Multivariate Change of Variables for Bijections

Why the Multivariate Formula Needs a Jacobian

For a one-dimensional transformation, the density changes by the absolute value of a derivative. In several dimensions, a transformation can stretch, compress, or shear a small region in different directions. The corresponding density adjustment is determined by a determinant: the Jacobian determinant of the inverse transformation. The inverse is central because the transformed density at a point $y$ must be evaluated using the original point that maps to $y$.

The formula applies to a one-to-one transformation between regions. If different original points can map to the same transformed point, their contributions must be combined; that is a different situation. Here, every point in the transformed region has exactly one preimage.

Definition: Let $U,V$ be open subsets of $\mathbb{R}^d$. A map $g:U\to V$ is a differentiable bijection with differentiable inverse if it is one-to-one and onto, $g$ is continuously differentiable, and $g^{-1}:V\to U$ is continuously differentiable. Write $Dg(x)$ for the $d\times d$ derivative matrix at $x$, and $\det Dg(x)$ for its Jacobian determinant.

The Density Transformation Formula

Let $\mathbf X$ be a random vector with density $f_{\mathbf X}$, and suppose $P(\mathbf X\in U)=1$. For a differentiable bijection $g:U\to V$ with differentiable inverse, define $\mathbf Y=g(\mathbf X)$. The density is evaluated at $g^{-1}(y)$, while the inverse Jacobian accounts for the change in volume.

Theorem: Suppose $U,V$ are open subsets of $\mathbb{R}^d$, $g:U\to V$ is a differentiable bijection with differentiable inverse, and $\mathbf X$ has density $f_{\mathbf X}$ with $P(\mathbf X\in U)=1$. Then $\mathbf Y=g(\mathbf X)$ has density $$ f_{\mathbf Y}(y)= \begin{cases} f_{\mathbf X}\bigl(g^{-1}(y)\bigr)\left|\det Dg^{-1}(y)\right|, & y\in V,\\ 0, & y\notin V. \end{cases} $$ Equivalently, for $x\in U$, $$ f_{\mathbf Y}\bigl(g(x)\bigr) =\frac{f_{\mathbf X}(x)}{\left|\det Dg(x)\right|}. $$

The inverse derivative is taken at $y$, and the forward derivative in the equivalent expression is taken at $x$. These are reciprocal absolute determinants at corresponding points, not generally the same determinant.

Result: For every $x\in U$, $$ Dg^{-1}\bigl(g(x)\bigr)Dg(x)=I_d, \qquad \left|\det Dg^{-1}\bigl(g(x)\bigr)\right| =\frac{1}{\left|\det Dg(x)\right|}. $$

Proof. The identity $g^{-1}(g(x))=x$ holds for every $x\in U$. Differentiate both sides with respect to $x$. The chain rule gives $Dg^{-1}(g(x))Dg(x)=I_d$. Taking determinants yields $$ \det Dg^{-1}\bigl(g(x)\bigr)\det Dg(x)=1. $$ Thus neither determinant is zero, and taking absolute values gives the reciprocal relation. $\square$

Proof of the density formula. The multivariate change-of-variables theorem for integrals states that, for a nonnegative measurable function $h$ on $V$, $$ \int_V h(y)\,dy =\int_U h(g(x))\left|\det Dg(x)\right|\,dx. $$ For any Borel set $A\subseteq\mathbb{R}^d$, the event $\mathbf Y\in A$ is the event $\mathbf X\in g^{-1}(A\cap V)$. Therefore, $$ P(\mathbf Y\in A) =\int_{g^{-1}(A\cap V)} f_{\mathbf X}(x)\,dx. $$ Apply the integral change-of-variables theorem to this integral, using the inverse map on $V$. It becomes $$ P(\mathbf Y\in A) =\int_{A\cap V} f_{\mathbf X}\bigl(g^{-1}(y)\bigr) \left|\det Dg^{-1}(y)\right|\,dy. $$ This is the defining integral property of the stated density, with density zero outside $V$. The reciprocal determinant result gives the equivalent forward-Jacobian expression. $\square$

The transformed density is concentrated on $V$, but its positive-density region depends on the original density as well as on $g$. More precisely, within $V$, the displayed formula determines where the density is positive, up to changes on sets of Lebesgue measure zero. When tracking support, map the region where the original density is positive through $g$; do not assume the whole target region must have positive density.

Worked Applications

Worked Example: A Uniform Square Mapped to a Diamond

Let $X_1,X_2$ be independent uniform random variables on $(0,1)$. Their joint density is $f_{\mathbf X}(x_1,x_2)=1$ on the unit square and zero elsewhere. Define $$ Y_1=X_1+X_2,\qquad Y_2=X_1-X_2. $$ Find the joint density of $\mathbf Y=(Y_1,Y_2)$.

Solving the two equations for the original coordinates gives $$ x_1=\frac{y_1+y_2}{2},\qquad x_2=\frac{y_1-y_2}{2}. $$ Thus the inverse map is $g^{-1}(y_1,y_2)=((y_1+y_2)/2,(y_1-y_2)/2)$, whose derivative matrix and determinant are $$ Dg^{-1}(y_1,y_2)= \begin{pmatrix} 1/2&1/2\\ 1/2&-1/2 \end{pmatrix}, \qquad \det Dg^{-1}(y_1,y_2)=-\frac12. $$ The absolute inverse Jacobian is $1/2$.

