Recovering Atoms from Characteristic Functions

Why Average the Characteristic Function?

A characteristic function contains information about the entire distribution, but extracting one particular probability, such as $P(X=x)$, is not usually done by evaluating the characteristic function at a single frequency. Instead, we average it over a growing interval of frequencies. The oscillations contributed by values other than $x$ cancel in the limit, while the probability concentrated exactly at $x$ remains.

This method applies whether the distribution is discrete, continuous, or a mixture of the two. It does not require the characteristic function to be integrable, and it does not require $X$ to have a density.

Definition: A point $x\in\mathbb{R}$ is an atom of the distribution of $X$ if $P(X=x)>0$. The probability $P(X=x)$ is the mass of that atom. A distribution may have atoms and also assign positive probability to sets of values that are not atoms.

Throughout, the characteristic function of a real-valued random variable $X$ is $\phi_X(t)=E[e^{itX}]$, for $t\in\mathbb{R}$. For a fixed candidate point $x$, we center the characteristic function by multiplying by $e^{-itx}$. This gives $E[e^{it(X-x)}]$, so the centered frequency average responds to whether $X-x$ is exactly zero.

The Symmetric-Average Formula

For $T>0$, define the symmetric average at $x$ by

A_T(x)=\frac{1}{2T}\int_{-T}^{T} e^{-itx}\phi_X(t)\,dt.

The integral is well defined for each finite $T$: a characteristic function is bounded in absolute value by $1$, and the interval of integration is finite.

Theorem: For every real-valued random variable $X$ and every $x\in\mathbb{R}$, $$ P(X=x)=\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T} e^{-itx}\phi_X(t)\,dt. $$ No density, moment, or integrability assumption on $\phi_X$ is required.

Proof. Define $\operatorname{sinc}(u)=\sin(u)/u$ when $u\ne 0$, and $\operatorname{sinc}(0)=1$. Since $\phi_X(t)=E[e^{itX}]$, boundedness of the integrand and Fubini's theorem give

\begin{aligned} A_T(x) &=E\left[\frac{1}{2T}\int_{-T}^{T}e^{it(X-x)}\,dt\right]\\ &=E\left[\operatorname{sinc}\bigl(T(X-x)\bigr)\right]. \end{aligned}

Indeed, for a fixed real $y\ne0$, integrating $e^{ity}$ from $-T$ to $T$ gives $2\sin(Ty)/y$; when $y=0$, the integral is $2T$. Dividing by $2T$ yields the displayed sinc expression in both cases.

For each outcome, if $X=x$, then $\operatorname{sinc}(T(X-x))=1$ for every $T$. If $X\ne x$, then $T(X-x)$ grows in absolute value and the sinc term tends to zero. Also, $|\operatorname{sinc}(u)|\le1$ for every $u$. The dominated convergence theorem therefore applies, with the constant $1$ as an integrable bound on the probability space:

\lim_{T\to\infty}A_T(x) =E[\mathbf{1}_{\{X=x\}}] =P(X=x).

This proves the formula. $\square$

The symmetric average can also be written using only real quantities. The defining property of a characteristic function gives $\phi_X(-t)=\overline{\phi_X(t)}$. Thus the integrand at $-t$ is the complex conjugate of the integrand at $t$, and

A_T(x)=\frac{1}{T}\int_0^T \operatorname{Re}\!\left(e^{-itx}\phi_X(t)\right)\,dt.

This is an equivalent formula, not an additional assumption: its limit is again $P(X=x)$. In particular, the limiting value is real and nonnegative even though the integrand in the original formula can be complex.

Worked Applications

Worked Example: Finding the Atom in a Mixed Normal Distribution

Let $X$ equal $2$ with probability $1/3$, and otherwise have a standard normal distribution. More precisely, its distribution is the mixture of a point mass at $2$, of weight $1/3$, and $N(0,1)$, of weight $2/3$. Find its atoms using its characteristic function.

The standard normal characteristic function is $e^{-t^2/2}$, and the characteristic function of a point mass at $2$ is $e^{2it}$. By taking the weighted average of these two characteristic functions,

\phi_X(t)=\frac{1}{3}e^{2it}+\frac{2}{3}e^{-t^2/2}.

At the candidate point $x=2$, the centered symmetric average is

A_T(2)=\frac{1}{3} +\frac{2}{3}\frac{1}{2T}\int_{-T}^{T}e^{-2it}e^{-t^2/2}\,dt.

The absolute value of the second term is at most $\frac{2}{3}(2T)^{-1}\int_{\mathbb{R}}e^{-t^2/2}\,dt$. This upper bound tends to zero because the Gaussian integral is finite and the denominator grows without bound. Consequently, $P(X=2)=1/3$.

For any $x\ne2$, the point-mass contribution to $A_T(x)$ is $\frac{1}{3}\operatorname{sinc}(T(2-x))$, which tends to zero. The normal contribution also tends to zero by the same integrable bound. Hence $P(X=x)=0$ for every $x\ne2$: the only atom is at $2$, with mass $1/3$. The normal component adds no atoms.

