What a Rejection Establishes
An independence test addresses a specific population question: does the joint distribution of \(X\) and \(Y\) factor into their marginal distributions? A rejection is evidence against that factorization. It does not, by itself, name the relationship that replaces independence. The same decision can arise from a linear trend, a curved pattern, changing conditional variability, or another departure.
This distinction matters for distance covariance and kernel tests. Their ability to detect broad classes of dependence does not make a rejection a description of the dependence. Interpretation requires a second step: examine the paired observations using summaries and plots suited to plausible forms of association. Those diagnostics help describe what the data show; they do not turn an omnibus test into a test of a particular mechanism.
Suppose a test is calibrated at level \(\alpha\) under independence and rejects when its p-value is at most \(\alpha\). Its formal conclusion is that the observed data provide evidence against the independence null at that test level. A level-\(\alpha\) procedure limits the probability of rejection under the null to at most \(\alpha\), under the assumptions used for its calibration. This is not the probability that the null is true, nor does it quantify the strength or practical importance of the relationship.
Even when a test is designed to be consistent against a broad class of fixed dependent alternatives, that property concerns its long-run ability to reject independence. It does not guarantee that a particular finite sample will reveal the pattern clearly, and it does not identify which alternative generated the observations.
Covariance Is Not a Complete Description
Covariance is a useful summary of linear association when its moments exist, but it is not an independence test in general. The following result explains one direction of the relationship between independence and covariance; its converse fails.
Proof. Finite second moments imply finite first moments. By the Cauchy–Schwarz inequality, \(E[|XY|]\leq \sqrt{E[X^2]E[Y^2]}<\infty\). Independence therefore gives \(E[XY]=E[X]E[Y]\), and hence \(\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=0\). To see why the converse fails, the next example gives dependent variables with zero covariance. \(\square\)
Worked Example: A Curved Relationship With Zero Covariance
Let \(X\sim N(0,1)\) and let \(\varepsilon\sim N(0,0.25)\) be independent of \(X\). Define \(Y=X^2-1+\varepsilon\). This is an explicit paired distribution that can be simulated by generating independent \(X\) and \(\varepsilon\) values and forming each pair \((X,Y)\).
The normal moments give \(E[X]=0\), \(E[X^2]=1\), and \(E[X^3]=0\). Thus \(E[Y]=E[X^2]-1+E[\varepsilon]=0\), and \[ \operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y] =E[X^3-X+X\varepsilon]=0. \] Here \(E[X\varepsilon]=E[X]E[\varepsilon]=0\) by independence. Nevertheless, \(E[Y\mid X]=X^2-1\), which varies with \(X\). The conditional distribution of \(Y\) therefore changes with \(X\), so the variables are dependent.
A scatterplot would tend to show a noisy U-shaped pattern, while a straight-line correlation summary could be near zero. If an independence test rejects for data generated this way, the justified conclusion is evidence of dependence—not evidence of a linear trend. A scatterplot and a smooth estimate of the conditional mean are natural descriptive follow-ups.
The same point applies when \(X\) or \(Y\) is a vector: a covariance matrix records pairwise linear summaries, not the full joint distribution. A zero cross-covariance matrix is not, without additional assumptions, a certificate of independence.
Follow-Up Summaries for Different Patterns
A useful follow-up starts with the structure of the variables and the scientific question, rather than with a search for a single replacement test statistic. For scalar pairs, inspect a scatterplot with suitable axis scales. For vector-valued data, examine relevant coordinate pairs and, where meaningful, prespecified projections or summaries. A plot can reveal patterns that a single numerical summary conceals, but a collection of pairwise plots may still miss dependence involving several coordinates jointly.
The conditional distribution offers a useful organizing idea. If \(Y\) is scalar, ask whether its conditional mean, spread, or shape changes across values of \(X\). A mean summary is informative for some patterns; conditional variances or quantiles may be more revealing for others. These summaries are diagnostic descriptions, not automatic proof that a particular feature is the sole source of a test rejection.