The original support requires $0<x_1<1$ and $0<x_2<1$. Substituting the inverse coordinates gives $$ 0<y_1+y_2<2,\qquad 0<y_1-y_2<2. $$ Hence the transformed density is $$ f_{\mathbf Y}(y_1,y_2)= \begin{cases} 1/2, & 0<y_1+y_2<2\ \text{and}\ 0<y_1-y_2<2,\\ 0, & \text{otherwise}. \end{cases} $$ These inequalities describe the image of the unit square, a diamond-shaped region. The area of the original square is $1$; the transformation has forward determinant of absolute value $2$, so the image has area $2$. The resulting constant density $1/2$ therefore integrates to $1$, as required.

Worked Example: Taking Logarithms of Two Positive Variables

Let $X_1,X_2$ be independent exponential random variables with rate $1$, so their joint density is $e^{-x_1-x_2}$ for $x_1>0,x_2>0$. Set $Y_1=\log X_1$ and $Y_2=\log X_2$.

The map $g(x_1,x_2)=(\log x_1,\log x_2)$ is a differentiable bijection from $U=(0,\infty)^2$ to $V=\mathbb{R}^2$, with inverse $g^{-1}(y_1,y_2)=(e^{y_1},e^{y_2})$. The inverse derivative matrix is diagonal: $$ Dg^{-1}(y_1,y_2)= \begin{pmatrix} e^{y_1}&0\\ 0&e^{y_2} \end{pmatrix}, \qquad \left|\det Dg^{-1}(y_1,y_2)\right|=e^{y_1+y_2}. $$ Substitution into the formula gives, for every $(y_1,y_2)\in\mathbb{R}^2$, $$ f_{\mathbf Y}(y_1,y_2) =e^{-e^{y_1}-e^{y_2}}e^{y_1+y_2}. $$ The transformed support is all of $\mathbb{R}^2$, since the original density is positive throughout the positive quadrant and the logarithm maps that quadrant onto $\mathbb{R}^2$. As a check, the density factors as $$ \left(e^{y_1-e^{y_1}}\right)\left(e^{y_2-e^{y_2}}\right). $$ Each factor integrates to $1$: with $x=e^y$, $dy=dx/x$, so $$ \int_{-\infty}^{\infty}e^{y-e^y}\,dy =\int_0^\infty e^{-x}\,dx=1. $$ The product density therefore integrates to $1$.

Worked Example: A Linear Transformation of a Standard Normal Pair

Let $X_1,X_2$ be independent $N(0,1)$ random variables, with joint density $$ f_{\mathbf X}(x_1,x_2)=\frac{1}{2\pi}\exp\left(-\frac{x_1^2+x_2^2}{2}\right). $$ Define $Y_1=X_1$ and $Y_2=X_1+X_2$. The forward map has matrix $$ Dg= \begin{pmatrix} 1&0\\ 1&1 \end{pmatrix}, \qquad \det Dg=1. $$ It is a bijection from $\mathbb{R}^2$ to $\mathbb{R}^2$; its inverse is $x_1=y_1,\ x_2=y_2-y_1$, and the absolute inverse determinant is also $1$.

The original density is positive everywhere, and the transformation maps all of $\mathbb{R}^2$ onto itself. Thus the transformed support is $\mathbb{R}^2$, and $$ f_{\mathbf Y}(y_1,y_2) =\frac{1}{2\pi}\exp\left(-\frac{y_1^2+(y_2-y_1)^2}{2}\right), \qquad (y_1,y_2)\in\mathbb{R}^2. $$ The exponent uses the inverse coordinates, not $y_1^2+y_2^2$. Since the inverse determinant is $1$, no additional multiplicative factor appears.

Support and Calculation Checks

A reliable calculation separates three tasks: find the inverse, find its determinant, and translate the support. The following sequence makes the dependencies explicit.

Invert the transformation.
Express each original coordinate as a function of the transformed coordinates. Verify that the stated inverse maps the target region back into the source region.

Compute the inverse Jacobian.
Differentiate the inverse map, take its determinant, and use the absolute value in the density formula.

Transform the support.
Apply the inverse coordinates to the original support restrictions, or map the original positive-density region forward.

Assemble and check.
Evaluate the original density at the inverse coordinates, multiply by the inverse determinant, set the density to zero outside the image, and check that the result integrates to one when feasible.

A frequent error is to use $\left|\det Dg(y)\right|$ as the multiplier. This is generally wrong twice over: $Dg$ is evaluated at a source point $x$, not at $y$, and the density multiplier is the absolute determinant of the inverse derivative. The equivalent forward form is division by $\left|\det Dg(x)\right|$, where $x=g^{-1}(y)$. Another common error is to give the formula on all of $\mathbb{R}^d$ without restricting it to $V=g(U)$. Outside that image, the transformed density is zero.

Key takeaway: For a differentiable bijection, evaluate the original density at the unique inverse point and multiply by the absolute determinant of the inverse derivative. The transformed density lives on the image of the original region, with the original support restrictions carried through the map.

Check Your Understanding

Use the inverse map, its determinant, and the image of the original support to organize each calculation.

In the density formula, at which point is $f_{\mathbf X}$ evaluated when the transformed value is $y$?
Explain why the absolute determinants of $Dg^{-1}(g(x))$ and $Dg(x)$ are reciprocals.
If $g(x_1,x_2)=(2x_1,x_2)$, what is the absolute determinant of the inverse derivative?
Why must the transformed density be zero outside $V$, the image of $U$?
For a transformation with inverse $x_1=y_1+y_2,\ x_2=y_2$, write the conditions on $y_1,y_2$ corresponding to $0<x_1<1$ and $0<x_2<1$.

Multivariate Change of Variables for Bijections

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