Factorization Criteria for Independence

Why Factorization Detects Independence

A joint characteristic function records the distribution of a random vector through expectations of complex exponentials. Independence has a particularly simple signature in this transform: the joint characteristic function is the product of the marginal characteristic functions. The converse is equally important. If that product identity holds for every vector of arguments, the coordinates really are independent.

The phrase for every vector of arguments is essential. An equality at a single point, or even a comparison of moments such as covariance, is not the factorization criterion. We will state the criterion for any finite-dimensional random vector and then apply it to concrete bivariate distributions.

Definition: For a random vector $X=(X_1,\ldots,X_d)$ with values in $\mathbb{R}^d$, its joint characteristic function is $$ \varphi_X(t)=E\!\left[\exp\!\left(i\sum_{j=1}^d t_jX_j\right)\right],\qquad t=(t_1,\ldots,t_d)\in\mathbb{R}^d. $$ The marginal characteristic function of coordinate $X_j$ is $\varphi_{X_j}(t_j)=E[e^{it_jX_j}]$. The coordinates are independent if, for all Borel sets $A_1,\ldots,A_d\subseteq\mathbb{R}$, $P(X_1\in A_1,\ldots,X_d\in A_d)=\prod_{j=1}^dP(X_j\in A_j)$.

The Factorization Criterion

The theorem gives an equivalence, not just a useful implication. In the bivariate case, write the joint characteristic function as $\varphi_{X,Y}(s,t)=E[e^{i(sX+tY)}]$. Factorization means equality with $\varphi_X(s)\varphi_Y(t)$ at every $(s,t)\in\mathbb{R}^2$.

Theorem: The coordinates $X_1,\ldots,X_d$ of a random vector are independent if and only if $$ \varphi_X(t_1,\ldots,t_d)=\prod_{j=1}^d\varphi_{X_j}(t_j) \quad\text{for every }(t_1,\ldots,t_d)\in\mathbb{R}^d. $$

For the forward implication, independence allows expectations of products of bounded measurable functions of separate coordinates to factor. The reverse implication uses the fact that a probability distribution is determined by its characteristic function. Since that fact is central to the converse, the following proof establishes the needed determination result by Gaussian smoothing.

Independence implies factorization.
If the coordinates are independent, apply factorization of expectations to the bounded functions $x\mapsto e^{it_jx}$.

Factorization identifies the joint transform.
The product of the marginal characteristic functions is the characteristic function of the product of the marginal laws.

Equal transforms identify the laws.
Gaussian smoothing shows that two probability laws with the same characteristic function must be equal. The joint law is therefore the product law.

Proof of the Theorem

If $X_1,\ldots,X_d$ are independent, the random variables $e^{it_jX_j}$ are independent and bounded. Hence

\varphi_X(t)=E\!\left[\prod_{j=1}^d e^{it_jX_j}\right] =\prod_{j=1}^d E[e^{it_jX_j}] =\prod_{j=1}^d\varphi_{X_j}(t_j).

For the converse, let $\mu$ be the joint probability law of $X$, and let $\nu$ be the product of its marginal laws. By iterated integration, the characteristic function of $\nu$ is $\prod_{j=1}^d\varphi_{X_j}(t_j)$. The assumed factorization therefore says that $\mu$ and $\nu$ have the same characteristic function.

We verify that two probability measures on $\mathbb{R}^d$ with the same characteristic function are equal. Let $\alpha$ and $\beta$ be such measures. For $\varepsilon>0$, let $\gamma_\varepsilon$ be the $N(0,\varepsilon^2 I_d)$ law, with density $g_\varepsilon$. The convolutions $\alpha*\gamma_\varepsilon$ and $\beta*\gamma_\varepsilon$ have densities $p_{\alpha,\varepsilon}(x)=\int g_\varepsilon(x-y)\,\alpha(dy)$ and $p_{\beta,\varepsilon}(x)=\int g_\varepsilon(x-y)\,\beta(dy)$.

The Gaussian Fourier integral gives

g_\varepsilon(z)=\frac{1}{(2\pi)^d} \int_{\mathbb{R}^d}e^{-it\cdot z}e^{-\varepsilon^2|t|^2/2}\,dt.

Substitute this expression into the density formula. The integral of $e^{-\varepsilon^2|t|^2/2}$ over $\mathbb{R}^d$ is finite, and $\alpha$ is a probability measure, so Fubini's theorem applies to the absolutely integrable expression. With $\varphi_\alpha(t)=\int e^{it\cdot y}\alpha(dy)$, this gives

p_{\alpha,\varepsilon}(x)=\frac{1}{(2\pi)^d} \int_{\mathbb{R}^d}e^{-it\cdot x}\varphi_\alpha(t)e^{-\varepsilon^2|t|^2/2}\,dt.

The same representation holds for $\beta$. Their characteristic functions agree, so their smoothed densities agree at every $x$, for each $\varepsilon>0$.

Now let $f$ be any bounded continuous real-valued function on $\mathbb{R}^d$, and let $Z\sim N(0,I_d)$ be independent of a vector with law $\alpha$. The integral of $f$ under $\alpha*\gamma_\varepsilon$ is $E[f(Y+\varepsilon Z)]$, where $Y$ has law $\alpha$. As $\varepsilon$ decreases to zero, $f(Y+\varepsilon Z)$ converges pointwise to $f(Y)$, and its absolute value is bounded by $\sup_x|f(x)|$. Dominated convergence shows that these integrals tend to $\int f\,d\alpha$. The same argument applies to $\beta$. Since the smoothed laws agree for every $\varepsilon>0$, it follows that $\int f\,d\alpha=\int f\,d\beta$ for every bounded continuous $f$.

