Deriving the Jacobian Density Formula

Why a Density Changes Under a Transformation

When a transformation stretches a small region of the plane, the same probability is spread over a larger area, so the density becomes smaller. If the transformation compresses a region, the density becomes larger. The Jacobian determinant measures this local area change. The central point is that a density in the new coordinates uses the area change for the inverse map.

This tutorial derives that adjustment for a one-to-one transformation of two continuous random variables. The argument uses local area scaling and the fact that a change of coordinates cannot change the probability assigned to an event.

Definition: Let $(X,Y)$ have joint density $f_{X,Y}(x,y)$, and let $U=g_1(X,Y)$, $V=g_2(X,Y)$. Suppose the transformation is one-to-one between the relevant open regions, has a continuously differentiable inverse, and its Jacobian determinant is nonzero there. Write the inverse as $x=x(u,v)$, $y=y(u,v)$. The absolute inverse Jacobian determinant is $$ \left|\frac{\partial(x,y)}{\partial(u,v)}\right| = \left| \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{matrix} \right|. $$ Under these conditions, the transformed joint density is $$ f_{U,V}(u,v) = f_{X,Y}\bigl(x(u,v),y(u,v)\bigr) \left|\frac{\partial(x,y)}{\partial(u,v)}\right| $$ on the transformed region, and is zero outside it.

The formula is stated on the region where the inverse exists and the determinant is nonzero. Boundary curves are ordinarily irrelevant to the density when they have probability zero. If a transformation is not one-to-one, this single-inverse formula does not apply as written: contributions from distinct inverse points must be accounted for separately.

Derivation: Local Area Scaling and Probability Conservation

Consider a small region around $(x,y)$, and let $(u,v)$ be its image. For a differentiable transformation, sufficiently close to that point the map is approximated by its linear derivative matrix. A linear map in two dimensions changes area by the absolute value of its determinant. Thus, to first order, the forward map satisfies

dudv = \left|\frac{\partial(u,v)}{\partial(x,y)}\right|\,dxdy.

Because the inverse derivative matrix is the matrix inverse of the forward derivative matrix, their determinants are reciprocals. Therefore,

dxdy = \left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,dudv.

Now take a small region $R$ in the $(u,v)$-plane and let its corresponding region in the $(x,y)$-plane be $Q$. Probability is unchanged by describing the same event in different coordinates. In the original coordinates, its probability is approximately $f_{X,Y}(x,y)\,dxdy$; in the transformed coordinates it is approximately $f_{U,V}(u,v)\,dudv$. Equating these and substituting the inverse area relation gives the density formula.

Use the inverse map.
Express $x$ and $y$ as functions of $u$ and $v$, since the original density must be evaluated at the original point corresponding to the new coordinates.

Calculate its determinant.
Find $\partial(x,y)/\partial(u,v)$ and take the absolute value. This converts area in the new coordinates to area in the original coordinates.

Multiply and restrict to the image region.
Evaluate the original density at $(x(u,v),y(u,v))$, multiply by the absolute inverse determinant, and use the resulting density only where the inverse point lies in the original support.

The absolute value is necessary because a transformation may reverse orientation. A negative determinant represents orientation reversal, not negative area or negative probability. The absolute determinant records area scale alone.

Worked Applications

Worked Example: Rescaling a Uniform Rectangle

Let $X$ and $Y$ be independent, with $X\sim\operatorname{Uniform}(0,2)$ and $Y\sim\operatorname{Uniform}(0,3)$. Define $U=2X$ and $V=Y/3$. Find the joint density of $(U,V)$, including its support, and verify that it integrates to one.

Method and original density. Independence gives a constant joint density on the rectangle $0<x<2,\ 0<y<3$. Its value is the reciprocal of the rectangle's area: $$ f_{X,Y}(x,y)=\frac{1}{2\cdot3}=\frac16. $$

Invert the transformation. From $u=2x$ and $v=y/3$, we obtain $x=u/2$ and $y=3v$. The inverse Jacobian determinant is $$ \left|\frac{\partial(x,y)}{\partial(u,v)}\right| = \left| \begin{matrix} 1/2&0\\ 0&3 \end{matrix} \right| =\frac32. $$ The original support becomes $0<u<4,\ 0<v<1$. Thus, on this region, $$ f_{U,V}(u,v)=\frac16\cdot\frac32=\frac14. $$

Normalization check. The new rectangle has area $4\cdot1=4$, so its total probability is $4\cdot(1/4)=1$. The density is correctly normalized.

Common mistake avoided. The forward determinant is $2(1/3)=2/3$, which describes how the transformation expands area. It is not the density multiplier. The inverse determinant is $3/2$, and multiplying $1/6$ by $3/2$ gives $1/4$.

Worked Example: A Nonconstant Jacobian

Let $X$ and $Y$ be independent $\operatorname{Uniform}(0,1)$ random variables. Define $U=X$ and $V=XY$. Find the joint density of $(U,V)$ and check its total probability.

