Checking Multivariate Transformation Calculations

Why Audit a Transformed Density?

A bivariate change of variables can look plausible and still be wrong. The inverse may be incorrect, the transformed support may be too large or too small, or the Jacobian determinant may be used in the wrong direction. Each of these errors changes the resulting density, sometimes without making the mistake obvious from its algebraic form.

For a one-to-one transformation, the density formula provides a systematic audit. We check that the inverse works, map the original support into the new coordinates, evaluate the inverse Jacobian, and verify that the proposed density integrates to one. These checks are especially effective when written as separate steps rather than combined into one calculation.

The Change-of-Variables Checks

Let \(X=(X_1,X_2)\) have density \(f\), and suppose its probability is concentrated on an open set \(U\subset\mathbb{R}^2\). Let \(T:U\to V\) be a continuously differentiable bijection onto an open set \(V\), with a continuously differentiable inverse. Assume the derivative matrix is nonsingular throughout \(U\). For \(Z=T(X)\), the density on \(V\) is

g(z)=f\!\left(T^{-1}(z)\right)\left|\det D T^{-1}(z)\right|,\qquad z\in V.

The transformed density is zero outside \(V\). If \(f\) is zero on parts of \(U\), then \(g\) is also zero at the corresponding points of \(V\). Thus, the geometric image \(T(U)\) describes where the transformation can take values, while the density itself also depends on where the original density is positive.

Audit checklist: (1) Write the proposed inverse and verify both compositions. (2) Translate every original support restriction into transformed coordinates. (3) Differentiate the inverse and take the absolute determinant. (4) Include the transformed support in the density statement. (5) Integrate the result and confirm that its total is one.

The determinant in the formula is the determinant of the inverse map, evaluated at the transformed point. If instead we calculate the forward determinant, then the reciprocal relationship must be used. The following result explains both why the density integrates to one and how the two determinants are related.

Theorem: Under the conditions above, \(g\) is a density on \(V\) and integrates to one. Moreover, for \(z=T(x)\), \[ \left|\det D T^{-1}(z)\right|=\frac{1}{|\det DT(x)|}. \]

Proof. The inverse identity \(T^{-1}(T(x))=x\), together with the chain rule, gives

D T^{-1}(T(x))\,DT(x)=I.

Taking determinants shows that \(\det D T^{-1}(T(x))\det DT(x)=1\). Since these determinants are nonzero, taking absolute values and rearranging proves the reciprocal formula. For normalization, apply the change-of-variables formula to the nonnegative integrand \(g\):

\int_V g(z)\,dz =\int_V f(T^{-1}(z))|\det D T^{-1}(z)|\,dz =\int_U f(x)\,dx =1.

The last equality holds because \(f\) is a density concentrated on \(U\). Nonnegativity ensures the integrals are well-defined, and the change-of-variables formula applies under the stated differentiability and bijection assumptions. Therefore \(g\) is normalized. \(\square\)

Normalization is a necessary check, not a complete proof that a proposed formula is correct: an incorrect function can also happen to integrate to one. The inverse, support, and determinant must still be checked.

Worked Bivariate Calculations

Worked Example: A Triangular Map and a Slanted Support

Let \(X\) and \(Y\) be independent Uniform\((0,1)\) random variables. Define \(U=X+Y\) and \(V=Y\). Find the joint density of \((U,V)\), including its support, and check its integral.

The forward map is \(u=x+y,\ v=y\). Solving gives the proposed inverse \(x=u-v,\ y=v\). Substitution verifies it: \(x+y=(u-v)+v=u\) and \(y=v\). The original support is \(0<x<1,\ 0<y<1\); in the new coordinates this becomes

0<v<1,\qquad 0<u-v<1,

or equivalently \(0<v<1\) and \(v<u<v+1\). The inverse derivative matrix and its determinant are

D T^{-1}(u,v)= \begin{pmatrix} 1&-1\\ 0&1 \end{pmatrix}, \qquad |\det D T^{-1}(u,v)|=1.

The original joint density is \(1\) on the unit square. Thus \(g(u,v)=1\) on the stated image and zero elsewhere. Its integral is

\int_0^1\int_v^{v+1}1\,du\,dv =\int_0^1\big((v+1)-v\big)\,dv =1.

The support is not the rectangle \(0<u<2,\ 0<v<1\): for example, \(u=0.2,\ v=0.9\) lies in that rectangle but would require \(x=u-v<0\). The inverse inequalities preserve the dependence between the transformed coordinates.

Worked Example: A Positive Nonlinear Transformation

Suppose \(X\) and \(Y\) are independent Exponential\((1)\) random variables, so their joint density is \(f(x,y)=e^{-x-y}\) for \(x>0,\ y>0\). Set \(U=XY,\ V=Y\). Derive the transformed density and verify its total integral.

