Jacobian Determinants for Affine Maps

Affine Maps and the Volume Factor

An affine transformation of a \(d\)-dimensional vector combines a linear map with a translation. The translation moves points without changing distances between them; the matrix can stretch, compress, rotate, reflect, or shear space. To transform a probability density correctly, we need the factor by which the matrix changes volume. That factor is the absolute value of its determinant.

Definition: An affine map on \(\mathbb{R}^d\) has the form \(T(x)=Ax+b\), where \(A\) is a \(d\times d\) matrix and \(b\in\mathbb{R}^d\). It is invertible precisely when \(A\) is nonsingular. In that case, \(T^{-1}(y)=A^{-1}(y-b)\).

For a measurable set \(B\subseteq\mathbb{R}^d\), write \(\operatorname{Vol}(B)\) for its \(d\)-dimensional Lebesgue measure. The determinant is signed: its sign records whether the linear map preserves or reverses orientation. Volume, however, cannot be negative, so the volume factor is \(|\det A|\).

Theorem: If \(A\) is nonsingular and \(b\in\mathbb{R}^d\), then for every Lebesgue-measurable set \(B\subseteq\mathbb{R}^d\), \[ \operatorname{Vol}(AB+b)=|\det A|\,\operatorname{Vol}(B). \] The equality includes sets of infinite volume.

Why the Determinant Gives the Volume Change

A direct way to establish the theorem is to reduce an invertible matrix to the identity using elementary row operations. Each such operation changes volume by a known factor, and changes the absolute determinant by the same factor. The volume statements below follow by taking one-dimensional sections and applying Fubini's theorem; they hold for measurable sets, not only for boxes.

Swapping two coordinates preserves volume and has determinant of absolute value \(1\).
Multiplying one coordinate by a nonzero constant \(c\) multiplies volume by \(|c|\), by one-dimensional scaling on each section. The determinant's absolute value is multiplied by \(|c|\).
Adding \(c\) times one coordinate to a different coordinate is a shear. Fixing all coordinates other than the changed coordinate, this operation translates each one-dimensional section. Its length is unchanged, and the operation has determinant \(1\).

Every nonsingular matrix can be reduced to the identity by finitely many of these row operations. Reversing that sequence expresses \(A\) as a product of their matrices. Applying the operations one at a time multiplies volume by the product of their absolute volume factors. Determinants multiply under matrix products, so this product is \(|\det A|\). Finally, translation by \(b\) preserves volume because it translates every set without changing its measure. This proves the volume formula.

Result: If \(A\) is nonsingular, then for every nonnegative measurable function \(h\), \[ \int_{AB+b} h\!\left(A^{-1}(y-b)\right)\,dy =|\det A|\int_B h(x)\,dx. \] This follows first for indicator functions from the volume formula, then for nonnegative simple functions by addition, and finally for nonnegative measurable functions by monotone convergence.

The factor in this integral identity is \(|\det A|\) because the integration is over the transformed region in \(y\)-space. When expressing a density with respect to \(dy\), the corresponding factor appears in the denominator.

Transforming a Multivariate Density

Let \(X\) be a random vector in \(\mathbb{R}^d\) with density \(f_X\), and define \(Y=AX+b\), where \(A\) is nonsingular. The support is transformed along with the values: if \(X\) lies in a set \(S\), then \(Y\) lies in \(AS+b\). For any measurable set \(C\), the event \(Y\in C\) is the event \(X\in A^{-1}(C-b)\). The volume factor then gives the density of \(Y\).

Theorem: Under these conditions, a density of \(Y=AX+b\) is \[ f_Y(y)=\frac{1}{|\det A|}\, f_X\!\left(A^{-1}(y-b)\right). \] The formula applies on the transformed support; outside it the density is zero. Densities are defined up to changes on sets of Lebesgue measure zero.

To verify the formula, let \(C\) be any measurable set. By the definition of \(Y\) and the density of \(X\), \[ P(Y\in C)=\int_{A^{-1}(C-b)} f_X(x)\,dx. \] Apply the integral identity with the set \(B=A^{-1}(C-b)\) and \(h=f_X\). Since \(AB+b=C\), it gives \[ \int_C \frac{1}{|\det A|}f_X\!\left(A^{-1}(y-b)\right)\,dy =\int_{A^{-1}(C-b)} f_X(x)\,dx =P(Y\in C). \] Thus the displayed function integrates to the probability of every measurable set \(C\), which is exactly the defining property of a density of \(Y\). \(\square\)

Worked Example: A Rectangle Mapped to a Parallelogram

Suppose \(X=(X_1,X_2)^\mathsf{T}\) is uniform on \(0\leq x_1\leq2,\ 0\leq x_2\leq1\). The rectangle has area \(2\), so \(f_X(x_1,x_2)=1/2\) there and is zero elsewhere. Let \[ Y=AX+b,\qquad A=\begin{pmatrix}2&1\\0&3\end{pmatrix}, \qquad b=\begin{pmatrix}1\\-1\end{pmatrix}. \]

Determinant and inverse. The determinant is \(2(3)-0(1)=6\). The equations are \(y_1=2x_1+x_2+1\) and \(y_2=3x_2-1\), giving \(x_2=(y_2+1)/3\) and \(x_1=(3y_1-y_2-4)/6\).

