Real Characteristic Functions and Symmetric Laws

Why Reality of a Characteristic Function Is Informative

A characteristic function records a distribution through the oscillations of \(e^{itX}\). Its real part averages cosines, while its imaginary part averages sines. Reflection about zero leaves the cosine unchanged and reverses the sine. This gives a natural question: if those sine averages vanish for every frequency \(t\), must the distribution itself be unchanged by reflection? The answer is yes.

The condition is global: the characteristic function must be real at every real argument. A real value at one frequency, or a zero mean, does not establish symmetry. The distinction matters because symmetry is a property of the entire distribution, not merely of a single moment.

Definitions and the Criterion

Definition: A random variable \(X\) has a law symmetric about zero if \(X\) and \(-X\) have the same distribution. Equivalently, if \(\mu\) is the probability law of \(X\), then \(\mu(A)=\mu(-A)\) for every Borel set \(A\subseteq\mathbb{R}\), where \(-A=\{-x:x\in A\}\). The characteristic function of \(X\) is \(\varphi_X(t)=E[e^{itX}]\), for \(t\in\mathbb{R}\).

Since \(e^{itX}=\cos(tX)+i\sin(tX)\), its expected value has the decomposition

\varphi_X(t)=E[\cos(tX)]+iE[\sin(tX)].

Consequently, \(\varphi_X\) is real-valued exactly when \(E[\sin(tX)]=0\) for every \(t\). That observation explains why symmetry implies reality. The reverse direction requires showing that vanishing sine averages at all frequencies determines the reflected law as well.

Theorem: A real-valued random variable \(X\) has a real-valued characteristic function, meaning \(\varphi_X(t)\in\mathbb{R}\) for every \(t\in\mathbb{R}\), if and only if \(X\) is symmetric about zero.

Worked Examples

Worked Example: A Continuous Symmetric Distribution

Let \(X\) have the uniform distribution on \([-a,a]\), where \(a>0\). Its density is \(f(x)=1/(2a)\) for \(-a\leq x\leq a\), and zero otherwise. The density satisfies \(f(x)=f(-x)\), so the law is symmetric about zero. We can check the characteristic function directly.

For \(t\ne 0\), integrate the density:

\varphi_X(t)=\frac{1}{2a}\int_{-a}^{a}e^{itx}\,dx =\frac{1}{2a}\left[\frac{e^{itx}}{it}\right]_{-a}^{a} =\frac{e^{ita}-e^{-ita}}{2ait} =\frac{\sin(at)}{at}.

This is real for every nonzero \(t\). At \(t=0\), the defining expectation gives \(\varphi_X(0)=1\); the displayed expression has the same limiting value because \(\lim_{u\to 0}\sin(u)/u=1\). Thus the characteristic function is real on all of \(\mathbb{R}\), as the theorem predicts.

Worked Example: A Continuous Nonsymmetric Distribution

Let \(X\) have the exponential density \(f(x)=e^{-x}\) for \(x\geq 0\), and zero for \(x<0\). Its support is not symmetric about zero, so its law is not symmetric. Direct integration confirms that its characteristic function is not real-valued:

\varphi_X(t)=\int_0^\infty e^{itx}e^{-x}\,dx =\int_0^\infty e^{-(1-it)x}\,dx =\left[-\frac{e^{-(1-it)x}}{1-it}\right]_0^\infty =\frac{1}{1-it} =\frac{1+it}{1+t^2}.

The integral converges because the integrand has absolute value \(e^{-x}\). For \(t=1\), \(\varphi_X(1)=(1+i)/2\), which is not real. More generally its imaginary part is \(t/(1+t^2)\), nonzero whenever \(t\ne 0\).

Worked Example: A Discrete Symmetric Distribution

Suppose \(X\) takes the values \(-2,0,2\) with probabilities \(0.2,0.6,0.2\), respectively. The probabilities at opposite values agree, and the mass at zero reflects to itself. For every Borel set, reflection therefore preserves its probability, so the law is symmetric about zero. Its characteristic function is

\varphi_X(t)=0.2e^{-2it}+0.6+0.2e^{2it} =0.6+0.2(e^{-2it}+e^{2it}) =0.6+0.4\cos(2t).

This expression is real for every \(t\), in agreement with the criterion. Pairing the masses at \(-2\) and \(2\) makes the cancellation of the imaginary parts explicit.

Proof of the Theorem

We prove both implications. The forward implication from symmetry is direct. For the reverse implication, we need a way to turn equality of Fourier transforms into equality of the two probability laws being compared. The following argument supplies exactly that step.

Fourier separation fact: If \(\nu\) is a finite signed Borel measure on \(\mathbb{R}\) and \(\int e^{ity}\,d\nu(y)=0\) for every \(t\in\mathbb{R}\), then \(\nu\) is the zero measure.

To verify the fact, take \(\sigma>0\) and let

k_\sigma(x)=\frac{1}{\sqrt{2\pi}\sigma} \exp\!\left(-\frac{x^2}{2\sigma^2}\right).

The Gaussian Fourier integral gives

k_\sigma(x-y)=\frac{1}{2\pi}\int_{\mathbb{R}} e^{-it(x-y)}e^{-\sigma^2t^2/2}\,dt.

