Worked Problems in One-to-One Distribution Transformations

A Reliable Method for Transformation Problems

A transformation problem is not complete when a derivative or determinant has been calculated. The transformed support must also be correct, and the resulting density should agree with the distribution obtained directly from a CDF or integrate to one. These checks are particularly useful when a transformation reverses order or maps a simple support to a nonrectangular region.

For a one-to-one transformation, a dependable sequence is: identify the original support, find the image support, solve for the inverse, calculate the derivative or Jacobian of that inverse, and only then write the transformed density. The examples below use this sequence in both one and two dimensions.

Definition: Let $X$ have density $f_X$ on an interval $I$, and let $g:I\to J$ be a one-to-one transformation with image $J$. Write $h=g^{-1}$. If $g$ is continuously differentiable and $g'(x)\ne 0$ on $I$, then the transformed variable $Y=g(X)$ has density $$ f_Y(y)=f_X(h(y))|h'(y)|,\qquad y\in J, $$ and density zero outside $J$. Equivalently, the derivative factor is $1/|g'(h(y))|$. Density values at individual endpoints do not affect the distribution.

The absolute value is essential: a decreasing transformation reverses the direction of intervals, but a density cannot be negative. The inverse derivative appears because the transformed density is evaluated at a value $y$, which must first be mapped back to $x=h(y)$.

Worked One-Dimensional Transformations

Worked Example: A Reciprocal Transformation

Suppose $X$ has density $f_X(x)=2x$ for $0<x<1$, and define $Y=1/X$. Find the support and density of $Y$, and check the answer using its CDF.

The original support is $0<x<1$. Taking reciprocals maps it to $1<y<\infty$. The inverse transformation is $h(y)=1/y$, with derivative $h'(y)=-1/y^2$. Therefore,

f_Y(y)=f_X(1/y)\left|-\frac{1}{y^2}\right| =\frac{2}{y}\frac{1}{y^2} =\frac{2}{y^3},\qquad y>1.

To check by a CDF calculation, if $y\leq 1$, then $P(Y\leq y)=0$. If $y>1$, the reciprocal is decreasing, so $1/X\leq y$ is equivalent to $X\geq 1/y$. Thus

F_Y(y)=P(X\geq 1/y) =\int_{1/y}^{1}2x\,dx =1-\frac{1}{y^2},\qquad y>1.

This CDF tends to zero as $y$ decreases to 1 and to one as $y$ increases without bound. Its derivative on $y>1$ is $2/y^3$, agreeing with the transformation formula. The density also integrates to one: $\int_1^\infty 2y^{-3}\,dy=1$.

Worked Example: The Logarithm of an Exponential Variable

Let $X\sim\operatorname{Exponential}(2)$, with density $f_X(x)=2e^{-2x}$ for $x>0$. Set $Y=\log X$. Find the support, density, and CDF of $Y$.

The logarithm maps the positive half-line onto the real line, so $Y$ has support $\mathbb{R}$. The inverse is $h(y)=e^y$, and $h'(y)=e^y$. Applying the inverse-derivative formula gives

f_Y(y)=2e^{-2e^y}e^y,\qquad -\infty<y<\infty.

For an independent CDF check, monotonicity of the logarithm gives

F_Y(y)=P(\log X\leq y) =P(X\leq e^y) =1-e^{-2e^y}.

This expression tends to zero as $y\to-\infty$ and to one as $y\to\infty$. It is differentiable for every real $y$, and its derivative is $2e^y e^{-2e^y}$, the density just obtained. In addition, integrating that density over the real line gives the CDF’s total increase, $1-0=1$.

Why the One-Dimensional Formula Is Correct

The inverse-derivative formula follows from how probabilities of sets transform. Suppose $X$ has an integrable density on $I$, and $g$ satisfies the conditions in the definition. For any Borel set $A\subseteq J$, the event $Y\in A$ is the event $X\in h(A)$. The one-dimensional change-of-variables theorem, applied on the interval $J$, gives

P(Y\in A) =\int_{h(A)}f_X(x)\,dx =\int_A f_X(h(y))|h'(y)|\,dy.

This establishes the proposed function as a density: it gives the probability of every Borel set, not merely a derivative calculation at selected points. Taking $A=J$ also verifies normalization. Since $h(J)=I$,

\int_J f_X(h(y))|h'(y)|\,dy =\int_I f_X(x)\,dx =1.

The CDF can likewise be obtained from probabilities of intervals. If $g$ is increasing, then $Y\leq y$ is equivalent to $X\leq h(y)$; if $g$ is decreasing, it is equivalent to $X\geq h(y)$. Differentiating these CDF expressions is another way to recover the density where the derivative exists, but such differentiation need only hold almost everywhere. A density need not be continuous, and its CDF need not be differentiable at every point. The set-based change-of-variables argument does not require pointwise differentiability of the CDF.

A Bivariate Transformation

Worked Example: A Sum Coordinate and Its Support

Let $X$ and $Y$ be independent $\operatorname{Exponential}(1)$ random variables. Define $U=X$ and $V=X+Y$. Find the joint density and verify it by calculating the joint CDF.

