Triangular Transformations and Jacobian Structure

Why Triangular Structure Helps

A multivariate change of variables can look complicated because each output coordinate may depend on several input coordinates. In a triangular transformation, that dependence is organized: each coordinate depends only on itself and coordinates earlier in the chosen ordering. This structure makes the Jacobian determinant a product rather than a general determinant, and can make a density transformation a sequence of one-dimensional substitutions.

There is an important distinction between a triangular derivative matrix and a globally triangular description of the inverse coordinates. The determinant calculation needs only the derivative matrix and the chain rule. A sequential coordinate-by-coordinate solution for the inverse requires additional conditions on the domain and on the coordinate maps.

Triangular Maps and Their Jacobians

Definition: Let \(U\) be an open subset of \(\mathbb{R}^d\), and let \(T:U\to\mathbb{R}^d\) have coordinates \(T(x)=(g_1(x),\ldots,g_d(x))\). The map is lower triangular if \(g_i(x)\) depends only on \(x_1,\ldots,x_i\), for each \(i\). It is upper triangular if \(g_i(x)\) depends only on \(x_i,\ldots,x_d\). When the component functions are continuously differentiable, the Jacobian matrix \(DT(x)\) is correspondingly lower or upper triangular.

For a lower triangular map, the diagonal entries of \(DT(x)\) are \(\partial g_i/\partial x_i\). The determinant of a triangular matrix is the product of its diagonal entries, so the determinant formula is

\det DT(x)=\prod_{i=1}^d \frac{\partial g_i}{\partial x_i}(x).

The same formula holds for an upper triangular map. The diagonal derivatives must all be nonzero at a point for the Jacobian matrix to be nonsingular there.

Theorem: If \(T\) is continuously differentiable and triangular on an open set \(U\), then at each \(x\in U\), \(\det DT(x)\) is the product of its diagonal entries. If all those entries are nonzero, then

\left|\det DT(x)\right| = \prod_{i=1}^d \left|\frac{\partial g_i}{\partial x_i}(x)\right|.

Proof. Consider first a lower triangular matrix \(A=(a_{ij})\). In the permutation formula for its determinant, a nonzero term can occur only for a permutation \(\pi\) such that \(a_{i,\pi(i)}\ne 0\) for every \(i\). Lower triangularity requires \(\pi(i)\le i\) for every \(i\). Since \(\pi\) is a permutation, \(\sum_i\pi(i)=\sum_i i\). The inequalities \(\pi(i)\le i\) can therefore all hold only if \(\pi(i)=i\) for every \(i\). Thus only the identity-permutation term remains, and \(\det A=\prod_i a_{ii}\). For an upper triangular matrix the same argument uses \(\pi(i)\ge i\), again leaving only the identity permutation. Applying these facts to \(DT(x)\) proves the formula. Taking absolute values gives the second display. \(\square\)

Inverse Jacobians: Use the Chain Rule

Suppose \(T:U\to V\) is a bijection between open sets, \(T\) is continuously differentiable, and its inverse \(T^{-1}:V\to U\) is continuously differentiable. At \(y=T(x)\), differentiating \(T^{-1}(T(x))=x\) gives

D(T^{-1})(y)\,DT(x)=I, \qquad D(T^{-1})(y)=[DT(x)]^{-1}.

If \(DT(x)\) is triangular and nonsingular, its matrix inverse is triangular with diagonal entries equal to the reciprocals of the original diagonal entries. This is a matrix statement; it does not require assuming that each inverse coordinate is globally a function only of the first or last few output coordinates.

Result: For a differentiable triangular bijection with differentiable inverse and nonzero diagonal derivatives,

\left|\det D(T^{-1})(y)\right| = \frac{1}{|\det DT(x)|} = \prod_{i=1}^d \left|\frac{\partial g_i}{\partial x_i}(x)\right|^{-1}, \qquad x=T^{-1}(y).

To see the diagonal statement directly for a lower triangular matrix \(A\), solve \(Az=e_j\), where \(e_j\) is the \(j\)th coordinate vector. In rows \(i<j\), the right side is zero. Inductively, those equations give \(z_i=0\), since each diagonal entry is nonzero. In row \(j\), the equation then gives \(z_j=1/a_{jj}\). These are the entries in column \(j\) of \(A^{-1}\): entries above its diagonal are zero and its \(j\)th diagonal entry is \(1/a_{jj}\). The upper triangular case follows by solving from the last row upward. Taking determinants in \(D(T^{-1})(y)=[DT(x)]^{-1}\) gives the displayed inverse determinant formula.

