Why a CDF Test for Independence Matters
Independence is a statement about an entire joint distribution, not merely about whether two variables appear unrelated or have a particular pair of summary measures. The cumulative distribution function (CDF) offers a direct way to compare the joint law with the two marginal laws. In particular, it gives an exact criterion that applies to discrete, continuous, and mixed distributions.
The key is to compare probabilities at every pair of thresholds. A mismatch at even one pair rules out independence. Agreement at only a handful of thresholds is not enough to establish it.
The Product-CDF Criterion
The inequalities in these definitions are non-strict: each CDF uses \(\leq\). This matters when a variable has an atom, meaning positive probability at a single value. For example, \(F_X(x)\) includes all probability at \(x\), not just probability strictly below \(x\).
Why the Criterion Holds
First suppose \(X\) and \(Y\) are independent. Apply the definition of independence to the events \(\{X\leq x\}\) and \(\{Y\leq y\}\). Their intersection is \(\{X\leq x,\;Y\leq y\}\), so
This proves the forward direction. For the converse, suppose the equality holds for every real \(x,y\). A rectangle probability can be recovered from four joint CDF values. For finite \(a<b\) and \(c<d\), inclusion-exclusion gives
Substitute the assumed factorization into all four terms and collect the factors:
The last expression is \(P(a<X\leq b)P(c<Y\leq d)\). Thus probabilities factor for these rectangles. More generally, the assumed CDF equality says that the joint probability measure and the product of the marginal probability measures agree on every lower-left rectangle \((-\infty,x]\times(-\infty,y]\). These rectangles generate the Borel sets of the plane. The uniqueness theorem for probability measures therefore implies that the joint measure is the product of the marginal measures. Consequently, probabilities factor for all Borel sets \(A\) and \(B\), which is independence.
The rectangle calculation also explains why the criterion is not restricted to continuous distributions. It uses CDF increments, not densities. The measure-uniqueness step is what extends the rectangle result to arbitrary Borel events.
Applying the Criterion to Discrete Variables
For discrete variables, a systematic check groups thresholds according to which support values they include. The CDF is constant between successive support points, but its value changes at a support point because the threshold includes that point. If factorization holds across all such threshold regions, it holds for every real pair.
Worked Example: Checking Independence on a Finite Support
A hypothetical model has \(X\) taking values \(0\) and \(2\), with probabilities \(P(X=0)=0.4000\) and \(P(X=2)=0.6000\). The variable \(Y\) takes values \(-1\) and \(1\), with probabilities \(P(Y=-1)=0.2500\) and \(P(Y=1)=0.7500\). The joint probabilities are:
| \(X=0\) | \(X=2\) | Marginal probability for \(Y\) | |
|---|---|---|---|
| \(Y=-1\) | 0.1000 | 0.1500 | 0.2500 |
| \(Y=1\) | 0.3000 | 0.4500 | 0.7500 |
| Marginal probability for \(X\) | 0.4000 | 0.6000 | 1.0000 |
Method. Compare the joint CDF with the product of the marginal CDFs for every possible threshold region. For \(X\), the regions are \(x<0\), \(0\leq x<2\), and \(x\geq2\). For \(Y\), they are \(y<-1\), \(-1\leq y<1\), and \(y\geq1\).
The marginal CDF values across those regions are respectively \(0.0000,\;0.4000,\;1.0000\) for \(X\), and \(0.0000,\;0.2500,\;1.0000\) for \(Y\). The joint CDF is zero if either threshold region includes no support value. In the four regions where both include at least one support value, its values are:
- For \(0\leq x<2\) and \(-1\leq y<1\), the joint CDF is \(P(X=0,Y=-1)=0.1000\). The marginal product is \(0.4000(0.2500)=0.1000\).
- For \(0\leq x<2\) and \(y\geq1\), the joint CDF is \(P(X=0)=0.4000\). The marginal product is \(0.4000(1.0000)=0.4000\).
- For \(x\geq2\) and \(-1\leq y<1\), the joint CDF is \(P(Y=-1)=0.2500\). The marginal product is \(1.0000(0.2500)=0.2500\).
- For \(x\geq2\) and \(y\geq1\), the joint CDF is \(1.0000\). The marginal product is \(1.0000(1.0000)=1.0000\).
The CDFs therefore factor for all real thresholds, so the theorem establishes that \(X\) and \(Y\) are independent. As a check at a particular pair, \(F_{X,Y}(1,0)=0.1000\), matching \(F_X(1)F_Y(0)=0.4000(0.2500)=0.1000\). The joint probabilities also sum to \(0.1000+0.1500+0.3000+0.4500=1.0000\), and their row and column sums match the stated marginals.
Common mistake avoided: A threshold of \(x=0\) includes the atom at \(X=0\). Treating \(X\leq0\) as \(X<0\) would incorrectly give a zero CDF value and would invalidate the comparison.
One Mismatch Is Enough
To disprove independence, there is no need to calculate the whole joint CDF. It is enough to find one pair of thresholds where the equality fails. A useful search is to choose thresholds that include only part of each variable’s support, rather than thresholds so large that both CDFs equal one.
