For a real-valued random variable $X$, the characteristic function $\phi \colon \mathbb{R} \to \mathbb{C}$ is defined as $$\phi(t) = \mathbb{E}[e^{itX}].$$ One can check that $\phi$ satisfies

  • $\phi(0) = 1$,
  • $\phi$ is uniformly continuous,
  • $\phi(t) \to 0$ as $|t| \to \infty$.

Conversely, if we have a $\phi$ satisfying these three properties, is there a distribution with characteristic functions $\phi$? In my specific case, I have a $\phi \in C^\infty \cap L^1 \cap L^\infty$.

  • $\begingroup$ The Taylor coefficients of $\phi$ (so, the moments of $X$) need to satisfy some delicate conditions: see en.wikipedia.org/wiki/Moment_problem for a discussion. As a simple example, if $\phi'(0) = \phi''(0) = 0$, then $X = 0$ a.e, and so $\phi(t) = 1$. $\endgroup$ Commented Nov 28, 2017 at 2:57
  • $\begingroup$ I know if $X \in L^p$ then $\phi \in C^p$, but the converse need not be true right? $\endgroup$
    – user217285
    Commented Nov 28, 2017 at 2:59
  • 1
    $\begingroup$ The third property does not hold for all characteristic functions of real random variables. For instance, if $X=\pm1$ with equal probabilities, then $\phi(t)=\cos(t)$. $\endgroup$ Commented Nov 28, 2017 at 3:05
  • $\begingroup$ @kimchilover hmm I agree but why is this not in contradiction with Riemann-Lebesgue? $\endgroup$
    – user217285
    Commented Nov 28, 2017 at 4:55
  • 1
    $\begingroup$ Riemann Lebesgue Lemma applies only when X has a probability density function. $\endgroup$ Commented Nov 28, 2017 at 6:22

1 Answer 1


You want Bochner's theorem for necessary and sufficient conditions. The conditions you give do not ensure non-negativity of the measure with characteristic function $\phi$. Bochner's theorem does, by requiring also that the matrices $(\phi(t_i-t_j))$ be positive semi-definite. The necessity is easy to see. Consider expectations of form $$E\left|\sum_{k=0}^na_k \exp(i t_k X)\right|^2 = \sum_{k=0}^n\sum_{l=0}^n E (\exp(i(t_k-t_l)X) a_k \overline a_l ) =\sum_{k=0}^n\sum_{l=0}^n \phi(t_k-t_l) a_k \overline a_l,$$ for arbitrary $a_k\in\mathbb C$. Such sums must be non-negative for all choices of the $a_k$, and hence $\phi$ must be positive definite.

By the way, the limit $\phi(t)\to0$ as $|t|\to\infty$ that you state, does not hold for all real random variables. If $P(X=1)=P(X=-1)=1/2$, then $\phi(t)=\cos(t)$, for example. Or even more simply: if $P(X=0)=1$ then $\phi(t)=1$ for all $t$.

In your case, where $\phi\in C^\infty\cap L^1 \cap L^\infty$, there is a function $f$, not necessarily non-negative, such that $\phi(t)=\int \exp(itx)f(x)\,dx$. But without checking something equivalent to Bochner's condition that the kernel $K(s,t)=\phi(s-t)$ is positive definite, you won't know that $f$ is non-negative.

For example, the function $\phi(t)=1/(1+t^2)$ satisfies your conditions, and is the characteristic function of a probability distribution. But the function $\psi(t)=2\phi(t)-\phi(2t)$ also satisfies the conditions and is not the characteristic function of any probability distribution.

  • $\begingroup$ Incredibly thorough, thank you. $\endgroup$
    – user217285
    Commented Nov 28, 2017 at 19:24

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