Worked Example: A Two-Coordinate Shear

Let \(X_1,X_2\) be independent Uniform\((0,1)\) random variables and set \(Y_1=X_1\), \(Y_2=X_1+X_2\). Find the joint density of \((Y_1,Y_2)\).

The transformation is lower triangular. Its Jacobian matrix and determinant are

DT(x_1,x_2)= \begin{pmatrix} 1&0\\ 1&1 \end{pmatrix}, \qquad \det DT(x_1,x_2)=1.

The inverse is \(x_1=y_1\), \(x_2=y_2-y_1\), with inverse-Jacobian determinant also equal to \(1\). The original support is \(0<x_1<1,\ 0<x_2<1\). Substituting the inverse coordinates gives the image support

0<y_1<1,\qquad y_1<y_2<y_1+1.

The original joint density is \(f_X(x_1,x_2)=1\) on the unit square and zero elsewhere. Therefore,

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

As a check, the image has area \(\int_0^1\int_{y_1}^{y_1+1}dy_2\,dy_1=1\), so the transformed density integrates to one.

Density Transformation and Sequential Substitution

The multivariate change-of-variables formula applies when \(T:U\to V\) is a continuously differentiable bijection between open sets with continuously differentiable inverse and nonzero Jacobian determinant. If \(X\) has density \(f_X\) on \(U\) and \(Y=T(X)\), then

f_Y(y)=f_X(T^{-1}(y))\left|\det D(T^{-1})(y)\right|, \qquad y\in V.

For a triangular map, the inverse determinant factor is the product of the reciprocals of the diagonal derivatives, evaluated at \(x=T^{-1}(y)\). The density is zero outside the image of the support. Thus the determinant may be simple while identifying the image support still requires care.

Sometimes the transformation can also be performed one coordinate at a time. Here is a useful sufficient setup for a lower triangular map: the domain is described successively by intervals for \(x_1\), then \(x_2\) given \(x_1\), and so on; and, after fixing \(x_1,\ldots,x_{i-1}\), the function \(x_i\mapsto g_i(x_1,\ldots,x_i)\) is a one-to-one differentiable map on its allowed interval with nonzero derivative. Under these conditions, one can solve recursively for each \(x_i\) from the earlier coordinates and \(y_i\), and apply the one-dimensional substitution formula at each step. For nonnegative integrands, Tonelli's theorem justifies the successive integrations; for integrable signed integrands, Fubini's theorem does so. The product of the one-dimensional derivative factors is exactly the triangular Jacobian factor.

The following example illustrates both the inverse determinant and the sequential route. Its support has exactly the interval structure just described.

Worked Example: Sequential Substitution for a Product Coordinate

Let \(X_1\) and \(X_2\) be independent exponential random variables with rates \(2\) and \(3\), respectively. Define \(Y_1=X_1\) and \(Y_2=X_1X_2\). Find the joint density of \((Y_1,Y_2)\).

On the support \(x_1>0,\ x_2>0\), the map is one-to-one onto \(y_1>0,\ y_2>0\), with inverse \(x_1=y_1,\ x_2=y_2/y_1\). The forward derivative matrix is triangular:

DT(x_1,x_2)= \begin{pmatrix} 1&0\\ x_2&x_1 \end{pmatrix}, \qquad \det DT(x_1,x_2)=x_1.

Thus the absolute inverse determinant is \(1/y_1\). Since \(f_X(x_1,x_2)=6e^{-2x_1-3x_2}\) for positive \(x_1,x_2\), the transformed density is

f_Y(y_1,y_2)= \frac{6}{y_1} \exp\left(-2y_1-\frac{3y_2}{y_1}\right), \qquad y_1>0,\ y_2>0.

