Independence Under Null Modifications

Why Null Modifications Do Not Change Independence

Independence is a statement about probabilities of events, not about the values random variables happen to take on every individual outcome. Consequently, changing a random variable on a set of probability zero does not change its independence relations with other random variables.

The same principle applies to sigma-algebras. When independent sigma-algebras are enlarged by adjoining events that differ from their original events only on null sets, their probability factorization remains valid. The key point in both arguments is that a finite collection of null exceptional sets still has a null union.

Almost-Sure Equality and Independence

Definition: Two random elements \(X\) and \(X'\), taking values in the same measurable space, are equal almost surely if they agree outside a measurable null set. We write \(X=X'\) almost surely.

For a random element \(X\) and a measurable set \(C\) in its state space, the event \(\{X\in C\}\) records whether \(X\) falls in \(C\). Independence of a finite collection of random elements means that for every choice of measurable sets \(C_i\), the probability of the intersection of these events is the product of their probabilities. For an infinite family, this requirement is imposed on every finite subfamily.

Theorem: Let \((X_i)_{i\in I}\) and \((X'_i)_{i\in I}\) be two families of random elements, where \(X_i\) and \(X'_i\) take values in the same measurable space for each \(i\). Suppose \(X_i=X'_i\) almost surely for every \(i\). Then \((X_i)_{i\in I}\) is independent if and only if \((X'_i)_{i\in I}\) is independent.

Proof. Fix any finite set of distinct indices \(i_1,\ldots,i_m\) and measurable sets \(C_1,\ldots,C_m\) in their respective state spaces. Let \(E_j=\{X_{i_j}\in C_j\}\) and \(E'_j=\{X'_{i_j}\in C_j\}\). By almost-sure equality, for each \(j\) there is a measurable null set \(N_j\) outside which \(X_{i_j}=X'_{i_j}\). Thus \(E_j\triangle E'_j\subseteq N_j\), and, since \(E_j\triangle E'_j\) is measurable, \(P(E_j\triangle E'_j)=0\). The symmetric difference between the two intersections is contained in the union of the corresponding symmetric differences: \[ \left(\bigcap_{j=1}^m E_j\right)\triangle \left(\bigcap_{j=1}^m E'_j\right) \subseteq \bigcup_{j=1}^m(E_j\triangle E'_j). \] The union is finite and null. Thus the two intersections have equal probabilities. Also \(P(E_j)=P(E'_j)\) for every \(j\). Therefore the factorization identity for the \(X_{i_j}\) events holds exactly when it holds for the \(X'_{i_j}\) events. Since the finite set of indices and measurable sets were arbitrary, the assertion follows for the whole family. \(\square\)

The theorem does not require a single exceptional set to work for every index in an infinite family. For each finite subfamily, only finitely many individual exceptional sets are used, and their union is null. This matches the definition of independence for an infinite family, which tests one finite subfamily at a time.

Worked Example: Changing Two Coordinates at One Outcome

Let \(\Omega=[0,1]^2\) with uniform area probability, and let \(U(u,v)=u\) and \(V(u,v)=v\) be the coordinate random variables. They are independent. Let \(N=\{(1,1)\}\), a measurable singleton of probability zero. Define \(U'=0\) on \(N\) and \(U'=U\) elsewhere; define \(V'=0\) on \(N\) and \(V'=V\) elsewhere. Both modified variables are measurable and equal almost surely to their originals.

For example, take the events \(\{U'\leq 1/2\}\) and \(\{V'\leq 1/3\}\). Since the point \((1,1)\) lies outside both original coordinate events, \[ \{U'\leq 1/2\}=\{u\leq 1/2\}\cup N,\qquad \{V'\leq 1/3\}=\{v\leq 1/3\}\cup N. \] Their probabilities are \(1/2\) and \(1/3\). Their intersection is \(\{u\leq1/2,\ v\leq1/3\}\cup N\), so its probability is \(1/6\). Hence the factorization holds: \(P(U'\leq1/2,\ V'\leq1/3)=(1/6)=(1/2)(1/3)\). The theorem gives the full independence assertion for all measurable sets, not just these thresholds.

Completing Independent Sigma-Algebras

Let \((\Omega,\mathcal F,P)\) be a probability space. Its completion \(\widehat{\mathcal F}\) contains \(\mathcal F\) and every subset of every \(\mathcal F\)-measurable null set. For a sigma-algebra \(\mathcal G\subseteq\mathcal F\), define its completion relative to this ambient completed space by \[ \widehat{\mathcal G} = \{A\in\widehat{\mathcal F}:\text{ there is a }B\in\mathcal G \text{ with }P(A\triangle B)=0\}. \] Thus an event in \(\widehat{\mathcal G}\) may differ from an event in \(\mathcal G\) on a null set. This collection is a sigma-algebra: complements preserve symmetric differences, and for a countable sequence of such events, the symmetric difference between their unions is contained in the countable union of the individual null symmetric differences.

For a finite collection \(\mathcal G_1,\ldots,\mathcal G_m\), independence means that for every choice \(B_j\in\mathcal G_j\), \(P(\bigcap_{j=1}^m B_j)=\prod_{j=1}^m P(B_j)\). An infinite collection is independent when every finite subcollection is independent.

