Why Generating Events Can Be Enough
Independence is defined by checking an identity over entire sigma-algebras, which can involve many events. In applications, however, a smaller collection of events often generates each sigma-algebra. The pi-system criterion explains when checking the identity on those collections suffices. For real-valued random variables, lower half-lines generate the Borel sigma-algebra, so independence can be tested using joint probabilities of lower-left rectangles.
Throughout, \((\Omega,\mathcal{F},\mathbb{P})\) is a probability space: \(\mathcal{F}\) is a sigma-algebra and \(\mathbb{P}(\Omega)=1\). Thus \(\mathbb{P}\) is finite (and therefore sigma-finite); completeness is not assumed. Events belong to \(\mathcal{F}\). A real-valued random variable \(X\) is a measurable map from \((\Omega,\mathcal{F})\) to \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\). Statements about event probabilities below are exact, not merely almost sure.
We use the pi-lambda theorem, established in the earlier material on uniqueness of probability laws on pi-systems: if a lambda-system contains a pi-system \(\mathcal{C}\), it contains \(\sigma(\mathcal{C})\). Here \(\sigma(\mathcal{C})\) denotes the sigma-algebra generated by \(\mathcal{C}\). The independence criterion below is the step that applies this theorem to two event collections.
The Pi-System Criterion for Independence
Proof. Fix \(A\in\mathcal{A}\), and define \( \mathcal{D}_A=\{F\in\sigma(\mathcal{B}):\mathbb{P}(A\cap F)=\mathbb{P}(A)\mathbb{P}(F)\}. \) This is a lambda-system on \(\Omega\). First, \(\Omega\in\mathcal{D}_A\), since \(\mathbb{P}(A\cap\Omega)=\mathbb{P}(A)=\mathbb{P}(A)\mathbb{P}(\Omega)\). If \(F_1\subseteq F_2\) and both belong to \(\mathcal{D}_A\), finite additivity gives \( \mathbb{P}(A\cap(F_2\setminus F_1))=\mathbb{P}(A\cap F_2)-\mathbb{P}(A\cap F_1)=\mathbb{P}(A)\big(\mathbb{P}(F_2)-\mathbb{P}(F_1)\big)=\mathbb{P}(A)\mathbb{P}(F_2\setminus F_1). \) So \(F_2\setminus F_1\in\mathcal{D}_A\). If \((F_n)\) are pairwise disjoint members, countable additivity gives \( \mathbb{P}\big(A\cap\bigcup_nF_n\big)=\sum_n\mathbb{P}(A\cap F_n)=\mathbb{P}(A)\sum_n\mathbb{P}(F_n)=\mathbb{P}(A)\mathbb{P}\big(\bigcup_nF_n\big). \) Thus their union is in \(\mathcal{D}_A\).
The hypothesis says \(\mathcal{B}\subseteq\mathcal{D}_A\). By the pi-lambda theorem, \(\sigma(\mathcal{B})\subseteq\mathcal{D}_A\). Therefore the required identity holds for every \(A\in\mathcal{A}\) and \(F\in\sigma(\mathcal{B})\). Now fix any such \(F\), and define \( \mathcal{E}_F=\{E\in\sigma(\mathcal{A}):\mathbb{P}(E\cap F)=\mathbb{P}(E)\mathbb{P}(F)\}. \) The same three checks show that \(\mathcal{E}_F\) is a lambda-system: it contains \(\Omega\), is closed under nested differences by subtraction, and under disjoint countable unions by countable additivity. The result just proved gives \(\mathcal{A}\subseteq\mathcal{E}_F\), so the pi-lambda theorem yields \(\sigma(\mathcal{A})\subseteq\mathcal{E}_F\). This proves the forward implication. The converse follows directly because \(\mathcal{A}\subseteq\sigma(\mathcal{A})\) and \(\mathcal{B}\subseteq\sigma(\mathcal{B})\). \(\square\)
In this argument, fixing one event turns the independence identity into an equality of two finite measures as the other event varies. The lambda-system properties preserve that equality under the operations needed for the pi-lambda theorem. It is essential to complete the extension twice: first in the second event, then in the first.
From Event Independence to Random Variables
For a random variable \(X\), define \(\sigma(X)=\{X^{-1}(C):C\in\mathcal{B}_{\mathbb{R}}\}\). Two real-valued random variables \(X,Y\) are independent when \(\sigma(X)\) and \(\sigma(Y)\) are independent; equivalently, the identity below holds for every pair of Borel sets \(C,D\). The equivalence follows because events in those generated sigma-algebras are precisely such inverse images.
Proof. The families \(\{X^{-1}(C):C\in\mathcal{C}\}\) and \(\{Y^{-1}(D):D\in\mathcal{D}\}\) are pi-systems of events and contain \(\Omega\). Their generated sigma-algebras are \(\sigma(X)\) and \(\sigma(Y)\), respectively, because \(\mathcal{C}\) and \(\mathcal{D}\) generate \(\mathcal{B}_{\mathbb{R}}\). The assumed rectangle identity is the hypothesis of the theorem for these two event pi-systems. The theorem therefore gives independence of \(\sigma(X)\) and \(\sigma(Y)\). \(\square\)
A useful choice for both \(\mathcal{C}\) and \(\mathcal{D}\) is \( \mathcal{H}=\{\varnothing,\mathbb{R}\}\cup\{(-\infty,t]:t\in\mathbb{R}\}. \) This is a pi-system: the intersection of two lower half-lines is the lower half-line with the smaller endpoint, and intersections with \(\varnothing\) or \(\mathbb{R}\) remain in the family. It generates \(\mathcal{B}_{\mathbb{R}}\). Indeed, \((-\infty,b)=\bigcup_{n\ge1}(-\infty,b-1/n]\), and then \((a,b)=(-\infty,b)\cap(-\infty,a]^c\). Since open intervals generate the Borel sigma-algebra, the claim follows.