This identity follows from the Gaussian integral: the integral of \(e^{-\sigma^2t^2/2}e^{-itv}\) over \(t\) equals \(\sqrt{2\pi}\sigma^{-1}e^{-v^2/(2\sigma^2)}\). For example, differentiating that integral with respect to \(v\) is justified by the integrable bound \(|t|e^{-\sigma^2t^2/2}\); integration by parts gives the differential equation \(I'(v)=-(v/\sigma^2)I(v)\), and \(I(0)=\sqrt{2\pi}/\sigma\) fixes the solution.

Define \(g_\sigma(x)=\int k_\sigma(x-y)\,d\nu(y)\). Substituting the integral representation and interchanging integrals is valid: the absolute integral is bounded by \(\|\nu\|\int e^{-\sigma^2t^2/2}dt\), where \(\|\nu\|\) is the finite total variation of \(\nu\). Thus

g_\sigma(x)=\frac{1}{2\pi}\int_{\mathbb{R}} e^{-itx}e^{-\sigma^2t^2/2} \left(\int_{\mathbb{R}}e^{ity}\,d\nu(y)\right)dt=0.

Now let \(h\) be any continuous function with compact support. Fubini's theorem applies because \(h\) is bounded and integrable and \(\nu\) is finite. It yields

0=\int_{\mathbb{R}}h(x)g_\sigma(x)\,dx =\int_{\mathbb{R}}\left(\int_{\mathbb{R}}h(x)k_\sigma(x-y)\,dx\right)d\nu(y).

The inner integral tends uniformly in \(y\) to \(h(y)\) as \(\sigma\downarrow 0\). Indeed, \(h\), extended by zero outside its support, is bounded and uniformly continuous. For any \(\delta>0\), the contribution to the difference from \(|x-y|<\delta\) is bounded by the modulus of continuity of \(h\) at \(\delta\); the remaining contribution is at most \(2\|h\|_\infty P(|Z_\sigma|\geq\delta)\), where \(Z_\sigma\) has density \(k_\sigma\). This tail probability tends to zero because \(Z_\sigma\) has the same distribution as \(\sigma Z_1\). Therefore \(\int h\,d\nu=0\) for every such \(h\). Finite Borel measures on \(\mathbb{R}\) are determined by these integrals (by regularity, continuous compactly supported functions approximate indicators of compact sets). Hence \(\nu=0\), proving the separation fact. \(\square\)

Let \(\mu\) be the law of \(X\), and let \(\mu^{-}\) be the law of \(-X\), so that \(\mu^{-}(A)=\mu(-A)\). If \(X\) is symmetric about zero, \(\mu=\mu^{-}\). For any \(t\), symmetry gives

E[\sin(tX)]=E[\sin(-tX)]=-E[\sin(tX)].

Thus \(E[\sin(tX)]=0\), and \(\varphi_X(t)=E[\cos(tX)]\) is real. This proves one direction.

Conversely, suppose \(\varphi_X(t)\) is real for every \(t\). The characteristic function of \(-X\) at \(t\) is \(\varphi_X(-t)\), and conjugate symmetry gives \(\varphi_X(-t)=\overline{\varphi_X(t)}\). Since \(\varphi_X(t)\) is real, these two values are equal. The finite signed measure \(\nu=\mu-\mu^{-}\) therefore satisfies

\int e^{itx}\,d\nu(x)=\varphi_X(t)-\varphi_X(-t)=0 \qquad\text{for every }t\in\mathbb{R}.

By the Fourier separation fact, \(\nu=0\), so \(\mu=\mu^{-}\). This is precisely symmetry of \(X\) about zero, proving the other direction and the theorem. \(\square\)

Reading the Criterion Correctly

For a continuous distribution with density \(f\), symmetry about zero is equivalent to \(f(x)=f(-x)\) for almost every \(x\). For a discrete distribution with point probabilities \(p(x)=P(X=x)\), it is equivalent to \(p(x)=p(-x)\) for every \(x\). In either case, the theorem says that this distributional balance is also equivalent to a real characteristic function at every frequency.

A common mistake is to treat zero mean as a substitute for symmetry. It is not. For example, let \(P(X=-1)=2/3\) and \(P(X=2)=1/3\). Then

E[X]=(-1)\frac{2}{3}+2\frac{1}{3}=0, \qquad \varphi_X(t)=\frac{2}{3}e^{-it}+\frac{1}{3}e^{2it}.

At \(t=\pi/2\), the imaginary part of the characteristic function is

\operatorname{Im}\varphi_X(\pi/2) =-\frac{2}{3}\sin(\pi/2)+\frac{1}{3}\sin(\pi) =-\frac{2}{3}\ne 0.

The law is not symmetric: it assigns positive probability to \(-1\) but none to \(1\). Its mean vanishes because the two weighted values balance, not because the entire distribution balances under reflection. The real-valued characteristic-function criterion tests the latter, stronger property.

Key takeaway: A characteristic function is real at every real argument exactly when reflection about zero leaves the law unchanged. The condition concerns all frequencies, and it is stronger than having mean zero.

Check Your Understanding

Use the symmetry criterion and the calculations in this tutorial to answer the following questions.

For a random variable symmetric about zero, explain why \(E[\sin(tX)]=0\) for each fixed real \(t\).
A discrete law assigns probabilities \(0.1,0.8,0.1\) to \(-3,0,3\). Write its characteristic function and verify that it is real.
Why does a real value of \(\varphi_X(t)\) at one nonzero \(t\) not establish symmetry?
A density \(f\) satisfies \(f(x)=f(-x)\) almost everywhere. What does the theorem imply about its characteristic function?
Give one reason the example with masses at \(-1\) and \(2\) has mean zero but is not symmetric about zero.