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Characteristic Functions · Tutorial 276 of 946

Characteristic Functions as Fourier Transforms

Learn how a probability law becomes a characteristic function through Fourier transformation, and how to interpret and calculate that transform.

Intermediate 10 min read

What You'll Learn

  • Define a real-valued random variable’s characteristic function through its probability law.
  • Identify the sign convention and the full real frequency domain.
  • Relate the characteristic function to expectations of cosine and sine.
  • Calculate transforms for point masses and finite discrete laws.
  • Compute a characteristic function directly from a probability density.

From a Probability Law to a Function of Frequency

A probability law describes how probability is distributed across the possible values of a random variable. A characteristic function recodes that law as a function of a real frequency parameter. This representation is useful because it turns a probability measure into an object that can be studied using Fourier analysis.

The central operation is to average the complex oscillation \(e^{itx}\) over the law of the random variable. Here \(x\) is a possible value of the random variable, and \(t\) is the frequency at which we examine the law. The sign in the exponent is a convention, so it is important to state it explicitly.

Definition and Sign Convention

Let \(X\) be a real-valued random variable on a probability space \((\Omega,\mathcal F,\mathbb P)\). Its law, denoted by \(\mu_X\), is the probability measure on the Borel subsets of \(\mathbb R\) defined by \(\mu_X(B)=\mathbb P(X\in B)\). For a Borel set \(B\), this assigns to \(B\) the probability that \(X\) takes a value in \(B\).

Definition: The characteristic function of \(X\) is the function \(\varphi_X:\mathbb R\to\mathbb C\) given, for every real \(t\), by
$$ \varphi_X(t)=\mathbb E[e^{itX}] =\int_{\mathbb R} e^{itx}\,\mu_X(dx). $$
The Fourier transform convention used here has a positive sign in the exponent.

The expectation is well-defined: for every real \(x\) and \(t\), the complex number \(e^{itx}\) has modulus \(1\). In particular, its real and imaginary parts are bounded measurable functions, so their expectations exist. The argument \(t\) ranges over the entire real line; the characteristic function is not restricted to positive frequencies.

Some Fourier analysis conventions define the transform using \(e^{-itx}\) instead. Either sign convention can be used, but a calculation must follow the convention chosen at the outset. With the convention above, a point mass at a positive value \(a\) contributes the phase \(e^{ita}\). With the negative-sign convention, that phase would instead be \(e^{-ita}\).

Expectation as Integration Against the Law

The two expressions in the definition describe the same operation: one averages over outcomes in the probability space, and the other integrates over the values of \(X\) using its law. The following standard pushforward-integration identity explains why this change of viewpoint is valid.

Theorem: If \(g:\mathbb R\to\mathbb C\) is bounded and Borel measurable, then
$$ \mathbb E[g(X)]=\int_{\mathbb R}g(x)\,\mu_X(dx). $$

Proof. First take \(g\) to be the indicator function of a Borel set \(B\), so \(g(x)=\mathbf 1_B(x)\). Then \(\mathbf 1_B(X)=\mathbf 1_{\{X\in B\}}\), and therefore

$$ \mathbb E[\mathbf 1_B(X)] =\mathbb P(X\in B) =\mu_X(B) =\int_{\mathbb R}\mathbf 1_B(x)\,\mu_X(dx). $$

By linearity, the identity follows for every Borel simple function. For a bounded Borel complex-valued \(g\), choose Borel simple functions \(g_n\) that converge pointwise to \(g\) and are uniformly bounded. The dominated convergence theorem applies to \(g_n(X)\) under \(\mathbb P\), and also to \(g_n(x)\) under the probability measure \(\mu_X\). Taking limits in the identity for \(g_n\) proves the identity for \(g\). \(\square\)

Taking \(g(x)=e^{itx}\) gives the law-integral formula for the characteristic function. This also shows directly that the characteristic function depends on the distribution of \(X\), not on the particular probability space on which \(X\) is defined.

Real and Imaginary Parts

Euler’s formula separates the oscillation into its cosine and sine components. Applying it inside the expectation gives an equivalent description of the characteristic function.

Proposition: For every real \(t\),
$$ \varphi_X(t) =\mathbb E[\cos(tX)] +i\,\mathbb E[\sin(tX)]. $$
Thus the real part records the average cosine and the imaginary part records the average sine at frequency \(t\).

Proof. For each outcome \(\omega\), Euler’s formula gives

$$ e^{itX(\omega)} =\cos(tX(\omega))+i\sin(tX(\omega)). $$

Both trigonometric terms are bounded, so their expectations exist. Taking expectations and using linearity for the real and imaginary parts yields the stated identity. \(\square\)

This decomposition gives a concrete interpretation: at a chosen frequency, each value of \(X\) contributes a unit-length complex number pointing in a direction determined by \(tX\). The characteristic function is the average of those contributions. Contributions with different phases can reinforce one another or partially cancel.