Consequently, it is enough to check the identity for every \(s,t\in\mathbb{R}\): \( \mathbb{P}(X\le s,\ Y\le t)=\mathbb{P}(X\le s)\mathbb{P}(Y\le t). \) The inequalities are non-strict, as dictated by the chosen half-lines. Checking only a single threshold pair, or only marginal probabilities, is not this criterion.
Worked Example: Two Independent Fair Bits
Let \(\Omega=\{(0,0),(0,1),(1,0),(1,1)\}\), let \(\mathcal{F}\) be its power set, and assign probability \(1/4\) to each outcome. Define \(X(i,j)=i\) and \(Y(i,j)=j\). For either coordinate, take the pi-system of all subsets of \(\{0,1\}\). It contains the whole space and is closed under intersections; it also generates the full power set.
For any \(A,B\subseteq\{0,1\}\), the event \(\{X\in A,Y\in B\}\) consists of exactly the pairs in \(A\times B\), hence has \(|A||B|\) outcomes. Therefore \( \mathbb{P}(X\in A,Y\in B)=\frac{|A||B|}{4},\qquad \mathbb{P}(X\in A)\mathbb{P}(Y\in B)=\frac{|A|}{2}\frac{|B|}{2}=\frac{|A||B|}{4}. \) The identities agree for every pair of generator sets, including an empty set (both sides are zero). The rectangle criterion proves that \(X\) and \(Y\) are independent. Check: for \(A=\{1\}\) and \(B=\{0\}\), the sole joint outcome is \((1,0)\), so the joint probability is \(1/4\); the product of the marginal probabilities is \((2/4)(2/4)=1/4\).
Worked Example: Testing Uniform Coordinates with Lower Rectangles
Let \(\Omega=[0,1]^2\), let \(\mathcal{F}\) be its Lebesgue sigma-algebra, and let \(\mathbb{P}\) be two-dimensional Lebesgue measure restricted to \(\Omega\). Its total mass is \(1\). Set \(X(u,v)=u\) and \(Y(u,v)=v\). For \(s\in\mathbb{R}\), define \(q(s)=0\) if \(s\le0\), \(q(s)=s\) if \(0<s<1\), and \(q(s)=1\) if \(s\ge1\).
The event \(\{X\le s\}\) cuts out a vertical strip in the square of width \(q(s)\), so \(\mathbb{P}(X\le s)=q(s)\). This includes the boundary cases: at \(s=0\) the strip has zero area, and at \(s=1\) it is the whole square. The same computation gives \(\mathbb{P}(Y\le t)=q(t)\). The joint event is the rectangle \([0,q(s)]\times[0,q(t)]\), up to boundary sets of area zero, and has area \(q(s)q(t)\). Thus for all real \(s,t\), \( \mathbb{P}(X\le s,Y\le t)=q(s)q(t)=\mathbb{P}(X\le s)\mathbb{P}(Y\le t). \) The lower-half-line criterion proves independence. Check: at \(s=1/2,t=3/4\), the joint area is \((1/2)(3/4)=3/8\), and the product of the marginal probabilities is also \((1/2)(3/4)=3/8\). Computing the rectangle's side lengths directly gives the same area \(3/8\).
Worked Example: Equal Marginals Do Not Imply Independence
Let \(\Omega=\{0,1\}\) with probability \(1/2\) on each point, and define \(X(\omega)=\omega\) and \(Y(\omega)=\omega\). Both variables have the same marginal probabilities: \(\mathbb{P}(X=0)=\mathbb{P}(Y=0)=1/2\), and likewise both equal \(1/2\) at \(1\). But for the lower half-lines at \(s=t=0\), \( \mathbb{P}(X\le0,Y\le0)=\mathbb{P}(\omega=0)=\frac12,\qquad \mathbb{P}(X\le0)\mathbb{P}(Y\le0)=\frac12\cdot\frac12=\frac14. \) The two sides differ, so the rectangle criterion detects dependence. As a second check, the joint event \(\{X=1,Y=1\}\) has probability \(1/2\), whereas the product of its marginal event probabilities is \(1/4\). The variables have identical marginal laws, but the joint probabilities are not products.
Applying the Criterion Carefully
The practical method is to identify a generating pi-system for each variable, compute joint probabilities for every pair of sets in those systems, and then invoke the criterion. For real-valued variables, the lower-half-line test is especially useful because it is exactly a statement about the joint distribution function and the marginal distribution functions. No assumption that the variables have densities is required.
A common mistake is to reason that because two variables have specified marginal distributions, they must be independent. The equal-bit example shows the error concretely: the marginal probabilities are both \(1/2\), but the joint probability at the lower rectangle is \(1/2\), not \(1/4\). Another mistake is to check only one rectangle and conclude independence. The theorem requires the factorization identity for every pair of sets in the chosen generating pi-systems. Once that full generator check is established, the pi-lambda theorem supplies the extension to all Borel events.
Check Your Understanding
For each question, identify the generating sets and state which event-probability identity is needed.
- Why must the collections used in the criterion be closed under finite intersections?
- In the theorem's proof, why is \(\mathcal{D}_A\) closed under nested differences?
- Show that \(\{\varnothing,\mathbb{R}\}\cup\{(-\infty,t]:t\in\mathbb{R}\}\) generates \(\mathcal{B}_{\mathbb{R}}\).
- For the equal-bit example, compute \(\mathbb{P}(X\le 1,Y\le 0)\) and compare it with the product of the marginals.
- Does checking factorization for every lower-left rectangle require \(X\) and \(Y\) to have densities? Explain.