Deriving the Multivariate Jacobian Formula

Why a Jacobian Appears in a Density

A density describes probability per unit volume. When a smooth transformation stretches or compresses space, the probability assigned to a region does not change, but the region’s volume does. The transformed density must therefore adjust to compensate. In several dimensions, the determinant of a derivative matrix records the local volume change; the density formula uses the determinant of the inverse transformation because it expresses the original volume in terms of the transformed coordinates.

The derivation has two parts. A local calculation explains the volume factor, while the multivariate change-of-variables theorem for integrals makes the resulting density formula rigorous for probabilities of arbitrary measurable sets.

Set-Up and Formula

Definition: Let $U$ and $V$ be open subsets of $\mathbb{R}^d$, and let $T:U\to V$ be a bijection that is continuously differentiable and has a continuously differentiable inverse. Such a map is a $C^1$ diffeomorphism. Write $DT(x)$ for its derivative matrix at $x$. Suppose a random vector $X$ satisfies $P(X\in U)=1$ and has density $f_X$ on $U$, and define $Y=T(X)$. Then $Y$ has density $$ f_Y(y)=f_X\bigl(T^{-1}(y)\bigr)\left|\det D T^{-1}(y)\right|,\qquad y\in V. $$ The density is zero outside $V$. The displayed expression specifies a density up to changes on sets of Lebesgue measure zero.

The derivative matrix of the inverse is evaluated at the transformed point $y$, not at the original point $x$. If we write $x=T^{-1}(y)$, the chain rule gives

D T^{-1}(y)=\bigl(DT(x)\bigr)^{-1}, \qquad \left|\det D T^{-1}(y)\right| =\frac{1}{|\det DT(x)|}.

Thus the equivalent expression at $y=T(x)$ is $f_Y(T(x))=f_X(x)/|\det DT(x)|$. The absolute value is necessary: densities are nonnegative whether the transformation preserves or reverses orientation.

From Local Volume to Density

At a point $x$, differentiability says that for sufficiently small displacements $h$, the transformation is approximated by its linear part:

T(x+h)=T(x)+DT(x)h+o(\|h\|).

A linear map with matrix $DT(x)$ scales volume by $|\det DT(x)|$. Consequently, a sufficiently small region around $x$ is transformed into a region whose volume is approximately $|\det DT(x)|$ times as large. Probability assigned to corresponding regions is exactly the same: if the original region is $A\subset U$, the transformed region is $T(A)$, and $P(X\in A)=P(Y\in T(A))$.

When $f_X$ is continuous at $x$, probability near $x$ is approximately $f_X(x)$ times the original volume. Dividing that probability by the transformed volume gives the local transformed density $f_X(x)/|\det DT(x)|$. This explains both the reciprocal and the absolute value. It is a local derivation: by itself it is not a proof for arbitrary regions or merely measurable densities. The integral argument below supplies that proof.

Theorem: Under the conditions in the definition, for every Borel set $A\subseteq V$, $$ P(Y\in A)=\int_A f_X\bigl(T^{-1}(y)\bigr) \left|\det D T^{-1}(y)\right|\,dy. $$ Therefore the integrand is a density of $Y$ on $V$.

Proof. Since $Y=T(X)$ and $T$ is a bijection, the events $\{Y\in A\}$ and $\{X\in T^{-1}(A)\}$ are equal. The density of $X$ therefore gives

P(Y\in A) =P(X\in T^{-1}(A)) =\int_{T^{-1}(A)} f_X(x)\,dx.

The multivariate change-of-variables theorem for integrals applies because $T$ is a $C^1$ diffeomorphism between open sets. With $y=T(x)$, that theorem transforms the last integral into

\int_A f_X\bigl(T^{-1}(y)\bigr) \left|\det D T^{-1}(y)\right|\,dy.

This proves the probability identity for every Borel $A\subseteq V$. Its integrand is nonnegative and measurable, and taking $A=V$ shows that its integral is $P(Y\in V)=1$. Hence it is a density of $Y$. $\square$

Worked Applications

Worked Example: A Product Transformation on the Positive Quadrant

Let $X_1$ and $X_2$ be independent, with $X_1\sim\operatorname{Exp}(1)$ and $X_2\sim\operatorname{Exp}(2)$. Their joint density is $f_X(x_1,x_2)=2e^{-x_1-2x_2}$ for $x_1,x_2>0$. Transform by $u=x_1,\ v=x_1x_2$. Find the joint density of $(U,V)$.

The map $T(x_1,x_2)=(x_1,x_1x_2)$ sends the positive quadrant bijectively to itself. Solving for the original coordinates gives $x_1=u,\ x_2=v/u$, so the inverse is $T^{-1}(u,v)=(u,v/u)$. Its derivative matrix and determinant are

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

Substitute the inverse coordinates into the original density and multiply by this factor:

f_{U,V}(u,v) =2e^{-u-2v/u}\frac{1}{u} =\frac{2}{u}e^{-u-2v/u}, \qquad u>0,\ v>0.

