Why a Population Characterization Matters
A dependence statistic is useful only if its population target separates independence from dependence. A statistic that happens to be small in one sample may reflect sampling variation; a population discrepancy that is strictly positive under every fixed dependent distribution gives a different guarantee. It supplies the separation that allows an empirical statistic, once it converges, to detect fixed alternatives with probability tending to one.
Here the comparison is between the joint characteristic function and the product of the marginal characteristic functions. The main points are that a suitable weighted squared discrepancy is zero exactly under independence, and that the empirical version converges to this population quantity. The weight is required to be integrable and strictly positive at every frequency, but it need not be continuous.
Population Discrepancy and Independence
Let \(X\) take values in \(\mathbb{R}^p\) and \(Y\) in \(\mathbb{R}^q\). Write their joint and marginal characteristic functions as \(\varphi_{XY}(s,t)=E[\exp\{i(s^\mathsf{T}X+t^\mathsf{T}Y)\}]\), \(\varphi_X(s)=E[\exp(i s^\mathsf{T}X)]\), and \(\varphi_Y(t)=E[\exp(i t^\mathsf{T}Y)]\), for \(s\in\mathbb{R}^p\) and \(t\in\mathbb{R}^q\). Define their difference by \(\Delta(s,t)=\varphi_{XY}(s,t)-\varphi_X(s)\varphi_Y(t)\).
Since characteristic functions have modulus at most one, \(|\Delta(s,t)|\leq 2\), so \(0\leq D_w\leq 4W<\infty\). The population characterization is exact.
Proof. If \(X\) and \(Y\) are independent, their joint characteristic function factors into the product of their marginal characteristic functions at every \((s,t)\). Thus \(\Delta(s,t)=0\) everywhere, and \(D_w=0\).
Conversely, suppose \(D_w=0\). Characteristic functions are continuous, so \(\Delta\), and hence \(|\Delta|\), is continuous. If \(\Delta(s_0,t_0)\neq 0\) at some point, continuity gives a ball \(B\) around that point and a constant \(c>0\) such that \(|\Delta(s,t)|\geq c\) throughout \(B\). The ball has positive Lebesgue measure. Because \(w\) is measurable and strictly positive at every point, write \(B=\bigcup_{k=1}^{\infty}\{(s,t)\in B:w(s,t)\geq 1/k\}\). At least one set in this union has positive Lebesgue measure; otherwise the countable union would have measure zero. Therefore \(\int_B w(s,t)\,ds\,dt>0\), and \(D_w\geq c^2\int_B w(s,t)\,ds\,dt>0\), a contradiction. It follows that \(\Delta=0\) everywhere. The joint characteristic function is therefore the characteristic function of the product of the marginal laws. By uniqueness of characteristic functions, the joint law equals that product law, which is independence. \(\square\)
The converse does not require a continuous weight. The argument needs measurability and strict positivity to ensure positive weight integral over the ball where the continuous discrepancy is bounded away from zero. If a weight were merely nonnegative and could vanish throughout that ball, this argument would not apply.
Worked Example: Independent Gaussian Margins
Suppose \(X\sim N(0,1)\) and \(Y\sim N(2,4)\), independently. Their characteristic functions are \(\varphi_X(s)=\exp(-s^2/2)\) and \(\varphi_Y(t)=\exp(2it-2t^2)\), since the second argument of \(N(\mu,\sigma^2)\) is the variance.
Independence gives \(\varphi_{XY}(s,t)=\exp(-s^2/2+2it-2t^2)=\varphi_X(s)\varphi_Y(t)\). Consequently \(\Delta(s,t)=0\) for every pair of frequencies, and \(D_w=0\) for every admissible weight. The result is not specific to a Gaussian weight or to Gaussian variables; independence makes the discrepancy vanish pointwise.
A Dependent Distribution with a Positive Discrepancy
A useful check is a distribution where dependence and the discrepancy can both be calculated explicitly. Let \(X,Y\) each take values in \(\{-1,1\}\), with \(P(X=x,Y=y)=(1+\rho xy)/4\) for \(x,y\in\{-1,1\}\), where \(0<|\rho|\leq 1\). The four probabilities are nonnegative and sum to one; each marginal is uniform on \(\{-1,1\}\). Also \(E[XY]=\rho\), so the variables are dependent when \(\rho\neq0\).
