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Characteristic Functions · Tutorial 550 of 994

Joint Characteristic Functions and Marginal Distributions

Learn to define a joint characteristic function, extract marginal and projection characteristic functions, and calculate them for several types of random vectors.

Advanced 10 min read

What You'll Learn

  • Define the joint characteristic function of a random vector using its vector of arguments.
  • Recover a coordinate’s marginal characteristic function by setting the other arguments to zero.
  • Write the joint characteristic function from a discrete joint probability mass function.
  • Integrate against a continuous joint density and identify the resulting marginals.
  • Calculate the joint characteristic function of a vector whose distribution has no joint density.
  • Obtain the characteristic function of any scalar linear combination by evaluating the joint function along a line.

From a Scalar to a Vector

A scalar characteristic function tests a random variable using the complex exponential \(e^{itX}\). For a random vector, the same idea uses one argument for each coordinate. This gives a single function that can be evaluated in different directions: setting all but one argument to zero isolates one coordinate, while moving along a line of arguments corresponds to a scalar linear combination of the coordinates.

The word joint matters. A joint characteristic function is defined from the distribution of the whole vector, not by multiplying the characteristic functions of its coordinates. The coordinates may be dependent, and the joint function retains information about how they vary together.

Definition: Let \(X=(X_1,\ldots,X_d)^\mathsf{T}\) be a random vector with values in \(\mathbb{R}^d\). Its joint characteristic function is the function \(\varphi_X:\mathbb{R}^d\to\mathbb{C}\) defined by
$$ \varphi_X(t_1,\ldots,t_d) =E\!\left[\exp\!\left(i\sum_{j=1}^d t_jX_j\right)\right], \qquad (t_1,\ldots,t_d)\in\mathbb{R}^d, $$
where \(i^2=-1\). Equivalently, writing \(t=(t_1,\ldots,t_d)^\mathsf{T}\), this is \(\varphi_X(t)=E[e^{i t^\mathsf{T}X}]\). The expectation always exists because the integrand has modulus one.

The vector is assumed to be measurable as a map into \(\mathbb{R}^d\), so \(t^\mathsf{T}X\) and its complex exponential are random variables. No density, mass function, or moment assumption is required by the definition. In particular, it applies to discrete vectors, vectors with a joint density, and vectors whose distributions are concentrated on lower-dimensional sets.

Marginal Characteristic Functions by Coordinate Restriction

For a particular coordinate \(X_j\), let \(\varphi_{X_j}(u)=E[e^{iuX_j}]\) be its marginal characteristic function. In the joint characteristic function, set the \(j\)th argument equal to \(u\) and every other argument equal to zero. Since \(e^{i0X_k}=1\), the resulting expectation is exactly the characteristic function of \(X_j\).

Theorem: For each \(j\in\{1,\ldots,d\}\) and every \(u\in\mathbb{R}\),
$$ \varphi_{X_j}(u) =\varphi_X(0,\ldots,0,u,0,\ldots,0), $$
where \(u\) occupies the \(j\)th coordinate. Thus the marginal characteristic function is obtained by restricting the joint characteristic function to the \(j\)th coordinate axis.

Proof. By the definition of the joint characteristic function, evaluating it at the indicated argument vector gives

$$ \varphi_X(0,\ldots,0,u,0,\ldots,0) =E\!\left[\exp\!\left(i\sum_{k=1}^d t_kX_k\right)\right] =E[e^{iuX_j}] =\varphi_{X_j}(u). $$

Every expectation in this calculation exists because its integrand has modulus one. The identity follows directly from the definition. \(\square\)

The coordinate-axis restriction is also useful when the vector has a joint mass function or density. In those cases, the joint characteristic function can be calculated first, and the restriction can then be checked against the marginal distribution.

Worked Example: A Discrete Vector and Its Marginals

Suppose \((X,Y)\) takes the four values shown below, with the indicated probabilities.

\((x,y)\)Probability
\((0,0)\)\(0.2\)
\((0,1)\)\(0.3\)
\((1,0)\)\(0.1\)
\((2,1)\)\(0.4\)

The probabilities sum to \(0.2+0.3+0.1+0.4=1\), so they specify a probability distribution. For a discrete vector, the expectation defining the joint characteristic function is a sum over its possible values:

$$ \begin{aligned} \varphi_{(X,Y)}(s,t) &=\sum_{(x,y)}P(X=x,Y=y)e^{i(sx+ty)}\\ &=0.2+0.3e^{it}+0.1e^{is}+0.4e^{i(2s+t)}. \end{aligned} $$