For positive \(x,y\), the map is one-to-one onto the positive quadrant. Its inverse is \(x=u/v,\ y=v\), defined for \(u>0,\ v>0\). Indeed, \((u/v)v=u\) and the second coordinate remains \(v\). The inverse derivative is

D T^{-1}(u,v)= \begin{pmatrix} 1/v&-u/v^2\\ 0&1 \end{pmatrix}, \qquad |\det D T^{-1}(u,v)|=\frac{1}{v}.

Both original coordinates are positive exactly when \(u>0,\ v>0\). Substituting the inverse into the original density gives

g(u,v)=\frac{1}{v}\exp\!\left(-\frac{u}{v}-v\right), \qquad u>0,\ v>0.

To check normalization, integrate first over \(u\). For each fixed \(v>0\), use \(x=u/v\), so \(du=v\,dx\):

\begin{aligned} \int_0^\infty\int_0^\infty \frac{1}{v}e^{-u/v-v}\,du\,dv &=\int_0^\infty\int_0^\infty e^{-x-v}\,dx\,dv\\ &=\left(\int_0^\infty e^{-x}\,dx\right) \left(\int_0^\infty e^{-v}\,dv\right)=1. \end{aligned}

The factor \(1/v\) is essential: it compensates for the change in area under this nonlinear map. The integral check confirms the calculation’s normalization, while the inverse and support checks establish that the integrand was constructed on the right region.

Worked Example: Finding a Missing Jacobian Factor

Again let \(X,Y\) be independent Uniform\((0,1)\), but now set \(U=X+Y,\ V=X-Y\). A proposed answer gives density \(1\) on the image. Check it and correct the calculation.

Solving \(u=x+y,\ v=x-y\) gives \(x=(u+v)/2,\ y=(u-v)/2\). Substitution verifies both equations. The inverse derivative and determinant are

D T^{-1}(u,v)= \begin{pmatrix} 1/2&1/2\\ 1/2&-1/2 \end{pmatrix}, \qquad \left|\det D T^{-1}(u,v)\right|=\frac12.

The support conditions \(0<x<1,\ 0<y<1\) become \(0<u+v<2,\ 0<u-v<2\). Equivalently, \(0<u<2\) and \[ -\min(u,2-u)<v<\min(u,2-u). \] For \(0<u\leq1\), the two bounds reduce to \(-u<v<u\); for \(1\leq u<2\), they reduce to \(u-2<v<2-u\). The image therefore has cross-sectional width \(2\min(u,2-u)\).

The original density is one on the unit square, so the correct transformed density is \(1/2\) on this image. Its integral is

\int_0^1 \frac12(2u)\,du+ \int_1^2 \frac12\big(2(2-u)\big)\,du =\int_0^1u\,du+\int_1^2(2-u)\,du =\frac12+\frac12=1.

The proposed density \(1\) has integral \(2\), because the image area is \(2\). The missing factor \(1/2\) is exactly the inverse Jacobian determinant. Both the area calculation and the general normalization check expose the error.

Common Mistakes and a Reliable Routine

A frequent error is to compute \(|\det DT|\) and multiply by it, even though the formula requires \(|\det DT^{-1}|\). The correction is to decide explicitly which map was differentiated. If the forward determinant is used, evaluate it at \(x=T^{-1}(z)\) and take its reciprocal; the chain-rule identity above justifies that step.

Other mistakes often appear in the support or the final integration:

Keeping the old support unchanged. The support must be expressed in the transformed coordinates by substituting the inverse into the original inequalities.
Replacing a curved or slanted image with a bounding rectangle. A rectangle may include points with no valid preimage. Check every inverse coordinate against its original range.
Checking only the formula inside the support. A density statement must also specify where the density is zero.
Treating total integral one as sufficient evidence. It checks normalization only; an incorrect inverse or support can still produce a normalized but incorrect function.

A disciplined audit makes these errors easier to locate. First verify the inverse by substitution; next translate all support inequalities; then calculate the inverse determinant; finally integrate over the exact image. If the integral is not one, revisit the support bounds and determinant before changing the algebraic density.

Key takeaway: A valid one-to-one transformed density must use a verified inverse, the correct image support, and the absolute inverse-Jacobian determinant. Its integral must equal one, but that final check complements rather than replaces the first three.

Check Your Understanding

For each question, identify which part of the transformation audit is being tested.

For \(u=x-y,\ v=y\), write the inverse map and state how you would verify it.
If the original support is \(0<x<1,\ 0<y<1\), what substitutions should be made to determine the transformed support?
Why does the density formula use \(|\det DT^{-1}(u,v)|\) rather than \(|\det DT(x,y)|\)?
A proposed transformed density is nonnegative and has integral one. Does this alone prove it is correct? Explain.
For the map \(u=x+y,\ v=x-y\), which factor would be missing if one incorrectly used density \(1\) on the image instead of \(1/2\)?

Checking Multivariate Transformation Calculations

What You'll Learn