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The Joint CDF Factorization Criterion

Learn to test independence by comparing the joint CDF with the product of its marginal CDFs, and understand why the comparison is conclusive.

Intermediate 12 min read

What You'll Learn

  • State the joint CDF factorization criterion with its full quantifiers.
  • Prove that independence implies factorization at every pair of thresholds.
  • Use the pi-lambda theorem to prove the converse for arbitrary Borel events.
  • Apply the criterion to independent and dependent examples.
  • Recognize why a single matching CDF value, or matching marginals, does not establish independence.

Why a CDF Factorization Criterion Matters

Independence is a statement about all events determined by two random variables, not just about a single probability or a correlation. Yet the joint CDF records probabilities for a structured family of events: lower-left rectangles of the form \(\{X\leq x,\;Y\leq y\}\). The joint CDF factorization criterion says that checking these probabilities at every pair of real thresholds is enough to establish independence, even when the distributions have atoms or no density.

Write \(F_X(x)=P(X\leq x)\) and \(F_Y(y)=P(Y\leq y)\) for the marginal CDFs, and \(F_{X,Y}(x,y)=P(X\leq x,\;Y\leq y)\) for the joint CDF. The criterion concerns equality for all real \(x\) and \(y\), not just selected values.

Definition: The random variables \(X\) and \(Y\) are independent if the sigma-fields they generate, \(\sigma(X)\) and \(\sigma(Y)\), are independent. Equivalently, for every pair of Borel sets \(A,B\subseteq\mathbb{R}\), $$ P(X\in A,\;Y\in B)=P(X\in A)P(Y\in B). $$

The Joint CDF Factorization Criterion

Theorem: Two real-valued random variables \(X\) and \(Y\) are independent if and only if $$ F_{X,Y}(x,y)=F_X(x)F_Y(y) $$ for every \(x,y\in\mathbb{R}\).

The forward direction follows directly from the definition: the events \(\{X\leq x\}\) and \(\{Y\leq y\}\) belong to the respective generated sigma-fields. The converse is the substantive part. It uses the fact that lower half-lines determine Borel events, together with the pi-lambda theorem.

A pi-system is a collection of sets closed under finite intersections. A lambda-system contains the whole space, is closed under complements, and is closed under countable disjoint unions. The pi-lambda theorem states that if a pi-system \(\mathcal{P}\) is contained in a lambda-system \(\mathcal{D}\), then \(\mathcal{D}\) contains the sigma-field generated by \(\mathcal{P}\). The collection $$ \mathcal{C}=\{\mathbb{R}\}\cup\{(-\infty,t]:t\in\mathbb{R}\} $$ is a pi-system and generates the Borel sigma-field on \(\mathbb{R}\).

Proof: If \(X\) and \(Y\) are independent, then \(\{X\leq x\}\) and \(\{Y\leq y\}\) are independent events. Therefore $$ F_{X,Y}(x,y)=P(X\leq x,\;Y\leq y) =P(X\leq x)P(Y\leq y) =F_X(x)F_Y(y). $$ This holds for every \(x,y\in\mathbb{R}\).

Conversely, suppose the displayed factorization holds for every real \(x,y\). Fix \(y\in\mathbb{R}\) and define $$ \mathcal{D}_y= \{A\subseteq\mathbb{R}: A\text{ is Borel and } P(X\in A,\;Y\leq y)=P(X\in A)P(Y\leq y)\}. $$ This is a lambda-system. It contains \(\mathbb{R}\) because both sides then equal \(P(Y\leq y)\). If \(A\in\mathcal{D}_y\), subtracting its defining equality from \(P(Y\leq y)\) proves the equality for \(A^c\). For pairwise disjoint \(A_1,A_2,\ldots\in\mathcal{D}_y\), countable additivity gives $$ P\left(X\in\bigcup_{k=1}^{\infty}A_k,\;Y\leq y\right) =\sum_{k=1}^{\infty}P(X\in A_k,\;Y\leq y) =P(Y\leq y)\sum_{k=1}^{\infty}P(X\in A_k), $$ which is the required equality for their union.

The assumed CDF factorization shows that \((-\infty,x]\in\mathcal{D}_y\) for every real \(x\). Also \(\mathbb{R}\in\mathcal{D}_y\), so \(\mathcal{C}\subseteq\mathcal{D}_y\). By the pi-lambda theorem, \(\mathcal{D}_y\) contains every Borel set. Thus the required probability identity holds for every Borel \(A\) and this fixed \(y\).

Now fix any Borel set \(A\) and define $$ \mathcal{E}_A= \{B\subseteq\mathbb{R}: B\text{ is Borel and } P(X\in A,\;Y\in B)=P(X\in A)P(Y\in B)\}. $$ The same complement and disjoint-union calculations show that \(\mathcal{E}_A\) is a lambda-system. By the preceding paragraph it contains every \((-\infty,y]\), and it contains \(\mathbb{R}\). Another application of the pi-lambda theorem gives every Borel \(B\in\mathcal{E}_A\). Since \(A\) was arbitrary, the identity holds for every Borel pair \(A,B\), so \(X\) and \(Y\) are independent. \(\square\)