In particular, the two measures assign equal probabilities to every open set $G$. For a proper open set, the bounded continuous functions $f_k(x)=\min\{1,k\,\operatorname{dist}(x,G^c)\}$ increase pointwise to its indicator; monotone convergence gives the claim. The whole space has probability one under both measures. Since open sets are closed under finite intersections and generate the Borel sets, the uniqueness theorem for measures implies $\alpha=\beta$.

Applying this conclusion to $\mu$ and $\nu$ shows that the joint law equals the product of its marginals. Thus the coordinates are independent. $\square$

Worked Bivariate Examples

Worked Example: An Independent Rademacher Pair

Let $U$ and $V$ be independent, each taking the values $-1$ and $1$ with probability $1/2$. Set $X=U$ and $Y=UV$. The map $(u,v)\mapsto(u,uv)$ is a bijection on the four possible sign pairs. Each pair $(X,Y)\in\{(-1,-1),(-1,1),(1,-1),(1,1)\}$ therefore has probability $1/4$; the coordinates are independent Rademacher variables.

For a Rademacher variable, its characteristic function is $E[e^{itU}]=(e^{it}+e^{-it})/2=\cos t$. Directly using the four equally likely pairs,

\varphi_{X,Y}(s,t) =\frac14\sum_{x\in\{-1,1\}}\sum_{y\in\{-1,1\}}e^{i(sx+ty)} =\left(\frac{e^{is}+e^{-is}}2\right) \left(\frac{e^{it}+e^{-it}}2\right) =\cos s\cos t.

The marginals have $\varphi_X(s)=\cos s$ and $\varphi_Y(t)=\cos t$, so the joint characteristic function factors at every $(s,t)$, as required.

Worked Example: A Dependent Pair That Fails Factorization

Let $X=Y=U$, where $U$ is Rademacher. Both marginals have characteristic function $\cos t$, but the pair is dependent: for example, $P(X=1,Y=-1)=0$, whereas $P(X=1)P(Y=-1)=1/4$.

The joint characteristic function is $\varphi_{X,Y}(s,t)=E[e^{i(s+t)U}]=\cos(s+t)$. At $s=t=\pi/2$, the joint value is $\cos\pi=-1$, while the product of marginal values is $\cos(\pi/2)\cos(\pi/2)=0$. The identity fails, correctly ruling out independence.

Worked Example: Zero Covariance Does Not Ensure Independence

Let $X$ take $-1,0,1$, each with probability $1/3$, and set $Y=X^2$. Then $E[X]=0$, $E[Y]=2/3$, and $E[XY]=E[X^3]=0$. Thus $\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=0$. Nevertheless, $Y$ is determined by $X$, so the pair is dependent.

The joint characteristic function is $\varphi_{X,Y}(s,t)=(1+e^{i(-s+t)}+e^{i(s+t)})/3=(1+2e^{it}\cos s)/3$. The marginal functions are $\varphi_X(s)=(1+2\cos s)/3$ and $\varphi_Y(t)=(1+2e^{it})/3$. At $s=t=\pi$,

\varphi_{X,Y}(\pi,\pi)=\frac{1+2(-1)(-1)}3=1, \qquad \varphi_X(\pi)\varphi_Y(\pi) =\left(-\frac13\right)\left(-\frac13\right)=\frac19.

The functions do not factor, despite zero covariance. Covariance captures only a particular second-moment relationship; independence requires the full factorization identity.

Common Pitfalls and Uses

The criterion is useful because characteristic functions turn a question about a joint distribution into an identity between functions. It applies to discrete, continuous, and mixed distributions without requiring a joint density or finite moments. In particular, a student need not calculate covariance before using it.

Checking only selected arguments. Matching the two sides at $(0,0)$ proves nothing; both sides always equal one there. Independence requires the identity for all real arguments. A disagreement at even one point, however, is enough to establish dependence.
Reversing the implication without justification. Independence gives factorization directly, but factorization implies independence because characteristic functions determine probability laws. Merely writing the product identity without connecting it to equality of the joint and product laws leaves the converse unproved.
Treating zero covariance as independence. Zero covariance can rule out some forms of dependence but does not generally establish independence. The $Y=X^2$ example demonstrates this explicitly.

In practice, the strongest use is often negative: find a single pair of arguments where the joint characteristic function differs from the product of the marginals, and dependence follows immediately. To establish independence by this method, verify factorization as a functional identity over the entire argument space.

Check Your Understanding

Use the definition and proof above to answer each question.

State the factorization criterion for a random vector with three coordinates.
For $X=Y=U$, where $U$ is Rademacher, evaluate the joint and product characteristic functions at $s=t=\pi/2$.
Why does the reverse direction of the criterion require a result about determination of laws by characteristic functions?
In the zero-covariance example, verify $E[X]=0$, $E[Y]=2/3$, and $\operatorname{Cov}(X,Y)=0$.
Does checking factorization at $(0,0)$ provide evidence of independence? Explain.