The original support is $x>0,\ y>0$. Solving the transformation equations gives $x=u$ and $y=v-u$. Consequently, the transformed support is $0<u<v<\infty$: the condition $y>0$ becomes $v>u$, while $x>0$ becomes $u>0$.

The inverse map is $(u,v)\mapsto(u,v-u)$, whose Jacobian matrix and determinant are

\frac{\partial(x,y)}{\partial(u,v)} = \begin{pmatrix} 1&0\\ -1&1 \end{pmatrix}, \qquad \left|\det\frac{\partial(x,y)}{\partial(u,v)}\right|=1.

The original joint density is $f_{X,Y}(x,y)=e^{-x-y}$ on the positive quadrant, by independence. Substituting the inverse and its Jacobian gives

f_{U,V}(u,v) =e^{-u-(v-u)}\cdot 1 =e^{-v},\qquad 0<u<v<\infty.

This density integrates to one over its wedge-shaped support:

\int_0^\infty\int_u^\infty e^{-v}\,dv\,du =\int_0^\infty e^{-u}\,du =1.

For a CDF check, let $a>0$ and $b>0$, and put $m=\min(a,b)$. The event $U\leq a,\ V\leq b$, expressed in the original variables, is $0<X\leq a$ and $X+Y\leq b$. For a fixed $X=x$, this requires $0<x<m$ and $0<Y\leq b-x$. Independence therefore gives

\begin{aligned} P(U\leq a,V\leq b) &=\int_0^m e^{-x}\left(\int_0^{b-x}e^{-y}\,dy\right)dx\\ &=\int_0^m(e^{-x}-e^{-b})\,dx\\ &=1-e^{-m}-m e^{-b}. \end{aligned}

Integrating the transformed density over $0<u\leq a,\ u<v\leq b$ gives $\int_0^m\int_u^b e^{-v}\,dv\,du$, which is the same expression. If either $a\leq0$ or $b\leq0$, the joint CDF is zero. This check confirms both the transformed density and the nonrectangular support.

The Multivariate Formula and Its Proof

Theorem: Let $D,E$ be open subsets of $\mathbb{R}^k$, and let $g:D\to E$ be a continuously differentiable bijection with a continuously differentiable inverse $h:E\to D$. Suppose the random vector $X$ has density $f_X$ on $D$. Then $Z=g(X)$ has density $$ f_Z(z)=f_X(h(z))\left|\det Dh(z)\right|,\qquad z\in E, $$ and density zero outside $E$.

To prove the result, take any Borel set $A\subseteq E$. The event $Z\in A$ is $X\in h(A)$. The multivariate change-of-variables theorem applies because $g$ and $h$ are continuously differentiable inverses on open sets. It yields

P(Z\in A) =\int_{h(A)}f_X(x)\,dx =\int_A f_X(h(z))|\det Dh(z)|\,dz.

Thus the displayed function gives the probability of every Borel set and is a density of $Z$. Setting $A=E$ shows that its integral over $E$ is $\int_D f_X(x)\,dx=1$. This proves both the formula and its normalization. $\square$

The determinant in this result measures local volume change for the inverse map. It is the inverse Jacobian because the original density is expressed in $x$-coordinates while the transformed density is expressed in $z$-coordinates. As in one dimension, the support is the image of the original support; it must be determined from the transformation rather than guessed from the formula.

Common Mistakes and Practical Checks

Using the forward derivative in the inverse formula. In the reciprocal example, $h(y)=1/y$, so the correct factors are $f_X(h(y))=2/y$ and $|h'(y)|=1/y^2$, yielding $2/y^3$. Using the forward derivative at the inverse instead would give $|g'(h(y))|=y^2$, and the incorrect product $(2/y)y^2=2y$.
Keeping the original support unchanged. A transformation changes the range of possible values. Find the image of the support before writing the density, as the reciprocal and bivariate examples illustrate.
Dropping the absolute value. For a decreasing map, the inverse derivative is negative. The density uses its absolute value, not its signed value.
Assuming the support is rectangular. A bijection can map a simple support to a wedge or another constrained region. Integrating over the wrong region can invalidate both the density and its normalization check.
Claiming a CDF derivative exists everywhere. A density formula identifies a density up to sets of measure zero. When a CDF is not differentiable at some points, the density equality is interpreted almost everywhere, not as a pointwise identity at every point.

These checks matter because a plausible-looking algebraic expression can still assign probability to impossible values, have the wrong scale, or fail to integrate to one. Verifying a CDF in one dimension or integrating over the actual image support in several dimensions makes the calculation auditable.

Check Your Understanding

For each problem, identify the transformed support before calculating the density.

If $X$ has density $1$ on $0<x<1$ and $Y=1-X$, what are the inverse map and density of $Y$?
For $Y=\log X$ with $X\sim\operatorname{Exponential}(2)$, derive $F_Y(y)$ directly from a probability statement.
In a one-to-one transformation, why is the derivative factor evaluated at the inverse image of the transformed value?
For the bivariate transformation $U=X,\ V=X+Y$ with positive $X,Y$, explain why the transformed support requires $v>u>0$.
Why does checking that a proposed density integrates to one not, by itself, verify that its support and CDF are correct?