Applying the Criterion

Worked Example: Independent Uniform Variables

Let \(X\) and \(Y\) be independent, with \(X\sim\operatorname{Uniform}(0,1)\) and \(Y\sim\operatorname{Uniform}(-1,1)\). Their marginal CDFs are $$ F_X(x)= \begin{cases} 0,&x<0,\\ x,&0\leq x\leq 1,\\ 1,&x>1, \end{cases} \qquad F_Y(y)= \begin{cases} 0,&y<-1,\\ \dfrac{y+1}{2},&-1\leq y\leq 1,\\ 1,&y>1. \end{cases} $$ Because the variables are independent, the probability of the intersection of the two threshold events is the product of their probabilities: $$ F_{X,Y}(x,y)=F_X(x)F_Y(y) $$ for all real \(x,y\), including values outside either variable's support. For instance, at \(x=0.4\) and \(y=0\), $$ F_{X,Y}(0.4,0)=P(X\leq0.4)P(Y\leq0)=0.4\cdot\frac12=0.2. $$ If \(x<0\) or \(y<-1\), the corresponding marginal CDF is zero, so the joint CDF is zero as well. If \(x\geq1\), then \(F_X(x)=1\), and the joint CDF equals \(F_Y(y)\), as it must because \(X\leq x\) is certain. These boundary and exterior cases are included in the criterion.

Worked Example: Identical Uniform Variables Are Dependent

Let \(U\sim\operatorname{Uniform}(0,1)\), and set \(X=U\) and \(Y=U\). Each marginal CDF is zero below 0, equals its argument on \([0,1]\), and equals 1 above 1. But the joint threshold event is $$ \{X\leq x,\;Y\leq y\}=\{U\leq\min(x,y)\}. $$ Consequently, $$ F_{X,Y}(x,y)= \begin{cases} 0,&\min(x,y)<0,\\ \min(x,y),&0\leq\min(x,y)\leq1,\\ 1,&\min(x,y)>1. \end{cases} $$ At \(x=y=1/2\), the joint CDF is \(1/2\), whereas the product of the marginal CDFs is \((1/2)(1/2)=1/4\). The criterion fails, so \(X\) and \(Y\) are not independent. In event terms, \(X\leq1/2\) and \(Y\leq1/2\) are the same event, not two independent events.

Worked Example: Factorization With Different Supports

Let \(X\sim\operatorname{Exponential}(2)\) and \(Y\sim\operatorname{Uniform}(0,1)\), constructed independently. The marginal CDFs are $$ F_X(x)= \begin{cases} 0,&x<0,\\ 1-e^{-2x},&x\geq0, \end{cases} \qquad F_Y(y)= \begin{cases} 0,&y<0,\\ y,&0\leq y\leq1,\\ 1,&y>1. \end{cases} $$ Independence gives \(F_{X,Y}(x,y)=F_X(x)F_Y(y)\) at every pair of real thresholds. To check an interior value, take \(x=\frac{\log 2}{2}\) and \(y=\frac14\). Then $$ F_X(x)=1-e^{-2(\log 2/2)}=1-\frac12=\frac12, \qquad F_Y(y)=\frac14, $$ so $$ F_{X,Y}\left(\frac{\log 2}{2},\frac14\right)=\frac12\cdot\frac14=\frac18. $$ The supports differ, but that changes only the marginal CDF formulas; the factorization criterion is unchanged.

Why Threshold Rectangles Are Enough

A threshold event is a lower-left rectangle in the \((X,Y)\) plane. The theorem says that probabilities of all such rectangles determine whether the two variables are independent. The proof works by extending the probability identity from half-lines to all Borel sets, first in one coordinate and then in the other. This is why a condition stated using only CDF values can establish independence for much more general events.

The criterion makes no assumption that the joint law has a density, that either variable is continuous, or that the marginal CDFs are strictly increasing. Atoms and jumps are handled by requiring the equality at every real threshold. For example, the closed threshold events \(\{X\leq x\}\) include any mass at \(x\); changing to strict inequalities without accounting for boundary mass could change a probability.

Common Mistakes

  • Checking only one pair of thresholds. A match at one \((x,y)\) does not imply independence. The identical-uniform example has a particular value where a check might be made, but the criterion requires equality for every pair.
  • Checking only the marginals. Marginal CDFs describe each variable separately. Independence additionally requires the joint CDF to equal their product throughout the plane.
  • Assuming a density is necessary. The proof uses events and CDFs, not densities. It remains valid for discrete, continuous, and mixed distributions.
  • Forgetting the converse needs an argument. Factorization of threshold probabilities is not merely the definition of independence for all events; the pi-lambda argument explains why those threshold probabilities determine the probabilities for arbitrary Borel sets.
Key takeaway: \(X\) and \(Y\) are independent exactly when \(F_{X,Y}(x,y)=F_X(x)F_Y(y)\) for every real \(x,y\). The universal quantifier matters, and the criterion applies regardless of whether the distributions have densities.

Check Your Understanding

Use the criterion and its proof to answer the following questions.

  1. For independent \(X\) and \(Y\), why does the definition of independence apply to the two events used in \(F_{X,Y}(x,y)\)?
  2. If the factorization holds at \(x=0\) and \(y=0\), does that establish independence? Explain.
  3. In the example \(X=Y=U\) with \(U\sim\operatorname{Uniform}(0,1)\), calculate the joint CDF and the product of marginal CDFs at \(x=0.2,\;y=0.7\). Does this pair also reveal dependence?
  4. What role does the pi-lambda theorem play in extending the factorization from threshold events to all Borel events?
  5. Does the joint CDF factorization criterion require either variable to have a density? Why or why not?

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