Marginals and Independence from Characteristic Functions

Why Marginals and Independence Can Be Read from One Function

A joint characteristic function records information about an entire random vector in one object. Its arguments correspond to the coordinates of that vector, so choosing some arguments to be zero isolates the others. Comparing the resulting marginal characteristic functions with the joint one also gives a precise test for independence.

These tools apply whether the law has a density, consists of point masses, or has both discrete and continuous parts. The key facts are that a characteristic function determines its probability distribution uniquely, and that independence makes the joint characteristic function factor into marginal characteristic functions.

Recovering Marginal Characteristic Functions

Let $X=(X_1,\ldots,X_d)^T$ be a random vector in $\mathbb{R}^d$. Its characteristic function is $\varphi_X(t)=E[\exp(i t^T X)]$, for $t\in\mathbb{R}^d$, where $i^2=-1$. The characteristic function of a scalar random variable $X_j$ is $\varphi_{X_j}(u)=E[\exp(iuX_j)]$, for $u\in\mathbb{R}$.

Theorem: For any coordinate $j$, the marginal characteristic function is obtained from the joint one by setting every argument except the $j$th to zero: $$ \varphi_{X_j}(u) = \varphi_X(0,\ldots,0,u,0,\ldots,0). $$ More generally, for a set of coordinates $I\subseteq\{1,\ldots,d\}$, the characteristic function of the subvector $X_I$ is obtained by setting the arguments outside $I$ to zero.

To see the reason, the exponential in the definition of the joint characteristic function is $\exp(i\sum_{k=1}^d t_kX_k)$. If $t_k=0$ for every $k\ne j$, this becomes $\exp(iuX_j)$. Taking expectations gives the displayed identity. The same calculation works for a subvector: only terms with indices in $I$ remain.

This identity does not require a joint density or an integration over the other coordinates. It works directly from expectations and therefore applies to any probability law on $\mathbb{R}^d$.

Worked Example: Marginals and Dependence in a Two-Point Law

Suppose $(X,Y)$ equals $(0,0)$ with probability $1/2$ and $(1,1)$ with probability $1/2$. Find the joint and marginal characteristic functions, then use them to assess independence.

Joint characteristic function. By summing over the two possible outcomes, $$ \varphi_{X,Y}(s,t) =\frac12 e^{i(s\cdot 0+t\cdot 0)} +\frac12 e^{i(s\cdot 1+t\cdot 1)} =\frac{1+e^{i(s+t)}}{2}. $$

Marginal characteristic functions. Set $t=0$ to obtain the characteristic function of $X$, and set $s=0$ to obtain the one for $Y$: $$ \varphi_X(s)=\frac{1+e^{is}}{2}, \qquad \varphi_Y(t)=\frac{1+e^{it}}{2}. $$ Each marginal is a Bernoulli random variable with probability $1/2$ at 1.

Independence check. If $X$ and $Y$ were independent, the joint characteristic function would equal $\varphi_X(s)\varphi_Y(t)$ for every $s,t$. At $s=t=\pi$, the actual joint characteristic function is $$ \varphi_{X,Y}(\pi,\pi)=\frac{1+e^{i2\pi}}{2}=1, $$ whereas the product of the marginal characteristic functions is $$ \varphi_X(\pi)\varphi_Y(\pi) =\left(\frac{1+e^{i\pi}}{2}\right)^2=0. $$ The functions do not factor, so $X$ and $Y$ are not independent. Indeed, the two coordinates always agree.

Uniqueness: A Characteristic Function Identifies Its Law

The marginal identity produces a characteristic function, but a further result is needed to conclude that two random variables have the same distribution. The uniqueness theorem supplies that link.

Theorem (Uniqueness of characteristic functions): If two probability distributions on $\mathbb{R}^d$ have the same characteristic function at every $t\in\mathbb{R}^d$, then they are the same probability distribution. Equivalently, random vectors $X$ and $Y$ in $\mathbb{R}^d$ satisfy $X\overset{d}=Y$ if and only if $\varphi_X(t)=\varphi_Y(t)$ for all $t\in\mathbb{R}^d$.

