those conditions, if \(Y=T(X)\), then for every measurable set \(B\), \(P(Y\in B)=\int_B \sum_{x:T(x)=y,\ x\in S_X} f_X(x)/|T'(x)|\,dy\), where the sum includes only preimages in the original support. Consequently the sum is a density of \(Y\), up to changes on sets of Lebesgue measure zero.

Proof. Write the support, apart from probability-zero endpoints, as disjoint intervals \(I_1,\ldots,I_m\) on which \(T\) is one-to-one. Let \(J_j=T(I_j)\), and let \(x_j(y)\) be the inverse on \(J_j\). The events \(X\in I_j\) are disjoint, so

P(Y\in B) =\sum_{j=1}^m\int_{I_j}\mathbf{1}_{\{T(x)\in B\}}f_X(x)\,dx.

On \(I_j\), change variables \(y=T(x)\). The inverse derivative has absolute value \(1/|T'(x_j(y))|\). Thus the \(j\)th integral is

\int_{B\cap J_j} \frac{f_X(x_j(y))}{|T'(x_j(y))|}\,dy.

Summing these integrals adds one term for each interval whose image contains \(y\), equivalently one term for each preimage \(x\) of \(y\) in the original support. This gives the stated identity for every measurable \(B\), which is the defining integral property of a density. \(\square\)

Normalization as a Diagnostic

A correct transformed density must integrate to one. This is not a substitute for getting the support and branches right, but it is a powerful check: a result integrating to a value other than one is definitely wrong. Under the theorem's assumptions, normalization follows directly from the same branchwise substitution.

Derivation. Apply the probability identity just proved to the entire real line \(B=\mathbb{R}\). Since \(Y\) is a random variable, \(P(Y\in\mathbb{R})=1\). Therefore

\int_{\mathbb{R}} \sum_{x:T(x)=y,\ x\in S_X} \frac{f_X(x)}{|T'(x)|}\,dy =P(Y\in\mathbb{R})=1.

This argument also explains why omitting a valid branch loses probability, while including an invalid branch adds probability that was never present in the original distribution. Likewise, integrating over a set larger than \(T(S_X)\) is not justified merely because the algebraic expression for the inverse exists there. The inverse must correspond to an original point in the support.

Common Mistakes and a Practical Check

When auditing a change-of-variables calculation, work from the original support rather than guessing the new one from a sketch. For each proposed transformed value, solve the inverse equation, check every root against the original support, and calculate the Jacobian factor at each valid root.

Using the forward determinant as a multiplier. The density formula uses the absolute determinant of the inverse derivative. In an invertible transformation, this is the reciprocal of the absolute forward determinant.
Keeping only one root of a many-to-one map. List all solutions of \(T(x)=y\), then discard only those outside the original support.
Using every algebraic root without checking support. An inverse expression can exist even when it points to an original value that has zero density or lies outside the support.
Forgetting the transformed support. The density is zero outside the image of the original support, even if an inverse formula can be written there.
Treating a normalization check as proof of correctness. Different errors can sometimes offset numerically. Check the inverse, determinant, branches, and support separately, then verify the integral.

Invert the map.
Write the inverse coordinates and identify whether there is one inverse or several.

Check the original support.
Substitute each inverse branch into the original support restrictions before retaining it.

Use the right determinant.
For a one-to-one map, multiply by the absolute inverse Jacobian determinant; for multiple branches, do this for each valid branch.

Verify the result.
State the transformed support explicitly and check that the density integrates to one.

Key takeaway: A transformed density is the original density evaluated at valid inverse points, multiplied by the appropriate absolute inverse-Jacobian factor, and restricted to the image of the original support. All three parts are required.

Check Your Understanding

For each question, identify the relevant determinant, branch, or support check.

For a one-to-one transformation \(y=T(x)\), why is the density multiplier \(|(T^{-1})'(y)|\), rather than \(|T'(x)|\)?
If \(X\) is uniform on \((-2,1)\) and \(Y=X^2\), which of the two roots of \(x^2=y\) can contribute when \(1<y<4\)? Explain using the original support.
For \(U=X,\ V=XZ\) with \(0<X,Z<1\), derive the transformed support by substituting the inverse into both original inequalities.
A proposed density is nonnegative but integrates to \(3/2\). What does this establish, and which parts of the calculation should be checked first?
Why can a formula for an inverse coordinate fail to imply that the transformed density is positive at that point?