Worked Example: Computing the Discrepancy for a Binary Pair
For the uniform marginals, \(\varphi_X(s)=\cos s\) and \(\varphi_Y(t)=\cos t\). Using \(\exp(iu)=\cos u+i\sin u\), symmetry gives \(\varphi_{XY}(s,t)=\cos s\cos t-\rho\sin s\sin t\). Thus \(\Delta(s,t)=-\rho\sin s\sin t\).
Choose the integrable, everywhere positive weight \(w(s,t)=\exp\{-(s^2+t^2)/2\}\). Separating the integrals yields $$ D_w=\rho^2\left(\int_{-\infty}^{\infty}e^{-s^2/2}\sin^2(s)\,ds\right)^2. $$
Since \(\sin^2(s)=(1-\cos(2s))/2\), and the Gaussian Fourier integral gives \(\int_{-\infty}^{\infty}e^{-s^2/2}\cos(2s)\,ds=\sqrt{2\pi}e^{-2}\), the one-dimensional integral is \(\frac{\sqrt{2\pi}}{2}(1-e^{-2})\). Therefore $$ D_w=\rho^2\frac{\pi}{2}(1-e^{-2})^2>0. $$
This calculation verifies directly that the population discrepancy detects the dependence. It also illustrates why looking only near zero frequency, or only at covariance, is not the definition of the test: the discrepancy integrates the squared difference over all frequencies.
Empirical Convergence
Now suppose \((X_1,Y_1),\ldots,(X_n,Y_n)\) are independent and identically distributed copies of \((X,Y)\). Define \(\widehat{\varphi}_{XY}(s,t)=n^{-1}\sum_{j=1}^n\exp\{i(s^\mathsf{T}X_j+t^\mathsf{T}Y_j)\}\), \(\widehat{\varphi}_X(s)=n^{-1}\sum_{j=1}^n\exp(i s^\mathsf{T}X_j)\), and define \(\widehat{\varphi}_Y(t)\) analogously. Set \(\Delta_n(s,t)=\widehat{\varphi}_{XY}(s,t)-\widehat{\varphi}_X(s)\widehat{\varphi}_Y(t)\), and \(D_{n,w}=\int|\Delta_n(s,t)|^2w(s,t)\,ds\,dt\).
Proof. At each fixed frequency, the strong law of large numbers applied to the real and imaginary parts of the bounded summands gives convergence of each empirical characteristic function to its population counterpart almost surely. Hence \(\Delta_n(s,t)\to\Delta(s,t)\) almost surely at every fixed \((s,t)\).
To use this pointwise statement inside an integral, let \(\nu\) be the finite measure with density \(w\) relative to Lebesgue measure. The set of pairs consisting of a sample outcome and a frequency where \(\Delta_n\) fails to converge is measurable. For every fixed frequency it has probability zero by the strong law. Tonelli's theorem applied to its indicator shows that, with probability one, convergence holds for \(\nu\)-almost every frequency. For all frequencies, \(|\Delta_n|\leq2\), so \(|\Delta_n|^2\leq4\); the constant \(4\) is integrable under the finite measure \(\nu\). The dominated convergence theorem, on each sample outcome in this probability-one event, now gives \(D_{n,w}\to D_w\). \(\square\)
This argument does not claim simultaneous pointwise convergence at every frequency on one probability-one event. It establishes the almost-everywhere-in-frequency convergence needed for the weighted integral, which is enough for dominated convergence.
Worked Example: A Nonlinear Fixed Alternative
Let \(X\sim N(0,1)\) and \(Y=X^2\). These variables are dependent: if they were independent, the finite moments would satisfy \(E[X^2Y]=E[X^2]E[Y]\). But \(E[X^2Y]=E[X^4]=3\), whereas \(E[X^2]E[Y]=1\cdot1=1\). Thus independence is impossible.