Here is a proof of the direction that requires justification. Let $\mu$ and $\nu$ be probability distributions with the same characteristic function. For $\varepsilon>0$, convolve each distribution with the centered Gaussian distribution having covariance matrix $\varepsilon I_d$. The resulting distributions have densities $$ p_{\mu,\varepsilon}(x)=\int_{\mathbb{R}^d}g_\varepsilon(x-y)\,d\mu(y), \qquad p_{\nu,\varepsilon}(x)=\int_{\mathbb{R}^d}g_\varepsilon(x-y)\,d\nu(y), $$ where $g_\varepsilon$ is the density of $N(0,\varepsilon I_d)$.

The Gaussian Fourier identity gives $$ g_\varepsilon(z) =(2\pi)^{-d}\int_{\mathbb{R}^d} e^{-it^Tz}e^{-\varepsilon\lVert t\rVert^2/2}\,dt. $$ Substitute this expression into $p_{\mu,\varepsilon}(x)$. Interchanging the integrals is valid because $e^{-\varepsilon\lVert t\rVert^2/2}$ is integrable and $\mu$ is a probability measure, so the absolute integral is finite. The result is $$ p_{\mu,\varepsilon}(x) =(2\pi)^{-d}\int_{\mathbb{R}^d} e^{-it^Tx}e^{-\varepsilon\lVert t\rVert^2/2}\varphi_\mu(t)\,dt. $$ The same formula holds with $\nu$ in place of $\mu$. Since $\varphi_\mu=\varphi_\nu$, the smoothed densities agree for every $x$ and every $\varepsilon>0$.

Let $h$ be any bounded continuous function. If $Z\sim N(0,I_d)$ is independent of a random vector with law $\mu$, then $$ \int h\,d(\mu*N(0,\varepsilon I_d)) =E[h(X+\sqrt{\varepsilon}Z)]\longrightarrow E[h(X)] =\int h\,d\mu $$ as $\varepsilon\downarrow0$. For every fixed pair of finite vectors $X,Z$, continuity gives $h(X+\sqrt{\varepsilon}Z)\to h(X)$; boundedness permits the use of the dominated convergence theorem. The same reasoning applies to $\nu$. Since the smoothed distributions agree for each $\varepsilon>0$, their integrals of $h$ agree, and passing to the limit shows $\int h\,d\mu=\int h\,d\nu$.

For any closed set $F$, the functions $h_k(x)=\max\{0,1-k\,\operatorname{dist}(x,F)\}$ are bounded and continuous and converge pointwise to the indicator of $F$. Dominated convergence therefore gives $\mu(F)=\nu(F)$. Closed sets are closed under finite intersections and generate the Borel sets, so the probability measures agree on all Borel sets. Thus $\mu=\nu$. Conversely, equal distributions have equal expectations of the bounded function $x\mapsto e^{it^Tx}$, and hence equal characteristic functions. $\square$

Worked Example: Identifying a Distribution from Its Characteristic Function

Suppose a real-valued random variable $W$ is known only to have characteristic function $\varphi_W(t)=e^{-t^2/2}$ for every real $t$. Identify its distribution.

A standard normal random variable $Z\sim N(0,1)$ has characteristic function $\varphi_Z(t)=e^{-t^2/2}$. Thus $\varphi_W(t)=\varphi_Z(t)$ for every $t$. By the uniqueness theorem, $W\overset{d}=Z$, so $W\sim N(0,1)$. No density assumption on $W$ was needed: the characteristic function alone identifies its law.

Factorization Characterizes Independence

For a partition of the coordinates into two disjoint blocks $I$ and $J$, write $X_I$ and $X_J$ for the corresponding subvectors. Independence means that their joint distribution is the product of their marginal distributions.

Theorem: The subvectors $X_I$ and $X_J$ are independent if and only if, for every pair of argument vectors $s$ and $t$ of the corresponding dimensions, $$ \varphi_{X_I,X_J}(s,t)=\varphi_{X_I}(s)\varphi_{X_J}(t). $$ In particular, the coordinates $X_1,\ldots,X_d$ are mutually independent if and only if $$ \varphi_X(t_1,\ldots,t_d)=\prod_{j=1}^d\varphi_{X_j}(t_j) $$ for all $t_1,\ldots,t_d\in\mathbb{R}$.