Transformed support. The conditions \(0\leq x_2\leq1\) become \(-1\leq y_2\leq2\). The conditions \(0\leq x_1\leq2\) become \(0\leq3y_1-y_2-4\leq12\), or \(y_2+4\leq3y_1\leq y_2+16\). Together these describe the image parallelogram.

Density. On that parallelogram, \[ f_Y(y_1,y_2)=\frac{1/2}{6}=\frac{1}{12}. \] The image area is \(6\cdot2=12\), so the density integrates to \(12(1/12)=1\), as required.

Worked Example: A Negative Determinant and a Reflection

Let \(X\) be uniform on the unit square \(0\leq x_1,x_2\leq1\), with density \(1\) there. Define \[ Y_1=3-2X_2,\qquad Y_2=X_1-1. \] Here \(A=\begin{pmatrix}0&-2\\1&0\end{pmatrix}\), \(b=(3,-1)^\mathsf{T}\), and \(\det A=2\). In particular, this example has a positive determinant; reversing the sign of the first row would give a negative determinant with the same absolute volume factor.

Solving the equations gives \(x_2=(3-y_1)/2\) and \(x_1=y_2+1\). Thus the unit-square support maps to \(1\leq y_1\leq3,\ -1\leq y_2\leq0\). The density on this rectangle is \(1/|\det A|=1/2\), and zero outside. Its area is \(2\), so its total probability is \(2(1/2)=1\). The density calculation uses the absolute determinant, not the signed determinant.

Worked Example: Transforming a Bivariate Normal Density

Suppose \(X_1\) and \(X_2\) are independent \(N(0,1)\) random variables. Their joint density is \(f_X(x_1,x_2)=(2\pi)^{-1}\exp[-(x_1^2+x_2^2)/2]\) on \(\mathbb{R}^2\). Set \(Y_1=X_1+X_2+2\) and \(Y_2=2X_2-1\).

The matrix is \(A=\begin{pmatrix}1&1\\0&2\end{pmatrix}\), so \(|\det A|=2\). Inverting gives \(x_2=(y_2+1)/2\) and \(x_1=y_1-2-(y_2+1)/2=y_1-y_2/2-5/2\). The map is invertible, so the support remains all of \(\mathbb{R}^2\).

Substitution into the density formula yields \[ f_Y(y_1,y_2)=\frac{1}{4\pi} \exp\!\left\{-\frac12\left[\left(y_1-\frac{y_2}{2}-\frac52\right)^2 +\left(\frac{y_2+1}{2}\right)^2\right]\right\}, \qquad (y_1,y_2)\in\mathbb{R}^2. \] The normalizing constant is \((2\pi)^{-1}/2=1/(4\pi)\). The exponent is the original sum of squared coordinates evaluated at the inverse image, rather than at \(y\) itself.

Checks, Scope, and a Common Pitfall

The formula has two separate ingredients: evaluate the original density at the inverse-transformed point, and divide by the absolute determinant. Both are needed. The inverse tells us which original point produced \(y\); the determinant corrects for the change in volume around that point. As a practical check, integrate the resulting density over the transformed support. It must give one.

A frequent mistake is to multiply the density by \(|\det A|\) because the map expands volume. But an expanded region contains more volume per unit of original probability, so the probability density per unit of new volume must decrease. For a one-dimensional map \(Y=3X\), for instance, an interval three times as long must have density one-third as large if it carries the same probability. The multivariate denominator is the same principle, with \(3\) replaced by \(|\det A|\).

The nonsingularity condition matters. If \(\det A=0\), the image lies in a lower-dimensional subset of \(\mathbb{R}^d\), and \(A^{-1}\) and the density formula above do not exist. Such transformations require a different treatment; assigning an infinite factor \(1/|\det A|\) is not valid.

Key takeaway: For an invertible affine map \(y=Ax+b\), the volume multiplier is \(|\det A|\), while the transformed density is the original density evaluated at \(A^{-1}(y-b)\), divided by \(|\det A|\).

Check Your Understanding

Use the inverse map, determinant, and transformed support to organize each calculation.

For \(Y=4X\) with \(X\) uniform on \([0,1]\), what is the support and density of \(Y\)?
Why does a negative determinant reverse orientation but not produce a negative volume?
If \(A=\begin{pmatrix}1&2\\0&3\end{pmatrix}\), what is the volume multiplier?
What feature of the support signals that the affine density formula cannot be used when \(A\) is singular?
For an invertible affine transformation, which argument of the original density is used to calculate \(f_Y(y)\)?