Worked Examples

Worked Example: A Constant Random Variable

Suppose \(X=c\) with probability \(1\), where \(c\in\mathbb R\). Its law places all its probability at \(c\), so \(\mu_X\) is the point mass \(\delta_c\). Integrating against this law evaluates the integrand at \(c\):

$$ \varphi_X(t) =\int_{\mathbb R}e^{itx}\,\delta_c(dx) =e^{itc}, \qquad t\in\mathbb R. $$

The same result follows directly from the expectation because \(e^{itX}=e^{itc}\) with probability \(1\). The sign convention is visible in the phase: for \(c>0\), the exponent is \(itc\), not \(-itc\). At frequency \(t=0\), this example gives \(e^0=1\).

Worked Example: A Three-Point Law

Suppose \(X\) takes the values \(-1\), \(0\), and \(2\) with probabilities \(1/4\), \(1/2\), and \(1/4\), respectively. These probabilities are nonnegative and sum to \(1\), so they specify a probability law. Integrating the exponential separately at each atom gives

$$ \begin{aligned} \varphi_X(t) &=\frac14 e^{-it}+\frac12 e^{i0t}+\frac14 e^{2it}\\ &=\frac14 e^{-it}+\frac12+\frac14 e^{2it}. \end{aligned} $$

Using Euler’s formula, the real and imaginary parts can also be displayed explicitly:

$$ \varphi_X(t) =\frac12+\frac14\cos(t)+\frac14\cos(2t) +i\left(-\frac14\sin(t)+\frac14\sin(2t)\right). $$

Each term corresponds to one possible value: the atom at \(-1\) contributes \(e^{-it}\), the atom at \(0\) contributes \(1\), and the atom at \(2\) contributes \(e^{2it}\). Substituting \(t=0\) gives \(1/4+1/2+1/4=1\), as it should for this probability law.

Worked Example: A Density on the Unit Interval

Suppose \(X\) has density \(f(x)=2x\) for \(0\le x\le 1\), and \(f(x)=0\) otherwise. This is a probability density because it is nonnegative and

$$ \int_0^1 2x\,dx=1. $$

The characteristic function is the Fourier integral of the density:

$$ \varphi_X(t)=\int_0^1 2x e^{itx}\,dx. $$

At \(t=0\), the integral is \(1\). If \(t\ne 0\), an antiderivative of \(2xe^{itx}\) is \(e^{itx}\left(\frac{2x}{it}+\frac{2}{t^2}\right)\). Indeed, differentiating this expression gives \(2xe^{itx}\), since its additional terms \(2i e^{itx}/t\) and \(2e^{itx}/(it)\) cancel. Evaluating at the endpoints yields

$$ \varphi_X(t) =e^{it}\left(\frac{2}{it}+\frac{2}{t^2}\right)-\frac{2}{t^2} =\frac{2e^{it}}{it}+\frac{2(e^{it}-1)}{t^2}, \qquad t\ne 0. $$

The value at zero is stated separately because the displayed expression has denominators involving \(t\). Directly from the defining integral, \(\varphi_X(0)=\int_0^1 2x\,dx=1\).

Why the Fourier Viewpoint Matters

The definition gives a systematic way to turn a probability law into a function of frequency. For a discrete law, the transform is a weighted sum of phases; for a law with a density, it is an integral of those phases against the density. The same definition covers both cases, as well as laws that have neither a finite support nor a density.

The Fourier viewpoint makes the law’s contributions visible at different frequencies. A low frequency changes the phase slowly as \(x\) varies, while a higher frequency rotates the phase more rapidly. The resulting average reflects how the probability mass is distributed relative to that oscillation. This interpretation is useful in later results about characteristic functions and limit theorems.

A common pitfall is to lose track of the sign in the exponent or to integrate against the wrong measure. The variable \(X\) belongs inside the expectation, while \(x\) is the integration variable when using \(\mu_X\). Once the sign convention and the law are fixed, both expressions must give the same value.

Key takeaway: Under the positive-exponent convention, the characteristic function of a real-valued random variable is the Fourier transform \(t\mapsto\int e^{itx}\,\mu_X(dx)\) of its probability law, defined for every real frequency \(t\).

Check Your Understanding

Use the definition and the examples above to check that you can move between a probability law and its Fourier representation.

  1. What is the probability law \(\mu_X\) of a real-valued random variable \(X\)?
  2. Under the sign convention used here, what phase does a point mass at \(a\) contribute at frequency \(t\)?
  3. Why is the expectation in the definition of \(\varphi_X(t)\) well-defined for every real \(t\)?
  4. Write \(\varphi_X(t)\) in terms of the expectations of \(\cos(tX)\) and \(\sin(tX)\).
  5. For a random variable with finitely many possible values, how do the probabilities and values enter its characteristic function?
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