For the Gaussian weight used in the binary example, \(w(s,t)=e^{-(s^2+t^2)/2}\), the population theorem implies \(D_w>0\). The empirical convergence theorem then gives \(D_{n,w}\to D_w\) almost surely. In particular, for any deterministic threshold sequence \(\tau_n\to0\), the event \(D_{n,w}>\tau_n\) occurs eventually almost surely: once \(D_{n,w}\) is sufficiently close to the positive limit, it exceeds the shrinking threshold. Therefore the rejection probability tends to one against this fixed nonlinear alternative.
A Shrinking-Threshold Test
The convergence theorem gives a direct consistency result. For any fixed dependent distribution, \(D_w>0\), while under independence \(D_w=0\). A threshold tending to zero is therefore eventually below the empirical statistic under each fixed alternative. A threshold alone does not automatically provide a useful null error rate, so it is helpful to verify one simple bound as well.
Under independence, write \(\mu=\varphi_X(s)\), \(\nu=\varphi_Y(t)\), and let \(\bar A,\bar B,\bar C\) denote the empirical means of the joint, \(X\)-marginal, and \(Y\)-marginal exponential terms at fixed \((s,t)\). Then \(E[A]=\mu\nu\). Since \(|\bar C|\leq1\) and \(|\mu|\leq1\), \(|\bar B\bar C-\mu\nu|\leq|\bar B-\mu|+|\bar C-\nu|\). It follows that \(|\Delta_n|\leq|\bar A-E[A]|+|\bar B-\mu|+|\bar C-\nu|\). Using \((a+b+c)^2\leq3(a^2+b^2+c^2)\), and the identity \(E|\bar Z-E[Z]|^2=E|Z-E[Z]|^2/n\leq1/n\) for an average of iid complex variables with \(|Z|=1\), gives \(E|\Delta_n(s,t)|^2\leq9/n\). Tonelli's theorem then yields \(E[D_{n,w}]\leq9W/n\) under independence.
Worked Example: A Threshold with Vanishing Null Rejection Probability
Take the same Gaussian weight \(w(s,t)=e^{-(s^2+t^2)/2}\). Its total integral is \(W=2\pi\), because each one-dimensional Gaussian integral equals \(\sqrt{2\pi}\). Use the rule that rejects independence when \(D_{n,w}>n^{-1/2}\).
Under independence, Markov's inequality and the bound just derived give $$ P(D_{n,w}>n^{-1/2})\leq\frac{E[D_{n,w}]}{n^{-1/2}} \leq\frac{9(2\pi)}{\sqrt n}. $$ This upper bound tends to zero, so the rule has null rejection probability tending to zero. Under any fixed dependent distribution, \(D_{n,w}>0\) and \(D_{n,w}^{\text{empirical}}\to D_{n,w}\) almost surely, while \(n^{-1/2}\to0\); hence its power tends to one. The bound is conservative, but it verifies both claims without requiring a null limit distribution.
Common Pitfall and Practical Meaning
A common mistake is to infer that a zero integral forces \(\Delta\) to vanish everywhere just because the integrand is nonnegative. A zero integral first implies only that the integrand is zero almost everywhere with respect to the weighted measure. The additional steps matter: continuity of the characteristic-function difference rules out an isolated nonzero value, and strict positivity of the measurable weight gives positive weighted mass to any ball where the difference stays away from zero. Continuity of the weight is not needed.
The other essential distinction is between fixed and local alternatives. The proof gives power tending to one for each fixed dependent distribution, whose population discrepancy is a fixed positive number. It does not by itself describe power against alternatives that change with \(n\) and approach independence. The characterization and convergence results establish consistency, not a rate of detection for such changing alternatives.
Check Your Understanding
Use the definitions and arguments above to answer the following.
- Why does independence make the population discrepancy zero at every frequency, rather than merely making its integral zero?
- In the converse proof, what roles are played by continuity of \(\Delta\) and strict positivity of \(w\)?
- Why is pointwise almost-sure convergence at each fixed frequency not, by itself, the same as convergence at every frequency on one probability-one event?
- Under a fixed dependent alternative, which two limits ensure that a threshold \(\tau_n\to0\) is eventually crossed?
- Does the shrinking-threshold argument establish power against alternatives that themselves approach independence as \(n\) increases? Explain.