Mapping Supports Under Bivariate Transformations

Why the Support Must Be Mapped

A bivariate transformation changes more than the formula for a density: it changes the region on which that density is defined. The original variables $(X,Y)$ may be restricted to a rectangle, triangle, or curved region. After transforming to $(U,V)$, the possible values of $(U,V)$ are exactly the image of that original region. Getting this image right is essential; a correct Jacobian applied on the wrong region does not give the correct joint distribution.

The reliable method is to solve for $X$ and $Y$ in terms of $U$ and $V$, then substitute those inverse expressions into every inequality that defines the original support. Mapping the boundary curves provides a useful geometric check, but the transformed inequalities determine the region precisely.

Definition: Let $(X,Y)$ have joint density $f_{X,Y}(x,y)$ on a support $D$, and suppose $(U,V)=T(X,Y)$ is one-to-one and differentiable on $D$, with a differentiable inverse $(x,y)=T^{-1}(u,v)$ and nonzero inverse Jacobian determinant. The transformed support is $$ D_{U,V}=\{(u,v): (x(u,v),y(u,v))\in D\}. $$ On this region, the transformed density is $$ f_{U,V}(u,v)=f_{X,Y}(x(u,v),y(u,v)) \left|\frac{\partial(x,y)}{\partial(u,v)}\right|. $$ Outside the image region, $f_{U,V}(u,v)=0$. For continuous distributions, whether boundary points are included usually does not affect probabilities.

A Practical Boundary-Mapping Method

Write the support as a list of inequalities before doing any transformation. For example, a triangular support might require both $0<x<1$ and $0<y<1-x$; the second condition cannot be replaced by the weaker condition $0<y<1$. After finding the inverse transformation, substitute into each inequality separately. This process handles curved boundaries and restrictions that depend on another coordinate.

Record the original support.
List every inequality that describes where the original joint density is positive.

Find the inverse transformation.
Express $x$ and $y$ as functions of $u$ and $v$, and check that the inverse is valid on the region being considered.

Translate all constraints.
Substitute the inverse expressions into every original support inequality. Their intersection is the transformed support.

Map boundaries and calculate the density.
Check boundary curves or vertices against the inequalities, then multiply the original density by the absolute inverse Jacobian determinant.

Worked Transformations

Worked Example: A Rectangle with a Curved Image Boundary

Suppose $X$ and $Y$ have joint density $f_{X,Y}(x,y)=1/2$ on $0<x<1,\ 0<y<2$, and density zero elsewhere. Define $U=X$ and $V=Y+X^2$. Find the transformed support and density.

Method and inverse. From $u=x$ and $v=y+x^2$, the inverse is $x=u,\ y=v-u^2$. The inverse Jacobian matrix and its determinant are $$ \frac{\partial(x,y)}{\partial(u,v)} = \begin{pmatrix} 1&0\\ -2u&1 \end{pmatrix}, \qquad \left|\frac{\partial(x,y)}{\partial(u,v)}\right|=1. $$

Translate the support. The original constraint $0<x<1$ becomes $0<u<1$. The other constraint becomes $$ 0<y<2 \quad\Longleftrightarrow\quad 0<v-u^2<2 \quad\Longleftrightarrow\quad u^2<v<u^2+2. $$ Thus the image lies between the curved boundaries $v=u^2$ and $v=u^2+2$, for $0<u<1$.

Density and check. The density is the original constant multiplied by the absolute inverse determinant, so $f_{U,V}(u,v)=1/2$ on the stated region and zero elsewhere. For a check, the vertical width at each $u$ is $(u^2+2)-u^2=2$, and the total integral is $$ \int_0^1\int_{u^2}^{u^2+2}\frac12\,dv\,du =\int_0^1 1\,du=1. $$ The density is normalized. The common mistake this method avoids is retaining the rectangle’s old bounds $0<v<2$, even though the transformation shifts each vertical slice by $u^2$.

Worked Example: A Triangular Support under a Shear

Suppose $(X,Y)$ is uniform on the triangle $0<x<1,\ 0<y<1-x$. Its area is $1/2$, so its joint density is $2$ on that triangle. Define $U=X$ and $V=Y+2X$. Find the image region and transformed density.

Method and inverse. Solving $u=x,\ v=y+2x$ gives $x=u,\ y=v-2u$. Therefore $$ \frac{\partial(x,y)}{\partial(u,v)} = \begin{pmatrix} 1&0\\ -2&1 \end{pmatrix}, \qquad \left|\frac{\partial(x,y)}{\partial(u,v)}\right|=1. $$

Translate every constraint. The condition $0<x<1$ gives $0<u<1$. The lower bound on $y$ gives $v-2u>0$, or $v>2u$. The upper bound gives $$ v-2u<1-u \quad\Longleftrightarrow\quad v<1+u. $$ Hence the transformed support is $0<u<1,\ 2u<v<1+u$. The two sloping edges meet when $2u=1+u$, namely at $u=1$, so the region closes at that end.

