Many-to-One Transformations and Multiple Preimages

Why Multiple Preimages Matter

A differentiable transformation can send different input values to the same output. For example, squaring sends both $x$ and $-x$ to $x^2$. A density calculation that follows only one of those input values misses probability mass. The correct calculation adds the contributions from all valid preimages, weighting each by the absolute Jacobian determinant of its inverse branch.

The key is to organize the input space into regions on which the transformation is one-to-one. The usual change-of-variables formula applies on each such region; the transformed density is the sum of those branch-specific densities. The regions must be disjoint, so that an input point is counted exactly once.

Definition and Density Formula

Definition: Let $T:\Omega\to\mathbb{R}^d$ be a transformation on an open set $\Omega\subseteq\mathbb{R}^d$. A collection of inverse branches consists of restrictions of $T$ to pairwise disjoint open input regions $\Omega_j$, such that $T$ is one-to-one on each $\Omega_j$ and has a continuously differentiable inverse $g_j:T(\Omega_j)\to\Omega_j$. For an output $y\in T(\Omega_j)$, $g_j(y)$ is the preimage of $y$ in that branch. Because the input regions are disjoint, valid inverse points from different branches are distinct.

Here is a standard set of sufficient conditions for the density formula. Let $X$ be a random vector in $\mathbb{R}^d$ with density $f_X$, and let $Y=T(X)$. Suppose the input regions $\Omega_j$ are finite or countably many, pairwise disjoint open sets whose union omits only a set $N$ with $P(X\in N)=0$. Suppose also that, on each $\Omega_j$, $T$ is continuously differentiable, one-to-one, and has a continuously differentiable inverse $g_j$ on the open image $V_j=T(\Omega_j)$. In particular, the derivative matrix of each inverse is nonsingular on $V_j$.

Theorem: Under these conditions, $Y=T(X)$ has density $$ f_Y(y)=\sum_{j:\,y\in V_j} f_X(g_j(y))\left|\det Dg_j(y)\right| $$ for almost every $y\in\mathbb{R}^d$. Equivalently, one may include an indicator $1_{\{y\in V_j\}}$ in each term and sum over all branches. The sum includes every valid inverse point, with one contribution per branch.

Different image regions $V_j$ may overlap: that is precisely where multiple preimages occur. Their inverse points are nevertheless distinct because they belong to disjoint input regions. The formula is an almost-everywhere statement; values assigned to a density on an output set of probability zero do not change the distribution.

Why the Contributions Add

The proof applies the ordinary bijective change-of-variables formula separately on each input region, then adds the probabilities. It also shows why the partition condition matters: the input regions form disjoint pieces, so their probabilities add without overlap.

Start with an output event.
For a measurable set $A\subseteq\mathbb{R}^d$, calculate $P(Y\in A)=P(T(X)\in A)$.

Partition the input.
Up to the probability-zero set $N$, the event splits into disjoint pieces inside the regions $\Omega_j$.

Change variables on each branch.
Use the inverse $g_j$ and its absolute Jacobian determinant, then add the resulting integrals.

More explicitly, because $Y=T(X)$ and $P(X\in N)=0$,

P(Y\in A) =\sum_j\int_{\Omega_j}1_{\{T(x)\in A\}}f_X(x)\,dx.

On $\Omega_j$, set $y=T(x)$, so $x=g_j(y)$. The bijective change-of-variables formula gives

\int_{\Omega_j}1_{\{T(x)\in A\}}f_X(x)\,dx =\int_{V_j}1_{\{y\in A\}}f_X(g_j(y))|\det Dg_j(y)|\,dy.

The summands are nonnegative, so their countable sum can be interchanged with integration by Tonelli’s theorem. Therefore

P(Y\in A) =\int_A\sum_{j:\,y\in V_j}f_X(g_j(y))|\det Dg_j(y)|\,dy.

This is the defining integral property of the stated density. In particular, taking $A=\mathbb{R}^d$ shows that the formula integrates to one, since the input regions cover $X$ with probability one. $\square$

The Scalar Version

For a scalar random variable with density $f_X$, suppose the real line, up to a set of probability zero, is partitioned into intervals on which $T$ is continuously differentiable, one-to-one, and has nonzero derivative. If $g_j$ is the inverse on the image of interval $j$, then $Dg_j(y)=g'_j(y)$. Differentiating $T(g_j(y))=y$ gives $T'(g_j(y))g'_j(y)=1$, hence

f_Y(y)=\sum_{j:\,y\in T(\Omega_j)} \frac{f_X(g_j(y))}{|T'(g_j(y))|}.

The absolute value is essential: an inverse branch can be decreasing, but its density contribution must be nonnegative. Points where the derivative vanishes can be placed in the omitted set only when that set has probability zero; the density formula is then asserted almost everywhere, not necessarily at the corresponding output values.

Worked Example: Squaring a Standard Normal Variable

Let $X\sim N(0,1)$ and $Y=X^2$. Find the density of $Y$.

Partition and inverses. The map $T(x)=x^2$ is one-to-one on $\Omega_1=(-\infty,0)$ and on $\Omega_2=(0,\infty)$. The excluded point $0$ has probability zero. For $y>0$, the inverse points are $g_1(y)=-\sqrt y$ and $g_2(y)=\sqrt y$, with $|g'_j(y)|=1/(2\sqrt y)$ on both branches.

The standard normal density is $f_X(x)=(2\pi)^{-1/2}e^{-x^2/2}$. Thus each branch contributes

f_X(\pm\sqrt y)\frac{1}{2\sqrt y} =\frac{1}{2\sqrt{2\pi y}}e^{-y/2},\qquad y>0.