Set \(t=0\) to obtain the characteristic function of \(X\):

$$ \varphi_X(s)=\varphi_{(X,Y)}(s,0) =0.2+0.3+0.1e^{is}+0.4e^{2is} =0.5+0.1e^{is}+0.4e^{2is}. $$

This agrees with the marginal probabilities \(P(X=0)=0.2+0.3=0.5\), \(P(X=1)=0.1\), and \(P(X=2)=0.4\). Setting \(s=0\) instead gives

$$ \varphi_Y(t)=\varphi_{(X,Y)}(0,t) =0.2+0.3e^{it}+0.1+0.4e^{it} =0.3+0.7e^{it}, $$

which agrees with \(P(Y=0)=0.2+0.1=0.3\) and \(P(Y=1)=0.3+0.4=0.7\). The two different restrictions recover the two different marginal characteristic functions.

Calculating from a Joint Density

If \(X=(X_1,\ldots,X_d)^\mathsf{T}\) has a joint probability density function \(f\) with respect to \(d\)-dimensional Lebesgue measure, its joint characteristic function can be written as

$$ \varphi_X(t)=\int_{\mathbb{R}^d}e^{it^\mathsf{T}x}f(x)\,dx. $$

This integral is absolutely convergent: \(\lvert e^{it^\mathsf{T}x}f(x)\rvert=f(x)\), and a probability density integrates to one. The formula is the density version of the expectation in the definition.

Worked Example: A Dependent Vector with Uniform Marginals

Let \((X,Y)\) have joint density

$$ f(x,y)= \begin{cases} 1+\dfrac12(2x-1)(2y-1),&0\leq x\leq1,\;0\leq y\leq1,\\ 0,&\text{otherwise}. \end{cases} $$

Because each factor \(2x-1\) and \(2y-1\) lies between \(-1\) and \(1\), the density is at least \(1-\tfrac12=\tfrac12\) on the square, so it is nonnegative. Also,

$$ \int_0^1\int_0^1 f(x,y)\,dy\,dx =1+\frac12\left(\int_0^1(2x-1)\,dx\right) \left(\int_0^1(2y-1)\,dy\right) =1, $$

since both one-dimensional integrals in the product are zero. Thus \(f\) is a probability density.

Define the two one-dimensional integrals

$$ A(u)=\int_0^1e^{iux}\,dx, \qquad B(u)=\int_0^1(2x-1)e^{iux}\,dx. $$

Integrating the joint density against \(e^{i(sx+ty)}\), and separating the two terms of the density, gives

$$ \begin{aligned} \varphi_{(X,Y)}(s,t) &=\int_0^1\int_0^1 e^{i(sx+ty)} \left[1+\frac12(2x-1)(2y-1)\right]\,dy\,dx\\ &=A(s)A(t)+\frac12B(s)B(t). \end{aligned} $$

The integrals are over a bounded square and their integrands are bounded, so separating the product terms and evaluating the iterated integrals is valid. For \(u\ne0\),

$$ A(u)=\frac{e^{iu}-1}{iu}, \qquad A(0)=1. $$

For the other integral, integration by parts gives, when \(u\ne0\),

$$ \int_0^1 xe^{iux}\,dx =\frac{e^{iu}}{iu}+\frac{e^{iu}-1}{u^2}, \qquad B(u)=2\left[\frac{e^{iu}}{iu}+\frac{e^{iu}-1}{u^2}\right]-A(u). $$

At zero, \(B(0)=\int_0^1(2x-1)\,dx=0\). Consequently, restricting to either coordinate axis yields

$$ \varphi_X(s)=\varphi_{(X,Y)}(s,0)=A(s)A(0)+\frac12B(s)B(0)=A(s), $$
$$ \varphi_Y(t)=\varphi_{(X,Y)}(0,t)=A(0)A(t)+\frac12B(0)B(t)=A(t). $$

These are the characteristic functions of uniform random variables on \([0,1]\), as can also be checked by integrating \(f\) over the other coordinate. For example,

$$ \int_0^1 f(x,y)\,dy=1+\frac12(2x-1)\int_0^1(2y-1)\,dy=1 \quad (0\leq x\leq1). $$

The joint density is not merely the product of its marginal densities: its additional term varies with both coordinates. The marginal characteristic functions alone do not display that term; the joint function does.

Scalar Projections of a Random Vector

A coordinate is one particular linear combination of the coordinates of a vector. More generally, for a fixed vector \(a=(a_1,\ldots,a_d)^\mathsf{T}\), the scalar projection \(a^\mathsf{T}X=\sum_{j=1}^d a_jX_j\) has characteristic function that can be read directly from the joint function.

