Laws of Measurable Transforms of Joint Vectors

Introduction

A random vector contains joint information about several coordinates. A measurable function of that vector may retain some of this information, combine it, or collapse many possible input values to the same output. To find the resulting distribution, begin with the event that the output lies in a specified set, then describe that event as a preimage in the original space.

This viewpoint works whether the joint law has a density, is discrete, or has neither form. It also helps avoid a common mistake: a formula for transforming point values does not, by itself, tell us how to transform a probability density.

The Law of a Measurable Transform

Let $X=(X_1,\ldots,X_d)$ be a random vector with values in $\mathbb{R}^d$, and let $\mu_X$ be its joint law on the Borel sets of $\mathbb{R}^d$. Thus $\mu_X(A)=\mathbb{P}(X\in A)$ for each Borel set $A$. Let $T:\mathbb{R}^d\to\mathbb{R}^k$ be Borel measurable, and set $Y=T(X)$. Measurability ensures that $Y$ is a random vector.

Definition: The law of $Y=T(X)$, denoted by $\mu_Y$, is the pushforward of $\mu_X$ under $T$. For every Borel set $B\subseteq\mathbb{R}^k$, $$ \mu_Y(B)=\mu_X(T^{-1}(B)) =\mathbb{P}(T(X)\in B). $$

The preimage $T^{-1}(B)=\{x:T(x)\in B\}$ is a subset of the input space. The formula says that the transformed probability is obtained by measuring that preimage under the original joint law. In particular, when $\mu_X$ has density $f_X$ with respect to Lebesgue measure,

\mathbb{P}(T(X)\in B) = \int_{\mathbb{R}^d} \mathbf{1}_{\{T(x)\in B\}} f_X(x)\,dx.

This is an integration formula over the original space. It does not assume that the transformed law itself has a density.

Integrating Functions of the Transformed Vector

The preimage formula extends from probabilities to expectations. This is useful when a transformed density is difficult to find or does not exist: one can still compute expectations by integrating over the original joint law.

Theorem: Let $X$, $T$, and $Y=T(X)$ be as above. For every nonnegative Borel function $g:\mathbb{R}^k\to[0,\infty]$, $$ \int_{\mathbb{R}^k} g(y)\,\mu_Y(dy) = \int_{\mathbb{R}^d} g(T(x))\,\mu_X(dx). $$ The same identity holds for any real-valued Borel function $g$ integrable with respect to $\mu_Y$, and for any complex-valued Borel function whose absolute value is integrable.

Proof. First take $g=\mathbf{1}_B$, where $B$ is Borel. By the definition of $\mu_Y$, $$ \int \mathbf{1}_B(y)\,\mu_Y(dy) =\mu_Y(B) =\mu_X(T^{-1}(B)) =\int \mathbf{1}_B(T(x))\,\mu_X(dx). $$ Linearity gives the identity for nonnegative simple functions. For an arbitrary nonnegative Borel $g$, choose nonnegative simple functions $g_n$ increasing pointwise to $g$. The monotone convergence theorem applies to $g_n$ under $\mu_Y$ and to $g_n\circ T$ under $\mu_X$, giving the identity for $g$. For an integrable real-valued $g$, apply the nonnegative result to its positive and negative parts and subtract the resulting finite integrals. For a complex-valued $g$, apply it separately to the real and imaginary parts; integrability of $|g|$ ensures the required parts are integrable. $\square$

In probability notation, the theorem gives $\mathbb{E}[g(Y)]=\mathbb{E}[g(T(X))]$, with the right-hand expectation understood using the joint law of $X$. The theorem is sometimes called the change-of-variables formula for a pushforward measure; it requires no assumption that $T$ is one-to-one.

Distribution Functions from Preimages

For a real-valued transform $Y=T(X)$, its distribution function is obtained by taking the output set to be $(-\infty,t]$. For a vector-valued transform, use a lower orthant, the set of points whose every coordinate is at most its corresponding threshold.

