When Characteristic-Function Inversion Needs Care

One Transform, Several Different Questions

A characteristic function can encode an interval probability, the probability of a single point, or a density, but these are different inversion questions. Their formulas do not have identical hypotheses or identical meanings. In particular, writing down an inverse Fourier integral does not by itself show that a distribution has a density.

Throughout, let \(X\) be a real-valued random variable with characteristic function \(\varphi(t)=E[e^{itX}]\). The distribution may have a density, atoms, or both. The central distinction is this: interval inversion and atom inversion apply without assuming a density, whereas the standard absolutely convergent density inversion theorem requires \(\varphi\) to be integrable.

Interval Probabilities: Boundary Mass Must Be Counted

Lévy inversion recovers probabilities of intervals directly from \(\varphi\). The symmetric truncation in the formula matters: if the distribution has mass at an endpoint, that endpoint contributes one half. Thus the formula applies whether or not \(X\) has a density, but the endpoint convention must be respected.

Here is the justification, including why the expectation and the limiting operation are legitimate. For finite \(T\), the integrand can be written as an expectation and integrated over a bounded interval; the quotient has a finite limit at zero. For a fixed value \(x\), the resulting kernel is

The sine integral \(\int_0^T\sin(tu)/t\,dt\) is bounded by a constant independent of \(u\) and \(T\). As \(T\to\infty\), it tends to \(\frac{\pi}{2}\) for \(u>0\), to \(-\frac{\pi}{2}\) for \(u<0\), and is zero for \(u=0\). Therefore the displayed kernel is uniformly bounded and converges to 1 when \(a<x<b\), to \(1/2\) when \(x=a\) or \(x=b\), and to 0 otherwise. Taking expectations and applying bounded convergence proves the formula. \(\square\)

Atoms: A Separate Inversion Formula

An atom is a point \(x\) with \(P(X=x)>0\). Its mass can be recovered without assuming that \(X\) has a density or that \(\varphi\) is integrable.

To prove this, interchange expectation and integration on the finite interval. The right-hand expression before taking the limit becomes

where the ratio is defined as 1 when \(X=x\). The ratio has absolute value at most 1. As \(T\to\infty\), it tends to 1 if \(X=x\) and to 0 otherwise. Bounded convergence gives the claimed probability. \(\square\)

Density Inversion: What Integrability Guarantees

A useful sufficient condition for ordinary density inversion is absolute integrability of the characteristic function. It ensures both that the inverse integral exists and that the distribution is absolutely continuous.

A regularization argument proves the claim without assuming the desired density in advance. Let \(Z\sim N(0,1)\) be independent of \(X\), and for \(\varepsilon>0\) consider \(X+\varepsilon Z\). This variable has a density \(f_\varepsilon\), obtained by averaging the normal density centered at \(X\). Its characteristic function is \(\varphi(t)e^{-\varepsilon^2t^2/2}\), so Fourier inversion for the smoothed distribution gives

Because \(|\varphi|\) is integrable, dominated convergence shows that \(f_\varepsilon(x)\) converges uniformly in \(x\) to the inverse integral \(g(x)\): indeed, the supremum of the difference is at most

The inverse integral \(g\) is bounded and uniformly continuous, by integrability of \(\varphi\) and dominated convergence applied to translations in \(x\). Each \(f_\varepsilon\) is nonnegative, so the uniform limit \(g\) is nonnegative. For any continuous compactly supported function \(h\), uniform convergence on its support gives \(\int h(x)f_\varepsilon(x)\,dx\to\int h(x)g(x)\,dx\). On the other hand, \(E[h(X+\varepsilon Z)]\to E[h(X)]\) by bounded convergence, using the same \(X,Z\) for all \(\varepsilon\). Thus

Equality for all continuous compactly supported \(h\) identifies the distribution of \(X\) with the measure having density \(g\). Consequently \(g\) integrates to 1 and is the asserted continuous density. \(\square\)

When the Characteristic Function Is Not Integrable

Failure of \(\varphi\in L^1\) has a limited but important meaning: the theorem guaranteeing an absolutely convergent inverse integral and a continuous density cannot be applied. It does not prove that no density exists. Some distributions with densities have nonintegrable characteristic functions, and a symmetric or regularized inverse may still recover their density at suitable points.

Common Pitfalls and a Practical Choice

A recurring error is to treat every inverse Fourier expression as if it were a density. For example, the argument “the characteristic function has a Fourier inverse, so that inverse must be the density” skips the essential hypotheses. A point mass has a perfectly valid characteristic function but no density, and its formal symmetric inverse does not converge to an ordinary density. Conversely, the uniform example shows that \(\varphi\notin L^1\) alone does not imply that a density is absent.

Choose the inversion tool to match the question:

• For an interval probability, use Lévy inversion and account for half-mass at each endpoint.
• For the probability at one point, use atom inversion.
• For a density from an absolutely integrable characteristic function, use the ordinary inverse integral.
• If the characteristic function is not integrable, do not infer either the presence or absence of a density from that fact alone. Check the distributional structure and the convergence mode of any proposed inverse.

Absolute integrability is a clean sufficient condition, not a necessary condition for every possible density inversion formula. A conditional limit, a symmetric truncation, or a smoothing procedure needs its own justification and may encode boundary values differently. Keeping these distinctions explicit prevents a formal calculation from being mistaken for a theorem.