Density and check. The transformed density is $2$ on this region and zero elsewhere. Its integral is $$ \int_0^1\int_{2u}^{1+u}2\,dv\,du =\int_0^1 2(1-u)\,du =2\left(1-\frac12\right)=1. $$ This confirms the result. A common incorrect bound is $v<1+2u$, obtained by replacing the original upper restriction $y<1-x$ with $y<1$. The triangular constraint must be translated as written.

Worked Example: A Curved Lower Boundary on an Unbounded Support

Let $X$ and $Y$ be independent exponential random variables, each with rate $1$, so their joint density is $e^{-x-y}$ for $x>0,\ y>0$. Define $U=X$ and $V=Y-X^2$. Find the transformed support and density.

Method and inverse. The equations give $x=u,\ y=v+u^2$. The inverse Jacobian matrix is $$ \frac{\partial(x,y)}{\partial(u,v)} = \begin{pmatrix} 1&0\\ 2u&1 \end{pmatrix}, \qquad \left|\frac{\partial(x,y)}{\partial(u,v)}\right|=1. $$

Map the support. The original restrictions $x>0$ and $y>0$ become $u>0$ and $v+u^2>0$. Thus the transformed support is $$ u>0,\qquad v>-u^2. $$ For each positive $u$, $v$ ranges above the curved boundary $-u^2$ without an upper bound.

Density and check. Substituting the inverse into the original density gives $$ f_{U,V}(u,v)=e^{-u-(v+u^2)} $$ on the image region, and zero elsewhere. To check its integral, first integrate over $v$: $$ \int_0^\infty\int_{-u^2}^{\infty} e^{-u-v-u^2}\,dv\,du =\int_0^\infty e^{-u-u^2}e^{u^2}\,du =\int_0^\infty e^{-u}\,du=1. $$ Independence supplies the original joint density in this example; it does not remove the need to map the support. In particular, the transformed lower bound is not $v>0$: negative values of $V=Y-X^2$ are possible.

Why the Inequalities Give the Image

The support description follows directly from the inverse map. A point $(u,v)$ is attainable if and only if its inverse $(x(u,v),y(u,v))$ is an allowed point of the original support. In symbols, $(u,v)$ belongs to the image exactly when $$ (x(u,v),y(u,v))\in D. $$ Substituting the inverse into the inequalities defining $D$ is therefore not an approximation or a graphical guess; it is an equivalent test for membership in the image.

Once the image is identified, the change-of-variables formula gives the density there. The absolute value of the inverse determinant accounts for how areas change under the map. If the transformation is not one-to-one on the original support, a single inverse and this single-branch calculation are not sufficient; separate inverse branches must be handled. The examples here use one-to-one transformations.

Common Exam Traps

Dropping a constraint. A triangular support is not a rectangle. Translate each inequality, including bounds that depend on another variable.
Using a boundary image as the whole answer. Mapped edges help identify the region, but inequalities are needed to establish which side of each edge is included.
Using the forward determinant. The displayed density formula uses the absolute determinant $\left|\partial(x,y)/\partial(u,v)\right|$, not its reciprocal. If the forward and inverse determinants are both nonzero, their absolute values are reciprocals.
Forgetting that the density is zero off the image. A formula obtained by substitution is only the transformed density on points that correspond to allowed original values.
Failing to check the result. Integrating over the transformed region should give $1$. A result different from $1$ often signals a missing support restriction, incorrect bounds, or a Jacobian error.

Key takeaway: To map a support, invert the transformation, substitute into every original support inequality, and use their intersection as the transformed region. Then apply the absolute inverse Jacobian there and set the density to zero outside it.

Check Your Understanding

For each problem, find the transformed support and joint density. The answers provide a check on both the mapped inequalities and the Jacobian.

Let $(X,Y)$ be uniform on $0<x<2,\ 0<y<1$, and set $U=X,\ V=Y+3X$. State the image and transformed density. (Answer: $0<u<2,\ 3u<v<3u+1;\ f_{U,V}(u,v)=1/2$ there, zero elsewhere.)
Let $(X,Y)$ be uniform on $0<x<1,\ 0<y<1-x$, and set $U=X,\ V=Y+X$. State the image and transformed density. (Answer: $0<u<1,\ u<v<1;\ f_{U,V}(u,v)=2$ there, zero elsewhere.)
Let $X$ and $Y$ be independent exponential variables with rates $2$ and $1$, respectively, and set $U=X,\ V=Y+X^2$. Find the image and density. (Answer: $u>0,\ v>u^2;\ f_{U,V}(u,v)=2e^{-2u-v+u^2}$ there, zero elsewhere.)
Let $(X,Y)$ be uniform on $0<x<1,\ 0<y<x$, and set $U=X,\ V=Y-X$. Find the image and density. (Answer: $0<u<1,\ -u<v<0;\ f_{U,V}(u,v)=2$ there, zero elsewhere.)
Let $(X,Y)$ be uniform on $-1<x<1,\ 0<y<1$, and set $U=X,\ V=Y+X$. Find the image and density. (Answer: $-1<u<1,\ u<v<u+1;\ f_{U,V}(u,v)=1/2$ there, zero elsewhere.)

What You'll Learn