Bivariate Nonlinear Transformations With Inverses

Why the Image and Inverse Matter

A nonlinear transformation changes both the values of a random vector and the region in which those values can occur. To find the transformed density, it is not enough to differentiate the formula for the transformation: we must identify the image region, solve for the inverse map there, and account for how the inverse changes area. The result is the bivariate version of the change-of-variables method, applied here to one-to-one transformations.

For a map $T$, write $(U,V)=T(X,Y)$. The inverse tells us which original point produced a given $(u,v)$, while the absolute determinant of its derivative adjusts density for the change in area. A correct calculation keeps these two tasks, as well as the transformed support, together.

The Bivariate Density Formula

Definition: Let $D$ and $E$ be open subsets of $\mathbb{R}^2$. A map $T:D\to E$ is a $C^1$ diffeomorphism if it is bijective, continuously differentiable, and has a continuously differentiable inverse $T^{-1}:E\to D$. Write $J_T(x,y)$ for the determinant of the derivative matrix of $T$ at $(x,y)$.

Suppose $(X,Y)$ has density $f_{X,Y}$, with probability one in $D$, and $T$ is a $C^1$ diffeomorphism from $D$ onto $E$. For $(u,v)\in E$, the transformed density is

f_{U,V}(u,v) = f_{X,Y}\!\left(T^{-1}(u,v)\right) \left|\det D T^{-1}(u,v)\right|.

Outside $E$, set $f_{U,V}(u,v)=0$. The absolute value is essential: a determinant can be negative if the transformation reverses orientation, but a density cannot be negative.

Theorem: Under the conditions above, the displayed formula is a density for $(U,V)=T(X,Y)$. At every point of $E$, $$ \left|\det D T^{-1}(u,v)\right| = \frac{1}{\left|J_T\!\left(T^{-1}(u,v)\right)\right|}. $$

Proof. The identity map on $D$ is $T^{-1}\circ T$. By the chain rule,

D T^{-1}(T(x,y))\,D T(x,y)=I_2.

Taking determinants gives $\det D T^{-1}(T(x,y))\det D T(x,y)=1$. Since $T$ is onto $E$, this yields the stated reciprocal identity at every point of $E$. To verify the density formula, take any Borel set $A\subseteq\mathbb{R}^2$. By the change-of-variables theorem for integrals, applied to the diffeomorphism $T$,

\begin{aligned} P\big((U,V)\in A\big) &=\int_{T^{-1}(A\cap E)} f_{X,Y}(x,y)\,dx\,dy\\ &=\int_{A\cap E} f_{X,Y}\!\left(T^{-1}(u,v)\right) \left|\det D T^{-1}(u,v)\right|\,du\,dv. \end{aligned}

Thus integrating the proposed density over any Borel set gives the probability that $(U,V)$ belongs to that set. It is nonnegative, and taking $A=\mathbb{R}^2$ shows that it integrates to one. Therefore it is a density of $(U,V)$. $\square$

In practice, it is often simpler to differentiate the inverse than to differentiate the forward map and take a reciprocal. Either way, specify the region where the inverse is valid: the same algebraic expression can have a different image or fail to be one-to-one on another domain.

Worked Transformations

Worked Example: Coordinatewise Logarithms

Let $X$ and $Y$ be independent exponential random variables with rate $1$, so $f_{X,Y}(x,y)=e^{-x-y}$ for $x>0,y>0$, and zero elsewhere. Set $U=\log X$ and $V=\log Y$. Find the image, inverse, and joint density of $(U,V)$.

The original support is $D=(0,\infty)^2$. Taking logarithms maps each positive coordinate onto the whole real line, so $E=\mathbb{R}^2$. The inverse and its derivative matrix are

x=e^u,\qquad y=e^v,\qquad D T^{-1}(u,v)= \begin{pmatrix} e^u&0\\ 0&e^v \end{pmatrix}.

The absolute determinant is $e^{u+v}$. Substitution into the density formula gives, for all real $u,v$,

f_{U,V}(u,v) =e^{-e^u-e^v}e^{u+v} =\exp\!\left(u+v-e^u-e^v\right).

As a check, the density factors as $e^{u-e^u}e^{v-e^v}$. Each factor integrates to one: with $t=e^u$, $du=dt/t$, so $\int_{-\infty}^{\infty} e^{u-e^u}\,du=\int_0^\infty e^{-t}\,dt=1$; the same calculation applies to $v$. Thus the joint density integrates to one on $\mathbb{R}^2$.

