Symmetric Laws and Real Characteristic Functions

Symmetry and characteristic functions

A probability law can be symmetric even when it has no density: a pair of equally weighted atoms at opposite points is one simple example. Characteristic functions provide a single criterion that applies to both discrete and continuous laws: symmetry about zero is equivalent to the characteristic function being real at every real argument.

The forward implication reflects the cancellation of positive and negative values. The converse is more subtle. It compares the original law with its reflection and then uses the fact that a probability law is determined by its characteristic function.

Definitions and the reflection of a law

Definition: A probability measure $\mu$ on the Borel subsets of $\mathbb{R}$ is symmetric about zero if, for every Borel set $A\subseteq\mathbb{R}$, its probability agrees with that of the reflected set $-A=\{-x:x\in A\}$. In symbols, $$ \mu(A)=\mu(-A)\qquad\text{for every Borel set }A. $$ The characteristic function of $\mu$ is the function $\varphi_\mu:\mathbb{R}\to\mathbb{C}$ defined by $$ \varphi_\mu(t)=\int_{\mathbb{R}} e^{itx}\,\mu(dx),\qquad t\in\mathbb{R}. $$

The integral exists for every real $t$, since $|e^{itx}|=1$. If $X$ is a real-valued random variable with law $\mu$, this definition is also written $\varphi_X(t)=\mathbb{E}[e^{itX}]$. Its real and imaginary parts are $\mathbb{E}[\cos(tX)]$ and $\mathbb{E}[\sin(tX)]$, respectively.

Define the reflected law $\mu^{\mathrm r}$ by $\mu^{\mathrm r}(A)=\mu(-A)$ for every Borel set $A$. Equivalently, if $X$ has law $\mu$, then $-X$ has law $\mu^{\mathrm r}$. A change of variable under this reflection gives $$ \varphi_{\mu^{\mathrm r}}(t)=\int_{\mathbb{R}}e^{-itx}\,\mu(dx)=\varphi_\mu(-t). $$ Thus symmetry says that $\mu$ and $\mu^{\mathrm r}$ are the same law, while the corresponding characteristic functions are $\varphi_\mu(t)$ and $\varphi_\mu(-t)$.

The forward implication

For any probability measure, conjugate symmetry gives $\varphi_\mu(-t)=\overline{\varphi_\mu(t)}$. This also follows directly by conjugating the defining integral. If $\mu$ is symmetric, its reflection is itself, so the reflection identity also gives $\varphi_\mu(-t)=\varphi_\mu(t)$. Together these identities show that $\varphi_\mu(t)=\overline{\varphi_\mu(t)}$, which means its value is real.

Proposition: If a probability measure on $\mathbb{R}$ is symmetric about zero, then its characteristic function is real-valued on $\mathbb{R}$.

This implication can also be seen by considering the sine term in the characteristic function. For a symmetric law, the expectation of the odd function $x\mapsto\sin(tx)$ is zero. Hence $\varphi_\mu(t)=\int\cos(tx)\,\mu(dx)$, a real number.

Why the converse needs uniqueness

A real-valued characteristic function satisfies $\varphi_\mu(t)=\varphi_\mu(-t)$, because conjugate symmetry holds for every characteristic function. The reflection identity then says that $\mu$ and $\mu^{\mathrm r}$ have the same characteristic function. To conclude that the laws agree, we need uniqueness: two probability measures with identical characteristic functions must be equal.

For completeness, the uniqueness fact needed here is proved below. The argument smooths each measure by adding a small centered Gaussian. The resulting laws have densities that can be recovered directly from the characteristic function. Letting the Gaussian variance decrease to zero then recovers the original law.

Lemma (uniqueness of characteristic functions): If $\mu$ and $\nu$ are probability measures on $\mathbb{R}$ and $\varphi_\mu(t)=\varphi_\nu(t)$ for every $t\in\mathbb{R}$, then $\mu=\nu$.

Proof. For $\sigma>0$, let $\gamma_\sigma$ be the centered Gaussian probability measure with variance $\sigma^2$, with density $g_\sigma(z)=(\sqrt{2\pi}\sigma)^{-1}\exp(-z^2/(2\sigma^2))$. The Gaussian integral gives the Fourier representation $$ g_\sigma(z)=\frac{1}{2\pi}\int_{\mathbb{R}}e^{-itz}e^{-\sigma^2t^2/2}\,dt. $$ The convolution $\mu*\gamma_\sigma$, the law of the sum of independent variables with laws $\mu$ and $\gamma_\sigma$, has density $$ p_{\mu,\sigma}(x)=\int_{\mathbb{R}}g_\sigma(x-y)\,\mu(dy) =\frac{1}{2\pi}\int_{\mathbb{R}}e^{-itx}e^{-\sigma^2t^2/2}\varphi_\mu(t)\,dt. $$ The exchange of integrals is justified because $e^{-\sigma^2t^2/2}$ is integrable and $\mu$ is a probability measure. The same formula holds for $\nu$. The assumed equality of characteristic functions therefore gives $p_{\mu,\sigma}(x)=p_{\nu,\sigma}(x)$ for every $x$, so $\mu*\gamma_\sigma=\nu*\gamma_\sigma$.