The same factor appears from sequential substitutions. First set \(y_1=x_1\). With \(x_1\) fixed and positive, set \(y_2=x_1x_2\), so \(dx_2=dy_2/x_1=dy_2/y_1\). The joint differential factor is consequently \(dx_1dx_2=dy_1dy_2/y_1\). As a normalization check, integrating first over \(y_2\) gives

\int_0^\infty \frac{6}{y_1}e^{-2y_1-3y_2/y_1}\,dy_2 =2e^{-2y_1}, \qquad \int_0^\infty 2e^{-2y_1}\,dy_1=1.

Worked Example: A Three-Dimensional Nonlinear Map

Let \(X_1,X_2,X_3\) be independent \(N(0,1)\) random variables. Define

Y_1=e^{X_1},\qquad Y_2=X_1+X_2,\qquad Y_3=e^{X_2}X_3.

The map is lower triangular: each \(Y_i\) depends only on \(X_1,\ldots,X_i\). Its diagonal derivatives are \(e^{x_1},1,e^{x_2}\), giving

\det DT(x)=e^{x_1+x_2}.

The map is a bijection from \(\mathbb{R}^3\) to \((0,\infty)\times\mathbb{R}^2\). Indeed, for any point in that image, the inverse is

x_1=\log y_1,\qquad x_2=y_2-\log y_1,\qquad x_3=y_3e^{-x_2}=y_1y_3e^{-y_2}.

At this inverse point, \(e^{x_1+x_2}=e^{y_2}\), so the absolute inverse-Jacobian determinant is \(e^{-y_2}\). The standard normal joint density at \(x\) is \((2\pi)^{-3/2}\exp[-(x_1^2+x_2^2+x_3^2)/2]\). Substituting the inverse coordinates yields

f_Y(y_1,y_2,y_3) = \frac{e^{-y_2}}{(2\pi)^{3/2}} \exp\left[ -\frac{1}{2} \left\{ (\log y_1)^2 +(y_2-\log y_1)^2 +(y_1y_3e^{-y_2})^2 \right\} \right], \quad y_1>0,\ y_2,y_3\in\mathbb{R}.

The density is zero when \(y_1\le 0\). The image support follows from the inverse: every positive \(y_1\) and every real pair \((y_2,y_3)\) gives a real inverse point. The determinant factor can also be read from the inverse derivative matrix as \(e^{-y_2}\).

Why the Inverse-Coordinate Caution Matters

A common mistake is to infer that a lower triangular forward map automatically has a globally lower triangular inverse in the sense that \(x_i\) depends only on \(y_1,\ldots,y_i\). This inference can fail if the domain has disconnected pieces. For example, on

D=\bigl((-2,-1)\times(0,1)\bigr)\cup\bigl((1,2)\times(2,3)\bigr),

the map \(T(x_1,x_2)=(x_1^2,x_2)\) is one-to-one and has a lower triangular, nonsingular derivative matrix on \(D\). Yet at a fixed \(y_1\) in the appropriate overlapping squared ranges, the sign of \(x_1\) is determined by which image component contains \(y_2\); \(x_1\) is not globally determined by \(y_1\) alone. The derivative-based determinant result remains valid: use \(D(T^{-1})(y)=[DT(x)]^{-1}\) at the point \(x=T^{-1}(y)\), rather than assuming global inverse-coordinate dependence.

A second pitfall is to compute the product of diagonal derivatives and stop without transforming the support. The density formula requires both the inverse substitution and the absolute Jacobian factor, and it applies on the transformed support. Even a determinant equal to one does not mean the new density is one everywhere; it is zero off the image of the original support.

Key takeaway: For a triangular map, the Jacobian determinant is the product of the diagonal derivatives. For an invertible differentiable map, the inverse-Jacobian determinant is the reciprocal in absolute value. Use sequential coordinate substitutions only when the domain and coordinate maps support that recursive procedure; otherwise rely on the change-of-variables formula and the chain rule.

Check Your Understanding

Use the derivative structure, inverse map, and support together when answering.

For a lower triangular map in four dimensions, what expression gives its Jacobian determinant?
Why does the chain rule give a reliable inverse-Jacobian formula even if the inverse coordinates do not have a global triangular dependence?
For \(Y_1=X_1\) and \(Y_2=X_1+2X_2\), find the determinant and state the inverse-Jacobian determinant.
In a density transformation, why is it not enough to substitute the inverse coordinates and omit the support?
What extra domain and one-coordinate conditions make sequential substitutions valid for a lower triangular map?