Worked Example: Recovering a Poisson Probability

Suppose $X\sim\operatorname{Poisson}(2)$. Its characteristic function is $\phi_X(t)=\exp(2(e^{it}-1))$. The Poisson probabilities are $q_j=e^{-2}2^j/j!$, for integers $j\ge0$, and the characteristic function has the expansion $\phi_X(t)=\sum_{j=0}^{\infty}q_j e^{itj}$. Fix an integer $k\ge0$. Substituting this series into the symmetric average gives

A_T(k)=\sum_{j=0}^{\infty}q_j\operatorname{sinc}\bigl(T(j-k)\bigr).

The exchange of the sum and integral is justified because $\sum_j q_j=1$; equivalently, the series for $\phi_X(t)$ converges uniformly and absolutely. For $j=k$, the sinc factor is $1$ for every $T$. For each $j\ne k$, it tends to zero as $T\to\infty$, and every term has absolute value at most $q_j$. Since $\sum_j q_j$ is finite, dominated convergence for this weighted series gives $A_T(k)\to q_k$.

For example, at $k=3$,

P(X=3)=e^{-2}\frac{2^3}{3!} =e^{-2}\frac{8}{6} =\frac{4}{3}e^{-2} \approx 0.18045.

The same argument gives $P(X=k)=e^{-2}2^k/k!$ at every nonnegative integer $k$. At any real $x$ that is not a nonnegative integer, each term in the series average tends to zero; domination by the summable weights $q_j$ then gives $P(X=x)=0$.

Worked Example: Verifying That a Uniform Distribution Has No Atoms

Let $X$ be uniform on $[0,2]$. Its characteristic function is \[ \phi_X(t)=\frac{1}{2}\int_0^2 e^{itu}\,du =\frac{e^{2it}-1}{2it}\quad(t\ne0), \qquad \phi_X(0)=1. \] We show from the symmetric-average formula that no real number is an atom.

For $t\ne0$, the bounds $|e^{2it}-1|\le2$ and $|e^{2it}-1|\le2|t|$ imply \[ |\phi_X(t)|\le\min\left(1,\frac{1}{|t|}\right). \] The same upper bound, interpreted as $1$ at $t=0$, bounds the integrand in absolute value after multiplication by $e^{-itx}$, for every $x$. Therefore, for $T\ge1$,

|A_T(x)| \le\frac{1}{2T}\int_{-T}^{T}|\phi_X(t)|\,dt \le\frac{1}{2T}\left(2+2\int_1^T\frac{dt}{t}\right) =\frac{1+\log T}{T}.

The last expression tends to zero, so $P(X=x)=0$ for every real $x$. This establishes the absence of atoms directly from the characteristic function, without requiring a separate calculation at each candidate point.

Interpreting the Formula Correctly

The symmetric-average formula is a way to extract a point probability, not a claim that a single characteristic-function value identifies that probability. Each finite-$T$ average can involve contributions from the whole distribution; the atom emerges only after the averaging interval expands. The sinc representation explains why: values away from $x$ oscillate and their contribution vanishes in the limit.

Nor does the formula require the characteristic function itself to have a finite integral over the real line. The average is taken on a finite interval for every $T$, and its proof uses a bounded sinc function and dominated convergence on the probability space. This is useful in discrete and mixed models, where no density-based calculation need apply.

Common Mistake

Wrong reasoning: “If $X$ has a continuous component, then $P(X=x)=0$, so the whole distribution has no atoms.” A mixture can still put positive probability on a point.
Correction: Check the full distribution. In the mixed-normal example, the continuous component contributes no atom, but the point-mass component contributes $1/3$ at $2$. The formula recovers the mass of the whole mixture at the candidate point.
Another error: Treating $A_T(x)$ at a fixed, finite $T$ as if it already equaled $P(X=x)$. The theorem asserts equality only in the limit $T\to\infty$.

Key takeaway: Center the characteristic function at the candidate value $x$, average symmetrically over $[-T,T]$, and let $T$ grow. The limit is exactly $P(X=x)$, whether or not the distribution has a continuous component.

Check Your Understanding

Use the symmetric-average formula and its sinc representation to reason through these questions.

Why is $\operatorname{sinc}(T(X-x))$ bounded by an integrable random variable when proving the atom formula?
For a point mass at $a$, what is the centered average $A_T(x)$ when $x=a$, and what happens when $x\ne a$?
A distribution assigns probability $0.2$ to $5$ and probability $0.8$ to a continuous distribution with no atoms. What does the symmetric average converge to at $x=5$?
Why does the Poisson-series argument permit passage from the limit of each sinc term to the limit of the weighted sum?
Does the atom formula require $\int_{\mathbb{R}}|\phi_X(t)|\,dt<\infty$? Explain what condition the proof uses instead.