Proposition: If $Y=T(X)$ takes values in $\mathbb{R}^k$, then for every $t=(t_1,\ldots,t_k)\in\mathbb{R}^k$, its joint distribution function satisfies $$ F_Y(t) = \mathbb{P}(Y_1\le t_1,\ldots,Y_k\le t_k) = \int_{\mathbb{R}^d} \mathbf{1}_{\{T_1(x)\le t_1,\ldots,T_k(x)\le t_k\}}\,\mu_X(dx). $$

Proof. The event in the probability is exactly $\{T(X)\in B_t\}$, where $B_t=\prod_{j=1}^k(-\infty,t_j]$. Its preimage under $T$ is $\{x:T_j(x)\le t_j\text{ for every }j\}$. Applying the definition of the pushforward law to $B_t$, or integrating the indicator of this preimage, gives the stated identity. $\square$

Worked Example: The Maximum of Two Coordinates

Suppose $X=(X_1,X_2)$ has joint density $f_X(x_1,x_2)=1$ on the square $[0,1]^2$ and zero elsewhere. Define $M=\max(X_1,X_2)$. For $0\le t\le1$, the preimage of $(-\infty,t]$ is the part of the square where both coordinates are at most $t$: the square $[0,t]^2$. Therefore, integrating its indicator against the joint density gives

\mathbb{P}(M\le t) = \int_0^1\int_0^1 \mathbf{1}_{\{\max(x_1,x_2)\le t\}}\,dx_2\,dx_1 = \int_0^t\int_0^t 1\,dx_2\,dx_1 =t^2.

If $t<0$, the event is empty; if $t>1$, it is the entire square. Thus $F_M(t)=0$ for $t<0$, $F_M(t)=t^2$ for $0\le t\le1$, and $F_M(t)=1$ for $t>1$. This distribution function is continuous at both endpoints. Differentiating on $(0,1)$ gives the density $f_M(t)=2t$ there, and zero outside $[0,1]$. Its integral is $\int_0^1 2t\,dt=1$, as required.

Worked Example: Distance from the Origin

Let $X=(X_1,X_2)$ have density $1/\pi$ on the unit disk $D=\{(x_1,x_2):x_1^2+x_2^2\le1\}$, and zero outside it. Define $R=T(X)=\sqrt{X_1^2+X_2^2}$, the distance from the origin. For $0\le r\le1$, the preimage of $(-\infty,r]$ is the disk of radius $r$, contained in $D$. Its area is $\pi r^2$, so

\mathbb{P}(R\le r) = \int_D \mathbf{1}_{\{\sqrt{x_1^2+x_2^2}\le r\}}\frac{1}{\pi}\,dx_1\,dx_2 = \frac{\pi r^2}{\pi} =r^2.

For $r<0$ the probability is zero, and for $r>1$ it is one. Consequently $F_R(r)=0$ for $r<0$, $F_R(r)=r^2$ for $0\le r\le1$, and $F_R(r)=1$ for $r>1$. The density is $f_R(r)=2r$ on $(0,1)$, with zero density elsewhere. Its integral is one. The calculation uses the areas of preimage sets directly; it does not require treating the radius as if it were one of the original coordinates.

Worked Example: A Nonlinear Coordinate Transform

Let $X=(X_1,X_2)$ have density $1/2$ on the rectangle $[-1,1]\times[0,1]$, and zero elsewhere. Set $U=X_1^2$ and $V=X_2$. For any Borel set $E\subseteq\mathbb{R}^2$, the pushforward formula gives

\mathbb{P}((U,V)\in E) = \frac12\int_{-1}^{1}\int_0^1 \mathbf{1}_E(x_1^2,x_2)\,dx_2\,dx_1.

For each fixed $x_2$, split the $x_1$-integral at zero. On $0<x_1<1$, use $u=x_1^2$, so $dx_1=du/(2\sqrt{u})$. On $-1<x_1<0$, the same substitution reverses the endpoints and gives the same integral. The endpoints have Lebesgue measure zero. Hence

\begin{aligned} \mathbb{P}((U,V)\in E) &= \frac12\int_0^1\int_0^1 \mathbf{1}_E(u,x_2)\frac{1}{\sqrt{u}}\,du\,dx_2\\ &= \int_0^1\int_0^1 \mathbf{1}_E(u,v)\frac{1}{2\sqrt{u}}\,dv\,du. \end{aligned}

Thus $(U,V)$ has density $f_{U,V}(u,v)=1/(2\sqrt{u})$ for $0<u<1$ and $0<v<1$, and zero elsewhere. The density is unbounded near $u=0$, but it is integrable:

\int_0^1\int_0^1\frac{1}{2\sqrt{u}}\,dv\,du = \int_0^1\frac{1}{2\sqrt{u}}\,du =1.