The support is the positive quadrant because both inverse coordinates must be positive. As a check, integrate first with respect to $v$:

\int_0^\infty\int_0^\infty \frac{2}{u}e^{-u-2v/u}\,dv\,du =\int_0^\infty e^{-u} \left(\int_0^\infty \frac{2}{u}e^{-2v/u}\,dv\right)du =\int_0^\infty e^{-u}\,du =1.

In the inner integral, the substitution $s=2v/u$ gives $\int_0^\infty e^{-s}\,ds=1$. This confirms the normalization.

Worked Example: A Curved Image of a Uniform Density

Suppose $X$ has density $1$ on $U=(0,1)^2$ and zero elsewhere. Define $U_1=X_1$ and $V_1=X_1^2-X_2$. Find the density of $(U_1,V_1)$.

To avoid confusing the transformed first coordinate with the original domain, denote the transformed vector by $Y=(Y_1,Y_2)=T(X)$, where $T(x_1,x_2)=(x_1,x_1^2-x_2)$. Solving gives $x_1=y_1,\ x_2=y_1^2-y_2$. The image of the original square is therefore $0<y_1<1,\ y_1^2-1<y_2<y_1^2$.

The inverse derivative is

D T^{-1}(y_1,y_2)= \begin{pmatrix} 1&0\\ 2y_1&-1 \end{pmatrix}, \qquad \left|\det D T^{-1}(y_1,y_2)\right|=1.

The original density is $1$ throughout the square. Thus the transformed density is $1$ throughout its curved image and zero elsewhere:

f_Y(y_1,y_2)= \begin{cases} 1,&0<y_1<1,\quad y_1^2-1<y_2<y_1^2,\\ 0,&\text{otherwise}. \end{cases}

For each fixed $y_1$ between $0$ and $1$, the allowed interval for $y_2$ has length $1$. Hence the integral of this density is $\int_0^1 1\,dy_1=1$. The negative determinant before taking absolute value reflects orientation reversal; it does not produce a negative density.

Worked Example: Exponentiating a Bivariate Normal Vector

Let $X_1,X_2$ be independent standard normal random variables and set $Y_1=e^{X_1}$, $Y_2=e^{X_2}$. Derive the joint density of $Y$.

The transformation $T(x_1,x_2)=(e^{x_1},e^{x_2})$ is a smooth bijection from $\mathbb{R}^2$ to $(0,\infty)^2$. Its inverse is $T^{-1}(y_1,y_2)=(\ln y_1,\ln y_2)$, with derivative determinant $1/(y_1y_2)$. The original joint density is $f_X(x_1,x_2)=(2\pi)^{-1}\exp(-(x_1^2+x_2^2)/2)$.

Applying the formula gives

f_Y(y_1,y_2) =\frac{1}{2\pi y_1y_2} \exp\left[-\frac{(\ln y_1)^2+(\ln y_2)^2}{2}\right], \qquad y_1>0,\ y_2>0.

The support is positive in both coordinates; the density is zero elsewhere. To check its integral, substitute $x_i=\ln y_i$ in each coordinate, so $dy_i/y_i=dx_i$. The integral becomes $\int_{\mathbb R^2}(2\pi)^{-1}\exp(-(x_1^2+x_2^2)/2)\,dx_1\,dx_2=1$, the integral of the original joint normal density.

Why the Integral Proof Matters

The local picture is useful for remembering the formula, but probability statements concern sets, not infinitesimal regions. The set-based proof guarantees that the formula gives the correct probability for every Borel event, including regions with curved boundaries. It also makes normalization automatic: the transformation carries all probability from $U$ to $V$, and the integral substitution preserves that total probability.

In calculations, it is usually safest to proceed in this order: identify the image of the support, solve explicitly for the inverse, compute the inverse derivative determinant, and then substitute into the original density. A correct determinant paired with an incorrect image or inverse still gives an incorrect answer.

Common Mistake

A frequent error is to multiply by $|\det DT(x)|$ rather than divide by it. That factor measures how much the transformation expands a small original volume. When calculating density per transformed unit volume, expansion spreads the same probability over more space, so the density decreases. The inverse determinant in the formula performs exactly this adjustment. Another error is to omit the absolute value: an orientation-reversing map can have a negative determinant, but probabilities and densities cannot be negative.

Key takeaway: A bijective transformation preserves probability of corresponding events. Its derivative changes local volume, so the transformed density is the original density evaluated at the inverse point, multiplied by the absolute determinant of the inverse derivative.

Check Your Understanding

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

State the regularity conditions used in the density formula, and explain why having a bijection alone is not enough for the stated theorem.
If $T(x)=3x$ in one dimension, should the density at $y$ be multiplied by $3$ or by $1/3$? Explain using volume scaling.
Why must the absolute value of the determinant appear in the multivariate formula?
For the map $T(x_1,x_2)=(x_1,x_1x_2)$ on the positive quadrant, write its inverse and state the transformed support.
How does the set-based proof show that the transformed density integrates to one?