If $X_I$ and $X_J$ are independent, their joint law is the product of their marginal laws. Under a product law, the expectation of the product of bounded functions of the separate blocks factors. Applying this fact to the complex-valued functions $e^{is^TX_I}$ and $e^{it^TX_J}$ gives $$ \varphi_{X_I,X_J}(s,t) =E[e^{is^TX_I}e^{it^TX_J}] =E[e^{is^TX_I}]E[e^{it^TX_J}] =\varphi_{X_I}(s)\varphi_{X_J}(t). $$

For the converse, let $\mu_I$ and $\mu_J$ be the marginal laws. Their product measure $\mu_I\otimes\mu_J$ is a probability distribution on the joint space, and its characteristic function is $\varphi_{X_I}(s)\varphi_{X_J}(t)$. If the displayed factorization holds, this is also the characteristic function of the actual joint law of $(X_I,X_J)$. Uniqueness implies that the joint law is $\mu_I\otimes\mu_J$, which is precisely independence. Applying the two-block result repeatedly, or comparing with the product of all coordinate marginal laws, gives the stated criterion for mutual independence. $\square$

Worked Example: Checking Factorization for Independent Discrete Coordinates

Let $X$ and $Y$ be independent, with $P(X=1)=1/3$, $P(X=0)=2/3$, $P(Y=2)=3/4$, and $P(Y=0)=1/4$. Compute their characteristic functions and verify the joint factorization.

The marginal characteristic functions are $$ \varphi_X(s)=\frac23+\frac13e^{is}, \qquad \varphi_Y(t)=\frac14+\frac34e^{2it}. $$ By independence, the four joint probabilities are products of the marginal probabilities: $$ P(0,0)=\frac16,\quad P(0,2)=\frac12,\quad P(1,0)=\frac1{12},\quad P(1,2)=\frac14. $$ They are nonnegative and sum to $1/6+1/2+1/12+1/4=1$.

Summing over these outcomes gives $$ \begin{aligned} \varphi_{X,Y}(s,t) &=\frac16+\frac12e^{2it}+\frac1{12}e^{is} +\frac14e^{i(s+2t)}\\ &=\left(\frac23+\frac13e^{is}\right) \left(\frac14+\frac34e^{2it}\right)\\ &=\varphi_X(s)\varphi_Y(t). \end{aligned} $$ For instance, the coefficient of $e^{i(s+2t)}$ in the product is $(1/3)(3/4)=1/4$, matching $P(X=1,Y=2)$. Factorization holds for every $s,t$, consistently with independence.

Common Pitfall and Practical Use

A frequent error is to conclude that two coordinates are independent merely because their marginal characteristic functions are known. Marginal information alone does not determine the joint distribution: many different joint laws can have the same marginals. Independence requires the additional factorization identity for every possible pair of arguments.

The first worked example makes the distinction explicit. Both coordinates have Bernoulli $(1/2)$ marginals, but their joint characteristic function fails to factor, and they are dependent. Checking the identity at a single pair of arguments where it fails is enough to disprove independence; establishing independence by this method requires factorization for all arguments.

These results are useful when a joint characteristic function is easier to obtain than a joint density, or when a law is discrete, mixed, or concentrated on a lower-dimensional set. Setting arguments to zero extracts the marginal characteristic functions, while uniqueness turns equality of characteristic functions into equality of distributions. Factorization then converts a functional identity into an exact statement about independence.

Check Your Understanding

Answer these questions using the definitions and results above.

For $X=(X_1,X_2,X_3)^T$, write the joint characteristic-function expression that gives the marginal characteristic function of $(X_1,X_3)^T$.
Two random vectors have equal characteristic functions at every argument. What conclusion follows, and which theorem justifies it?
Why does knowing the marginal characteristic functions of $X$ and $Y$ not by itself establish independence?
If $X$ and $Y$ are independent, what identity must their joint characteristic function satisfy for every $s,t$?
If a random variable has characteristic function $e^{-4t^2}$, what normal distribution has that characteristic function? Use the convention that the second parameter of $N(\mu,\sigma^2)$ is the variance.