Proof. Independence implies that the conditional distribution of \(Y\) given \(X\) equals the marginal distribution of \(Y\), almost surely. Integrating \(y\) and \(y^2\) against these distributions gives \(E[Y\mid X]=E[Y]\) and \(E[Y^2\mid X]=E[Y^2]\) almost surely. Therefore \[ \operatorname{Var}(Y\mid X) =E[Y^2\mid X]-\{E[Y\mid X]\}^2 =E[Y^2]-\{E[Y]\}^2 =\operatorname{Var}(Y) \] almost surely. If either conditional moment is not almost surely constant, the conditional distribution cannot equal the same marginal distribution for almost every \(X\); independence fails. \(\square\)
Worked Example: A Linear Dependence Pattern
For a simulated paired distribution, take independent \(X,\varepsilon\sim N(0,1)\) and set \(Y=0.8X+0.6\varepsilon\). Both variables have mean zero. Since the two components of \(Y\) are independent, \[ \operatorname{Var}(Y)=0.8^2\operatorname{Var}(X)+0.6^2\operatorname{Var}(\varepsilon)=0.64+0.36=1, \] and \(\operatorname{Cov}(X,Y)=0.8\operatorname{Var}(X)+0.6\operatorname{Cov}(X,\varepsilon)=0.8\). Thus the correlation is \(0.8/\sqrt{1\cdot1}=0.8\).
The conditional mean is \(E[Y\mid X]=0.8X\), so a scatterplot should show a positive linear trend, with residual variation around it. If an independence test rejects on a simulated sample from this model, the test has detected evidence against independence; the plot and conditional-mean summary make the linear pattern interpretable. The rejection alone did not establish that the relationship was linear.
Worked Example: Dependence Through Changing Variability
Let \(X\sim\operatorname{Uniform}(-1,1)\), let \(\varepsilon\sim N(0,1)\) independently, and set \(Y=(1+2|X|)\varepsilon\). For every fixed \(X=x\), \(E[Y\mid X=x]=0\). Thus the conditional mean gives no trend. But \[ \operatorname{Var}(Y\mid X=x)=(1+2|x|)^2, \] which increases as \(|x|\) increases. This conditional variance is not constant, so the proposition shows that \(X\) and \(Y\) are dependent.
The marginal variance is finite and can also be calculated. Since \(E[|X|]=1/2\) and \(E[X^2]=1/3\), \[ \operatorname{Var}(Y)=E[(1+2|X|)^2] =1+4E[|X|]+4E[X^2] =1+2+\frac{4}{3}=\frac{13}{3}. \] A simulated scatterplot would tend to have wider vertical spread near both ends of the \(X\)-range, without a systematic mean trend. A plot of squared residuals or conditional spread summaries against \(X\) can help describe this pattern. A mean-only summary could overlook it.
Interpretation Without Overclaiming
An omnibus rejection narrows the population claim that is plausible: under the test’s assumptions and calibration, the data are inconsistent with independence at the stated level. It does not establish the sign, shape, strength, or scientific source of dependence. In particular, it does not show that changing \(X\) would change \(Y\), that the relationship is useful for prediction, or that a specific variable is responsible. Those claims require additional design, modeling, or evidence.
Treat follow-up plots and summaries as a way to understand the observed pairs. A practical sequence is to inspect the joint pattern, look for changes in conditional location and spread, and summarize features that are relevant to the variables and question. If the diagnostic exploration suggests a specific model or a new confirmatory claim, that claim needs an analysis with appropriate inferential support; a pattern noticed after seeing the data is not automatically confirmatory.
Identify the null as independence of the paired random variables, not as “no association of every possible kind.”
Say that the test provides evidence against independence at its stated level; do not name a pattern that the test did not test.
Use joint plots and relevant conditional summaries to investigate linear, curved, spread-changing, or more complex patterns.
Distinguish what the paired data display from causal, mechanistic, or confirmatory claims.
Common Pitfall
Wrong reasoning: “The independence test rejected, so the variables have a linear relationship.” A test of independence does not generally test linearity. The curved example has zero covariance but is dependent, and the variance-changing example has a constant conditional mean but is also dependent.
Correction: State only that the test found evidence against independence. Then use plots and summaries to assess the form of the dependence in the observed data. A visually plausible pattern is a descriptive finding, not a conclusion guaranteed by the rejection.
Check Your Understanding
Use the distinction between evidence against independence and evidence about dependence form to answer these questions.
- What population claim is rejected when a valid independence test rejects its null?
- Why does independence imply zero covariance when second moments are finite, and why does the reverse implication fail?
- For \(Y=X^2-1+\varepsilon\) with independent standard normal \(X\) and \(\varepsilon\), what feature of the conditional mean signals dependence even though covariance is zero?
- What follow-up summary is especially relevant when the conditional mean is constant but the conditional spread changes with \(X\)?
- Name two conclusions that a rejection of independence does not establish on its own.