Transformations and Joint Random Variables

From Random Variables to Transformations

Once a random variable is known to be measurable, it is natural to apply functions to it or to combine several random variables into one vector. Measurability is what ensures that events described by these new quantities remain events in the original probability space. The key facts are that Borel transformations preserve measurability and that, for a finite-dimensional vector, measurability can be checked coordinate by coordinate.

Throughout, the underlying probability space is \((\Omega,\mathcal{F},\mathbb{P})\). A real-valued random variable is a measurable map \(X:(\Omega,\mathcal{F})\to(\mathbb{R},\mathcal{B}_{\mathbb{R}})\), where \(\mathcal{B}_{\mathbb{R}}\) is the Borel sigma-algebra. Unless otherwise stated, random variables in this tutorial take finite real values everywhere, and measurability is an everywhere property of the map; probabilities of events are, of course, values of \(\mathbb{P}\). For \(d\ge1\), \(\mathcal{B}_{\mathbb{R}^d}\) denotes the Borel sigma-algebra of Euclidean space.

In particular, an event such as \(\{T\in B\}\) is shorthand for \(T^{-1}(B)\), and it belongs to \(\mathcal{F}\) whenever \(T\) is measurable and \(B\) is Borel. The condition concerns all Borel sets, not just intervals or open sets chosen one at a time.

Borel Transformations Preserve Measurability

Proof. Let \(C\in\mathcal{B}_{\mathbb{R}^k}\). Since \(g\) is Borel measurable, \(g^{-1}(C)\in\mathcal{B}_{\mathbb{R}^d}\). Since \(X\) is measurable, \(X^{-1}(g^{-1}(C))\in\mathcal{F}\). The identity \((g\circ X)^{-1}(C)=X^{-1}(g^{-1}(C))\) therefore shows that the preimage of every Borel \(C\) is in \(\mathcal{F}\). This is precisely measurability of \(g\circ X\). \(\square\)

Continuous maps between Euclidean spaces are Borel measurable, so the theorem applies to continuous transformations. It also applies to Borel maps that are not continuous. In the scalar case, sums and products are examples: the maps \((x,y)\mapsto x+y\) and \((x,y)\mapsto xy\) are continuous on \(\mathbb{R}^2\). Thus, once \((X,Y)\) is measurable, \(X+Y\) and \(XY\) are measurable real-valued random variables.

Joint Measurability Is Equivalent to Coordinate Measurability

Proof. Suppose first that \(T\) is measurable. For each \(j\), the coordinate projection \(\pi_j:\mathbb{R}^d\to\mathbb{R}\), defined by \(\pi_j(x_1,\ldots,x_d)=x_j\), is continuous and therefore Borel measurable. The composition theorem gives that \(\pi_j\circ T=X_j\) is measurable.

Conversely, suppose that every \(X_j\) is measurable. The Borel sigma-algebra \(\mathcal{B}_{\mathbb{R}^d}\) is generated by open boxes \(\prod_{j=1}^d(a_j,b_j)\), where the endpoints may be rational. Indeed, these boxes form a countable base for the Euclidean topology: every open set is a union of boxes from this base, and that union is countable because the base is countable. For any such box,

Each set on the right belongs to \(\mathcal{F}\), because \(X_j\) is measurable and \((a_j,b_j)\) is Borel. The finite intersection therefore belongs to \(\mathcal{F}\). The collection of subsets \(B\subseteq\mathbb{R}^d\) for which \(T^{-1}(B)\in\mathcal{F}\) is a sigma-algebra: preimages preserve complements and countable unions. It contains the open boxes, so it contains the sigma-algebra they generate, namely \(\mathcal{B}_{\mathbb{R}^d}\). Thus \(T\) is measurable. If \(d=1\), this is just the given measurability of \(X_1\), and the same conclusion holds. \(\square\)

This equivalence is specific to finite-dimensional Euclidean targets and their Borel sigma-algebras. It lets us establish measurability of a joint random vector without checking preimages of every Borel subset separately. Once the vector is measurable, the composition theorem then permits a Borel transformation of the entire vector.

A Borel Map Need Not Be Continuous

It is not necessary for a transformation to be continuous. For example, define \(g:\mathbb{R}\to\mathbb{R}\) by \(g(x)=1/x\) when \(x\ne0\), and \(g(0)=0\). This map is Borel measurable. To verify this, let \(O\subseteq\mathbb{R}\) be open. On the open set \(\mathbb{R}\setminus\{0\}\), the reciprocal map is continuous, so \(\{x\ne0:1/x\in O\}\) is relatively open in \(\mathbb{R}\setminus\{0\}\), hence Borel in \(\mathbb{R}\). The preimage \(g^{-1}(O)\) is this Borel set, with \(\{0\}\) added exactly when \(0\in O\). It is therefore Borel.

Measurability Does Not Imply a Product Rule

A common error is to confuse measurability of a joint vector with a rule for multiplying probabilities. Measurability says that the preimages of Borel sets are events; it does not say that the coordinates behave independently or that probabilities of intersections factor. For a concrete check, let \(\Omega=[0,1]\) with uniform probability and set \(X(\omega)=Y(\omega)=\omega\). Both coordinates, and hence \((X,Y)\), are measurable. The event \(\{X<1/2,\ Y<1/2\}\) is exactly \([0,1/2)\), so its probability is \(1/2\). A wrong calculation would multiply the two marginal probabilities, \(\mathbb{P}(X<1/2)=1/2\) and \(\mathbb{P}(Y<1/2)=1/2\), to get \(1/4\). That multiplication is not justified by measurability; the correct event preimage is the single interval \([0,1/2)\), of probability \(1/2\).

The practical method is therefore to identify the map and its codomain, establish measurability using the Borel-composition theorem or the coordinate criterion, and then compute an event by taking its actual preimage. Each step has a distinct purpose: measurability guarantees the preimage is an event, while the preimage calculation determines which event it is.