Theorem: If a finite or infinite collection of sigma-algebras \((\mathcal G_i)_{i\in I}\) is independent, then the collection of their completions \((\widehat{\mathcal G}_i)_{i\in I}\), relative to the same ambient probability space, is independent.

Proof. Choose any finite set of distinct indices \(i_1,\ldots,i_m\), and events \(A_j\in\widehat{\mathcal G}_{i_j}\). By the definition of completion, for each \(j\) there is an event \(B_j\in\mathcal G_{i_j}\) with \(P(A_j\triangle B_j)=0\). In particular, \(P(A_j)=P(B_j)\). If the indicators of membership in \(\bigcap_j A_j\) and \(\bigcap_j B_j\) differ at an outcome, then membership differs for at least one pair \(A_j,B_j\). Therefore \[ \left(\bigcap_{j=1}^m A_j\right)\triangle \left(\bigcap_{j=1}^m B_j\right) \subseteq \bigcup_{j=1}^m(A_j\triangle B_j). \] The right-hand side is a finite union of null events, so the intersections have equal probabilities. Independence of the original sigma-algebras gives \[ P\left(\bigcap_{j=1}^m A_j\right) =P\left(\bigcap_{j=1}^m B_j\right) =\prod_{j=1}^m P(B_j) =\prod_{j=1}^m P(A_j). \] This holds for every finite set of indices and every choice of events in the corresponding completions, which is precisely independence of the completed collection. \(\square\)

Worked Example: Shared Null Events in Two Completions

Again take \([0,1]^2\) with area probability. Let \(\mathcal G_U=\sigma(U)\) and \(\mathcal G_V=\sigma(V)\). These sigma-algebras are independent: for Borel sets \(S,T\subseteq[0,1]\), the events \(\{U\in S\}\) and \(\{V\in T\}\) have probabilities given by the lengths of \(S,T\), and their intersection has area equal to the product of those lengths.

The singleton \(N=\{(1,1)\}\) belongs to each completed sigma-algebra. Indeed, \(N\subseteq\{U=1\}\) and \(P(U=1)=0\); likewise \(N\subseteq\{V=1\}\) and \(P(V=1)=0\). Set \(A=\{u\leq1/2\}\cup N\) and \(B=\{v\leq1/3\}\cup N\). These are events in \(\widehat{\mathcal G}_U\) and \(\widehat{\mathcal G}_V\), respectively. As \(N\) is outside both coordinate events, \(P(A)=1/2\), \(P(B)=1/3\), and \[ A\cap B=\{u\leq1/2,\ v\leq1/3\}\cup N. \] Consequently \(P(A\cap B)=1/6=P(A)P(B)\). The same reasoning, formalized in the theorem, applies to every finite collection of completed events.

Worked Example: An Infinite Sequence Modified at a Single Null Outcome

Let \(\Omega=\{0,1\}^{\mathbb N}\) carry the product probability under which each coordinate is independently \(0\) or \(1\) with probability \(1/2\). Write \(X_n(\omega)=\omega_n\). The event \(N\) consisting of the all-zero sequence is measurable and null: it is contained, for every \(k\), in the event that the first \(k\) coordinates are zero, whose probability is \(2^{-k}\). Thus \(P(N)\leq2^{-k}\) for every \(k\), forcing \(P(N)=0\).

Define \(X'_n=1\) on \(N\) and \(X'_n=X_n\) outside \(N\), for every \(n\). Each \(X'_n\) is measurable and equals \(X_n\) almost surely. To check a finite-dimensional distribution directly, choose distinct indices \(n_1,\ldots,n_m\) and values \(a_1,\ldots,a_m\in\{0,1\}\). The event \(\{X'_{n_j}=a_j,\ 1\leq j\leq m\}\) differs from \(\{X_{n_j}=a_j,\ 1\leq j\leq m\}\) only on \(N\). The latter has probability \(2^{-m}\), so the former also has probability \(2^{-m}=\prod_{j=1}^m P(X'_{n_j}=a_j)\). Hence every finite subfamily remains independent, even though all modified coordinates take the value \(1\) together at the all-zero outcome.

What Is Preserved—and What Is Not

Almost-sure modification preserves the distribution of each random element and all joint distributions of finite subfamilies. It therefore preserves independence. Similarly, completing independent sigma-algebras does not alter the probability of any event, because each newly included event agrees almost surely with an event from the original sigma-algebra.

But almost-sure equality is not pointwise equality. In the infinite-sequence example, the original coordinates are all zero at one outcome, while the modified coordinates are all one there. Thus a null modification can change pointwise statements, including whether particular outcomes satisfy a given relation. Such changes do not affect independence because probabilities assign the exceptional set mass zero.

A common pitfall is to reason that adding the same null event to two events must break independence because it creates a shared outcome. It may change their set-theoretic intersection, but it does not change any of the probabilities in the factorization identity. The decisive test is always whether the changes occur on null sets, not whether the modified events intersect in a different way.

Check Your Understanding

Use the definitions and arguments above to answer the following questions.

Why does proving the random-element theorem require considering only finite subfamilies, even when the index set is infinite?
Show that if \(P(A\triangle B)=0\), then \(P(A)=P(B)\).
Why does the intersection of finitely many events modified on null sets still have the same probability as the original intersection?
In the coordinate example, why is the point \((1,1)\) outside each of the two original threshold events?
What changes, and what remains unchanged, when the all-zero sequence is modified in the infinite-sequence example?