Method and original density. Independence and the unit-square supports give $f_{X,Y}(x,y)=1$ for $0<x<1,\ 0<y<1$. The inverse equations are $x=u$ and $y=v/u$. They apply for $0<u<1$ and $0<v<u$, because $y$ must lie between zero and one.

The inverse Jacobian is $$ \left|\frac{\partial(x,y)}{\partial(u,v)}\right| = \left| \begin{matrix} 1&0\\ -v/u^2&1/u \end{matrix} \right| =\frac1u. $$ Consequently, $$ f_{U,V}(u,v)=\frac1u,\qquad 0<u<1,\quad 0<v<u. $$

Normalization check. Integrate first over $v$, whose range has length $u$, and then over $u$: $$ \int_0^1\int_0^u \frac1u\,dv\,du = \int_0^1 \frac{u}{u}\,du = \int_0^1 1\,du =1. $$

Common mistake avoided. Treating the Jacobian as a constant would be incorrect: the inverse area factor is $1/u$, which varies with the location. The shrinking range $0<v<u$ is also essential to the normalization check.

Worked Example: A Linear Map with Area Expansion

Let $X$ and $Y$ be independent $\operatorname{Uniform}(-1,1)$ random variables. Set $U=3X-Y$ and $V=X+2Y$. Find the transformed density and verify it using the area change.

Method and original density. The joint density is $1/4$ on the square $-1<x<1,\ -1<y<1$, whose area is $4$. The forward transformation matrix and its determinant are $$ \begin{pmatrix}3&-1\\1&2\end{pmatrix}, \qquad 3(2)-(-1)(1)=7. $$ Solving for the original variables gives $$ x=\frac{2u+v}{7},\qquad y=\frac{-u+3v}{7}. $$ The absolute inverse determinant is $1/7$. Hence the transformed density is $$ f_{U,V}(u,v)=\frac14\cdot\frac17=\frac1{28} $$ where $\left|(2u+v)/7\right|<1$ and $\left|(-u+3v)/7\right|<1$, and zero elsewhere.

Normalization check. A linear map with determinant $7$ multiplies the square's area $4$ by $7$, so the image region has area $28$. Its probability under the transformed density is $28\cdot(1/28)=1$.

Common mistake avoided. Using the forward determinant $7$ as the density multiplier would give $7/4$, which cannot be a density on a region of area $28$. The inverse determinant $1/7$ gives the correct value $1/28$.

Conditions, Checks, and Exam Traps

The formula is a change-of-coordinates result, not a shortcut that removes the need to check the model. Before applying it, establish that the original variables have a joint density, write the transformation and its inverse, and determine where that inverse corresponds to points in the original support. The determinant must be computed from the inverse map when using the displayed density formula.

Normalization is a useful error check: a valid transformed density must integrate to one over its transformed support. It does not replace careful support work. A correct determinant paired with an incorrect region can still produce an invalid answer. Conversely, an expression that looks plausible in the interior may fail normalization if the area multiplier or region is wrong.

Do not confuse area and density factors. The forward determinant scales area from $(x,y)$ to $(u,v)$; the inverse determinant multiplies the original density.
Do not omit the absolute value. Orientation reversal does not create negative density.
Do not evaluate at the wrong coordinates. Substitute the inverse expressions $x(u,v),y(u,v)$ into $f_{X,Y}$.
Do not assume the formula covers many-to-one maps. The one-to-one conditions matter; multiple inverse points require separate contributions.

Key takeaway: The transformed density is the original density evaluated at the inverse point, multiplied by the absolute inverse Jacobian determinant. This is exactly the area correction required to preserve probability.

Check Your Understanding

For each problem, identify the inverse area factor, apply it to the original joint density, and check the stated region where needed.

Let independent $X\sim\operatorname{Uniform}(0,2)$ and $Y\sim\operatorname{Uniform}(0,4)$. Set $U=3X$, $V=Y/2$. Find the transformed density and verify its integral. (Answer: $f_{U,V}(u,v)=1/12$ for $0<u<6,\ 0<v<2$; the integral is $12(1/12)=1$.)
Let independent $X,Y\sim\operatorname{Exponential}(2)$, where 2 is the rate parameter. Set $U=X,\ V=X+Y$. Find the joint density and support. (Answer: $f_{U,V}(u,v)=4e^{-2v}$ for $0<u<v$; its integral is $4\int_0^\infty ve^{-2v}\,dv=1$.)
Let independent $X,Y\sim\operatorname{Uniform}(0,1)$. Set $U=X+Y,\ V=X$. Find the transformed density and support. (Answer: $f_{U,V}(u,v)=1$ for $0<v<1,\ v<u<v+1$, and zero elsewhere.)
Let independent $X,Y\sim N(0,1)$, where the second parameter is the variance. Set $U=X+Y,\ V=2X-Y$. Find the transformed density. (Answer: $x=(u+v)/3,\ y=(2u-v)/3$, so $f_{U,V}(u,v)=\frac{1}{6\pi}\exp\left[-\frac{(u+v)^2+(2u-v)^2}{18}\right]$ for all real $u,v$.)

What You'll Learn