Theorem: For every fixed \(a\in\mathbb{R}^d\), the scalar random variable \(a^\mathsf{T}X\) has characteristic function
$$ \varphi_{a^\mathsf{T}X}(u)=\varphi_X(ua),\qquad u\in\mathbb{R}. $$

Proof. Apply the definition of the scalar characteristic function and then the definition of the joint characteristic function:

$$ \begin{aligned} \varphi_{a^\mathsf{T}X}(u) &=E[e^{iu(a^\mathsf{T}X)}]\\ &=E[e^{i(ua)^\mathsf{T}X}]\\ &=\varphi_X(ua). \end{aligned} $$

The expectations exist because each exponential has modulus one. This proves the identity for every \(u\in\mathbb{R}\). \(\square\)

Worked Example: A Vector Concentrated on a Line

Let \(U\) be uniform on \([0,1]\) and define \(X=U\), \(Y=2U\). The vector \((X,Y)\) is concentrated on the line segment \(y=2x\) in the plane, so it has no two-dimensional density. Its joint characteristic function is still defined by expectation:

$$ \begin{aligned} \varphi_{(X,Y)}(s,t) &=E[e^{i(sX+tY)}] =E[e^{i(s+2t)U}]\\ &=\int_0^1e^{i(s+2t)u}\,du = \begin{cases} \dfrac{e^{i(s+2t)}-1}{i(s+2t)},&s+2t\ne0,\\[6pt] 1,&s+2t=0. \end{cases} \end{aligned} $$

The second case follows because the integrand is identically one when \(s+2t=0\). Setting \(t=0\) gives the characteristic function of \(X=U\); setting \(s=0\) gives that of \(Y=2U\). The projection theorem also gives the characteristic function of \(X+Y\): take \(a=(1,1)^\mathsf{T}\), so

$$ \varphi_{X+Y}(u)=\varphi_{(X,Y)}(u,u) =\int_0^1e^{i3uu'}\,du' = \begin{cases} \dfrac{e^{i3u}-1}{i3u},&u\ne0,\\ 1,&u=0. \end{cases} $$

Here \(u'\) is the integration variable. This example shows why the expectation definition is more general than the density integral: a vector can have a perfectly well-defined joint characteristic function even when no joint density exists.

Common Mistake and Practical Use

A frequent error is to assume that a joint characteristic function must equal the product of the marginal characteristic functions. That reasoning treats the coordinates as if they were independent without checking the distribution. The definition instead gives

$$ \varphi_{(X,Y)}(s,t)=E[e^{i(sX+tY)}], $$

whereas the product of marginal characteristic functions is \(E[e^{isX}]E[e^{itY}]\). These are different expressions in general. For instance, in the line-supported example, \(Y=2X\), so the two coordinates are tied together; one cannot replace the joint expectation by a product merely because each marginal characteristic function is known. The precise factorization criterion for independence is a separate result.

Another common error is to set every argument to zero when trying to recover a marginal. That gives \(\varphi_X(0,\ldots,0)=1\), not a whole marginal function. To recover coordinate \(j\), retain its argument \(u\) and set only the other coordinates to zero.

Joint characteristic functions are useful because a single multivariate function supports several kinds of calculations. Coordinate-axis restrictions produce marginal characteristic functions, and line restrictions produce characteristic functions of linear combinations. These operations do not require a joint density or finite moments. In applications, the vector form also keeps the dependence structure visible rather than reducing the analysis to separate coordinate summaries.

Key takeaway: The joint characteristic function is \(\varphi_X(t)=E[e^{it^\mathsf{T}X}]\). Restricting it to a coordinate axis recovers that coordinate’s marginal characteristic function; restricting it to the line \(ua\) gives the characteristic function of \(a^\mathsf{T}X\).

Check Your Understanding

Use the definitions and results above to answer the following.

  1. For a random vector \((X_1,X_2,X_3)\), write the joint characteristic function and specify which arguments must be set to zero to obtain \(\varphi_{X_2}(u)\).
  2. A bivariate vector has mass \(0.25\) at \((0,0)\), \(0.25\) at \((1,0)\), and \(0.50\) at \((1,1)\). Write its joint characteristic function and obtain both marginal characteristic functions by restriction.
  3. Explain why the expectation defining a joint characteristic function exists even if the vector has no finite first moment.
  4. If \(W=2X_1-X_2\), state the joint characteristic-function argument at which \(\varphi_W(u)\) is evaluated.
  5. Why is setting both arguments to zero not enough to recover either marginal characteristic function?

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