There is no atom at $u=0$: the preimage $X_1=0$ is a line of probability zero under the original density. The singularity in the density is therefore an integrable one, not a point mass.

Worked Example: A Transform That Creates Atoms

Use the same uniform joint law on $[0,1]^2$ as in the maximum example, and define $W=\mathbf{1}_{\{X_1<X_2\}}$. The map is measurable and takes only the values zero and one. The preimage of $\{1\}$ is the triangle below the diagonal, whose area is $1/2$. Therefore

\mathbb{P}(W=1) = \int_0^1\int_0^1 \mathbf{1}_{\{x_1<x_2\}}\,dx_1\,dx_2 = \int_0^1 x_2\,dx_2 = \frac12.

The preimage of $\{0\}$ includes the other triangle and the diagonal. The diagonal has area zero, so $\mathbb{P}(W=0)=1/2$. The transformed law is discrete: it has mass $1/2$ at each of zero and one. In particular, a measurable transform of a vector with a density need not have a density with respect to Lebesgue measure on its output space.

What the Preimage Method Does—and Does Not—Say

The pushforward identity always determines the transformed law, but it does not promise that the law has a density. A map may collapse a region of positive probability to a single output, as in the indicator example. It may map a two-dimensional vector onto a curve or a point, producing a law concentrated on a set of lower dimension. Even when a density exists, as for $(X_1^2,X_2)$, it may be unbounded while remaining integrable.

Another common pitfall is to manipulate the input density using only the formula for $T$. A density is defined relative to a reference measure, and the output distribution must first be identified through preimages. If a density formula is then derived, it must account for all input points mapping to the output and must integrate to one. The square map illustrates why: almost every $u\in(0,1)$ has two preimages, $\sqrt{u}$ and $-\sqrt{u}$.

A reliable calculation follows this sequence:

Specify the output event.
Choose a Borel set $B$ in the transformed vector’s space.

Find its preimage.
Write $T^{-1}(B)=\{x:T(x)\in B\}$ explicitly in the input coordinates.

Integrate under the joint law.
Use the joint measure directly, or its density when one is available.

Check the resulting law.
For a claimed density, verify its support, integrability, and total mass; also check whether atoms or lower-dimensional concentration occur.

The central principle is simple but broad: probabilities of transformed events are probabilities of their preimages. It applies without independence assumptions and without requiring the transform to be invertible.

Check Your Understanding

Use preimages under the measurable map and the original joint law to answer these questions.

State the pushforward formula for $\mu_{T(X)}(B)$ and explain why the preimage must be measurable.
If $T:\mathbb{R}^d\to\mathbb{R}$, how would you express $\mathbb{P}(T(X)\le t)$ as an integral against $\mu_X$?
For the radius example, what geometric set is the preimage of $(-\infty,r]$ when $0\le r\le1$?
Why does the transformed vector $(X_1^2,X_2)$ have an unbounded density near $u=0$ but no atom there?
Give a reason that a measurable transform of a random vector with a density may fail to have a density on its output space.

Laws of Measurable Transforms of Joint Vectors

What You'll Learn

Introduction

The Law of a Measurable Transform

Integrating Functions of the Transformed Vector

Distribution Functions from Preimages

Worked Example: The Maximum of Two Coordinates

Worked Example: The Maximum of Two Coordinates

Worked Example: Distance from the Origin

Worked Example: Distance from the Origin

Worked Example: A Nonlinear Coordinate Transform

Worked Example: A Nonlinear Coordinate Transform

Worked Example: A Transform That Creates Atoms

Worked Example: A Transform That Creates Atoms

What the Preimage Method Does—and Does Not—Say

Check Your Understanding

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