Adding the two distinct preimage contributions gives

f_Y(y)= \begin{cases} \dfrac{1}{\sqrt{2\pi y}}e^{-y/2},&y>0,\\[6pt] 0,&y<0. \end{cases}

The value at $y=0$ can be assigned arbitrarily without changing the distribution. The density is unbounded as $y$ approaches zero from above, but it is integrable. Indeed, substituting $y=x^2$ shows $\int_0^\infty (2\pi y)^{-1/2}e^{-y/2}\,dy=1$.

Worked Example: Absolute Value of a Shifted Normal Variable

Let $X\sim N(1,1)$ and $Y=|X|$. Find the density of $Y$.

The branches are the negative and positive half-lines, excluding $0$, which has probability zero. For $y>0$, the two inverse points are $-y$ and $y$. The absolute derivative of either inverse is $1$. The normal density with mean $1$ and variance $1$ is $f_X(x)=(2\pi)^{-1/2}e^{-(x-1)^2/2}$, so

f_Y(y)=f_X(y)+f_X(-y) =\frac{1}{\sqrt{2\pi}}\left[ e^{-(y-1)^2/2}+e^{-(y+1)^2/2} \right],\qquad y>0.

For $y<0$, the density is zero, since an absolute value cannot be negative. To check the normalization, substitute $x=y$ in the first integral and $x=-y$ in the second:

\int_0^\infty f_X(y)\,dy+\int_0^\infty f_X(-y)\,dy =\int_0^\infty f_X(x)\,dx+\int_{-\infty}^0 f_X(x)\,dx=1.

The two terms need not be equal: the shifted normal distribution assigns different amounts of probability to positive and negative inputs.

A Bivariate Transformation

Worked Example: Squaring One Coordinate

Let $X_1$ and $X_2$ be independent standard normal variables, and define $Y_1=X_1^2$ and $Y_2=X_2$. Find the joint density of $Y=(Y_1,Y_2)$.

The joint input density is $f_X(x_1,x_2)=(2\pi)^{-1}e^{-(x_1^2+x_2^2)/2}$. Partition the input plane into $x_1<0$ and $x_1>0$; the omitted line $x_1=0$ has probability zero. For $y_1>0$, the two inverse branches are $g_\pm(y_1,y_2)=(\pm\sqrt{y_1},y_2)$. Their derivative matrices have determinants $\pm1/(2\sqrt{y_1})$, so both absolute inverse Jacobians are $1/(2\sqrt{y_1})$.

At either inverse point, $x_1^2=y_1$ and $x_2=y_2$. Each branch therefore contributes

\frac{1}{2\pi}e^{-(y_1+y_2^2)/2}\frac{1}{2\sqrt{y_1}}.

Adding both contributions yields

f_Y(y_1,y_2)= \begin{cases} \dfrac{1}{2\pi\sqrt{y_1}}e^{-(y_1+y_2^2)/2},&y_1>0,\ y_2\in\mathbb{R},\\[6pt] 0,&y_1<0. \end{cases}

Its integral is one. In fact, the integral factors as

\frac{1}{2\pi} \left(\int_0^\infty y_1^{-1/2}e^{-y_1/2}\,dy_1\right) \left(\int_{-\infty}^\infty e^{-y_2^2/2}\,dy_2\right) =\frac{1}{2\pi}\sqrt{2\pi}\sqrt{2\pi}=1.

For the first integral, setting $y_1=t^2$ gives $2\int_0^\infty e^{-t^2/2}\,dt=\sqrt{2\pi}$; the second is the standard normal integral. The output density has a singularity as $y_1$ approaches zero, but the calculation confirms that it is integrable.

Common Mistakes and Practical Checks

Counting a branch twice. It is not enough to list functions that happen to invert the transformation on some set. If two listed branches return the same input point, adding both terms double-counts that preimage. Define branches from disjoint input regions, as in the theorem, or otherwise establish that the inverse points are distinct.
Keeping only one preimage. For $x^2=y>0$, both $\sqrt y$ and $-\sqrt y$ contribute. Omitting one loses probability mass.
Using a forward Jacobian in the wrong direction. The density contribution uses $|\det Dg_j(y)|$, the inverse Jacobian. In the scalar case this is $1/|T'(g_j(y))|$, not $|T'(g_j(y))|$.
Ignoring the support. A transformation may only produce outputs in part of the real line or vector space. State the density as zero outside the attainable region, where appropriate.
Assuming a formula applies at a critical point. If an inverse derivative is unbounded at an output value, the density can still exist and be integrable, but the formula is generally an almost-everywhere statement. Check the assumptions on the branches and the probability of the excluded input set.

The branch method is useful whenever the transformed vector loses information about its input: several input configurations become indistinguishable after applying $T$. The Jacobian adjusts for local stretching or compression on each branch, and summing restores the total probability from every distinct input that can produce the observed output.

Check Your Understanding

For each question, identify the branches and account for every valid preimage.

For $Y=X^2$, with $X$ having a density supported on both sides of zero, what are the inverse branches for $y>0$, and what absolute inverse derivative does each branch contribute?
Why must branches be based on disjoint input regions, or otherwise have distinct inverse points?
If $T$ is decreasing on one branch, why does its density contribution remain nonnegative?
In the bivariate example, why does the output density equal zero when $y_1<0$?
Why can the density be unbounded near an output boundary and still integrate to one?