Now let $f:\mathbb{R}\to\mathbb{R}$ be bounded and continuous. If $X$ has law $\mu$ and $Z$ is an independent standard Gaussian, then $X+\sigma Z$ has law $\mu*\gamma_\sigma$. As $\sigma$ decreases to zero, $f(X+\sigma Z)$ converges pointwise to $f(X)$. Bounded convergence gives $$ \int f\,d\mu =\lim_{\sigma\downarrow0}\int f\,d(\mu*\gamma_\sigma) =\lim_{\sigma\downarrow0}\int f\,d(\nu*\gamma_\sigma) =\int f\,d\nu. $$ In particular, the two measures assign the same probability to $(-\infty,a]$ for every $a\in\mathbb{R}$: apply this equality to the bounded continuous functions that equal $1$ on $(-\infty,a]$, decrease linearly to $0$ on $[a,a+1/n]$, and equal $0$ thereafter, then let $n\to\infty$. Equality of these distribution functions implies $\mu=\nu$. $\square$

Theorem: A probability measure $\mu$ on $\mathbb{R}$ is symmetric about zero if and only if its characteristic function $\varphi_\mu$ is real-valued on $\mathbb{R}$.

Proof. The forward implication is the proposition above. Conversely, suppose $\varphi_\mu(t)$ is real for every $t$. Conjugate symmetry gives $\varphi_\mu(-t)=\overline{\varphi_\mu(t)}$, and reality makes the right-hand side equal to $\varphi_\mu(t)$. Since $\varphi_{\mu^{\mathrm r}}(t)=\varphi_\mu(-t)$, the characteristic functions of $\mu^{\mathrm r}$ and $\mu$ agree for every $t$. The uniqueness lemma implies $\mu^{\mathrm r}=\mu$, which means $\mu(A)=\mu(-A)$ for every Borel set $A$. Thus $\mu$ is symmetric about zero. \(\square

Worked examples

Worked Example: Equal masses at opposite points

Let $X$ take the values $-a$ and $a$, each with probability $1/2$, where $a>0$. Its characteristic function is $$ \varphi_X(t)=\frac12 e^{-iat}+\frac12 e^{iat}=\cos(at). $$ This is real for every real $t$, so the criterion proves that the law is symmetric about zero. Directly, the mass at $a$ equals the mass at $-a$; sets not containing either point have probability zero, and reflecting a set interchanges these two possible contributions.

Worked Example: An unbalanced two-point law

Suppose $X$ equals $1$ with probability $p$ and $-1$ with probability $1-p$, where $0\leq p\leq1$. Then $$ \varphi_X(t)=p e^{it}+(1-p)e^{-it} =\cos(t)+i(2p-1)\sin(t). $$ If $p=1/2$, this reduces to $\cos(t)$, and the law is symmetric. If $p\ne1/2$, take $t=\pi/2$. The imaginary part is $2p-1\ne0$, so the characteristic function is not real-valued. The law is not symmetric: the probabilities of $\{1\}$ and $\{-1\}$ differ.

Worked Example: The exponential distribution

Let $X$ have the exponential distribution with rate $1$, so it has density $e^{-x}$ for $x\geq0$ and zero for $x<0$. For real $t$, $$ \varphi_X(t)=\int_0^\infty e^{itx}e^{-x}\,dx =\int_0^\infty e^{-(1-it)x}\,dx =\frac{1}{1-it} =\frac{1+it}{1+t^2}. $$ For every $t\ne0$, the imaginary part $t/(1+t^2)$ is nonzero. Therefore the characteristic function is not real-valued, and the law is not symmetric about zero. This also follows from its support: it assigns probability one to $[0,\infty)$, unlike its reflection, which assigns probability one to $(-\infty,0]$.

Reading the criterion correctly

The theorem requires the characteristic function to be real at every real argument. Finding even one argument where its imaginary part is nonzero rules out symmetry, as in the unbalanced two-point and exponential examples. Checking only a few values that happen to give real outputs is not enough to establish symmetry.

Also, “real-valued” does not mean “nonnegative.” For example, if $X$ is equally likely to be $-1$ or $1$, then $\varphi_X(t)=\cos(t)$, which is negative at $t=\pi$. The theorem says that a symmetric law has a real characteristic function, not that its characteristic function must be nonnegative.

The equivalence is useful because symmetry can be difficult to see from a formula for a law, especially when the law has no density or has a complicated mixture of components. A characteristic function provides an exact test: reality on the whole real line identifies symmetry, and failure of reality at even one point disproves it.

Check Your Understanding

Use the reflection identity and the theorem to answer the following questions.

What does symmetry about zero mean in terms of $\mu(A)$ and $\mu(-A)$?
Why does a symmetric law have a real-valued characteristic function?
If $\varphi_\mu(t)$ is real for every $t$, why do $\mu$ and its reflected law have the same characteristic function?
For the law assigning probabilities $p$ and $1-p$ to $1$ and $-1$, respectively, which value of $t$ shows that the characteristic function is not real when $p\ne1/2$?
Does a symmetric law necessarily have a nonnegative characteristic function? Give an example supporting your answer.