Worked Example: A Ratio Coordinate

Let $X,Y$ again be independent exponential random variables with rate $1$. Define $U=X$ and $V=Y/X$. Find the transformed density and verify its integral.

Since $x>0,y>0$, the image is $u>0,v>0$. The inverse is $x=u,\ y=uv$, and its derivative matrix has determinant

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

The original joint density at the inverse point is $e^{-u-uv}$. Therefore

f_{U,V}(u,v)=u e^{-u(1+v)},\qquad u>0,\ v>0,

and the density is zero elsewhere. Integrating first with respect to $v$ gives

\int_0^\infty\int_0^\infty u e^{-u(1+v)}\,dv\,du =\int_0^\infty u e^{-u}\left(\int_0^\infty e^{-uv}\,dv\right)du =\int_0^\infty e^{-u}\,du =1,

because, for fixed $u>0$, $\int_0^\infty e^{-uv}\,dv=1/u$. The factor $u$ from the Jacobian cancels that integral’s $1/u$; omitting the Jacobian would not give the correct density.

Worked Example: A Polynomial Shear and Its Image

Suppose $X$ and $Y$ are independent Uniform$(0,1)$ variables, so their joint density is $1$ on $0<x<1,\ 0<y<1$. Transform them by $U=X,\ V=Y+X^2$. Determine the image and density.

From $u=x$ and $v=y+x^2$, the inverse is $x=u,\ y=v-u^2$. The original inequalities $0<x<1$ and $0<y<1$ become

0<u<1,\qquad u^2<v<u^2+1.

These inequalities describe the image $E$, not the entire plane. The inverse derivative and its determinant are

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

Consequently the transformed density is $1$ on $E$ and zero outside $E$. Its integral confirms the support and normalization:

\int_0^1\int_{u^2}^{u^2+1}1\,dv\,du =\int_0^1\big((u^2+1)-u^2\big)\,du =1.

Although the Jacobian factor is one here, the transformation still changes the support: at each $u$, $v$ lies in an interval shifted upward by $u^2$. A unit Jacobian does not mean the image region can be ignored.

A Reliable Calculation and a Common Pitfall

For a nonlinear bijection, keep the geometric and algebraic parts of the calculation in a fixed order. This makes support restrictions visible before the density is written down.

Transform the support.
Translate the original inequalities into conditions on $(u,v)$ to describe the image $E$.

Solve for the inverse.
Express $x$ and $y$ in terms of $u$ and $v$, and check that this inverse is valid throughout $E$.

Compute the inverse Jacobian.
Differentiate the inverse and take the absolute value of its determinant.

Substitute and restrict.
Evaluate the original density at the inverse point, multiply by the Jacobian factor, and set the result to zero outside $E$.

Check the result.
Integrate over $E$ when feasible; the result must be one.

A common mistake is to find a plausible inverse and Jacobian but then use the transformed formula on all of $\mathbb{R}^2$. In the polynomial-shear example, this would assign positive density to points with $u\leq 0$, $u\geq 1$, or $v$ outside $(u^2,u^2+1)$, none of which can occur. Another mistake is to multiply by $\left|\det D T\right|$ after already using the inverse: the formula uses $\left|\det D T^{-1}\right|$, or equivalently the reciprocal absolute determinant of the forward map evaluated at the inverse point.

Key takeaway: A one-to-one bivariate transformation requires three coordinated pieces: its image region, its inverse, and the absolute inverse-Jacobian determinant. The density formula is correct only when all three refer to the same domain.

Check Your Understanding

For each question, pay attention to both the transformation and the region where its inverse is valid.

For $U=\log X,\ V=Y$ with $X>0$ and $Y$ unrestricted, what is the image region?
If a transformation has inverse $x=u,\ y=v/u$ on its image, what absolute Jacobian factor belongs in the density formula?
For $U=X,\ V=Y+X^2$ with $0<X<1,\ 0<Y<2$, describe the image inequalities.
Why must a density obtained by transformation be set to zero outside the image of the original support?
In the ratio transformation $U=X,\ V=Y/X$, what is the inverse-Jacobian determinant and on what region is it positive?

Bivariate Nonlinear Transformations With Inverses

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