Worked Example: Detecting Dependence at an Atom
A hypothetical pair of binary variables \(X\) and \(Y\), each taking values \(0\) and \(1\), has the following joint probabilities:
| \(X=0\) | \(X=1\) | Marginal probability for \(Y\) | |
|---|---|---|---|
| \(Y=0\) | 0.2000 | 0.3000 | 0.5000 |
| \(Y=1\) | 0.3000 | 0.2000 | 0.5000 |
| Marginal probability for \(X\) | 0.5000 | 0.5000 | 1.0000 |
Method. Find a threshold pair with a nontrivial joint CDF and compare it with the marginal product. At \(x=0\) and \(y=0\), the non-strict inequalities include exactly the outcome \((X,Y)=(0,0)\). Hence
Since \(0.2000\ne0.2500\), the criterion fails at this pair, and \(X\) and \(Y\) are dependent. The probabilities are valid: the four joint probabilities sum to \(1.0000\), and each marginal probability is \(0.5000\).
Common mistake avoided: Checking only \(x=1,y=1\) would give \(F_{X,Y}(1,1)=1.0000=F_X(1)F_Y(1)\), but that equality provides no evidence of independence. The test requires equality at every pair; a threshold pair that exposes a difference is more informative.
The Criterion Also Covers Mixed Distributions
A density-based test can be unavailable when one variable has atoms, even though the product-CDF criterion still applies. Independence is about the joint probability law, not about whether a joint density exists.
Worked Example: A Discrete Variable and a Continuous Variable
In a hypothetical model, \(X\) is Bernoulli with \(P(X=0)=0.3000\) and \(P(X=1)=0.7000\). The variable \(Y\) is uniform on \([-1,2]\), so its density is \(1/3\) on that interval and zero elsewhere. Assume the joint law is the product of these two marginal laws. Verify the CDF criterion and evaluate it at \(x=0.5,\;y=0.5\).
Method. Write each marginal CDF, keeping the jump in \(F_X\) and the endpoints of the uniform interval:
Because the joint law is the product of the marginals, for any \(x,y\),
This equation verifies the criterion for every threshold pair, including those outside the support and those at the atom \(X=0\). At the requested pair, \(F_X(0.5)=0.3000\) and
As a boundary check, at \(x=0,\;y=-1\), the product is \(0.3000(0.0000)=0.0000\); indeed, a continuous uniform variable has zero probability at the single endpoint \(-1\). The atom at \(X=0\) is included by \(X\leq0\), while the endpoint event \(Y\leq-1\) has probability zero. This example shows why the same non-strict CDF definitions must be used for both variables.
Common mistake avoided: Requiring a joint density and trying to factor it would not be a suitable general test for this mixed model. The CDF criterion remains valid without assuming a joint density.
Exam Traps and Practical Use
The product-CDF criterion is useful because it is exact and applies uniformly across distribution types. In a problem, first identify the joint CDF and the two marginal CDFs. Then decide whether the task is to establish independence or to rule it out.
- To disprove independence: compute both sides at one carefully chosen threshold pair. A strict difference is conclusive.
- To establish independence: justify equality for every real pair. For finite discrete supports, partition the real line into threshold regions where each CDF is constant, then check all region combinations.
- Respect atoms and endpoints: \(F_X(x)=P(X\leq x)\), so a jump at \(x\) is included at that threshold.
- Do not infer independence from a few matches: equality at thresholds where both CDFs are zero or one is often automatic and may reveal little.
- Do not substitute a density condition without checking its assumptions: a joint density may not exist for discrete or mixed variables.
For the converse, the important logical step is not merely that rectangle probabilities factor in a few calculations. It is that CDF equality supplies those probabilities at all thresholds, and those rectangles determine the joint probability law. That is why a statement involving CDFs can characterize full independence.
Check Your Understanding
For each problem, use the product-CDF criterion and pay careful attention to which support values the thresholds include.
- Independent discrete variables have \(P(X=-1)=0.3000,\;P(X=2)=0.7000\) and \(P(Y=0)=0.6000,\;P(Y=4)=0.4000\). Find \(F_{X,Y}(0,1)\). (Answer: \(0.1800\))
- A binary pair has \(P(X=0,Y=0)=0.1000,\;P(X=0,Y=1)=0.4000,\;P(X=1,Y=0)=0.4000,\;P(X=1,Y=1)=0.1000\). Are the variables independent? (Answer: No; at \((0,0)\), the joint CDF is \(0.1000\), while the marginal product is \(0.2500\).)
- Independent variables \(U\sim\text{Uniform}(2,5)\) and \(V\sim\text{Uniform}(-1,3)\) are observed. Calculate \(F_{U,V}(3,1)\). (Answer: \(\frac{1}{6}=0.1667\))
- Independent \(X\) and \(Y\) have \(P(X=-1)=0.2000,\;P(X=2)=0.8000\), and \(Y\sim\text{Uniform}(0,4)\). Calculate \(F_{X,Y}(-1,3)\), including the atom at \(X=-1\). (Answer: \(0.2000(0.7500)=0.1500\))
- Suppose the product-CDF equality is known to hold whenever both thresholds are rational. Does it follow that \(X\) and \(Y\) are independent? (Answer: Yes. For any real \(x,y\), take rational sequences decreasing to \(x,y\); right-continuity of the joint and marginal CDFs gives equality at \((x,y)\), so the criterion holds everywhere.)