I am thinking about the statement if the characteristic function of a random variable $X$, $\Phi_X$, is always differentiable.

By definition, $$\Phi_X(t)=\int_{\Bbb{R}^d}e^{i\langle t,y \rangle}P_X(dy)$$ Hence, I think it has something to do with changing the integral and the derivative right? But my intuition tells me that there is a counterexample but I can't find one.

  • $\begingroup$ If $E|X| < \infty$, then $\Phi_X^\prime(t) = E(iXe^{itX})$ by LDCT, so you will have to look amongst RVs without that moment condition. $\endgroup$ May 9 at 18:16
  • $\begingroup$ @JoseAvilez So would it work if I consider $X=1$ $\endgroup$
    – Wave
    May 9 at 18:20
  • $\begingroup$ No. $X=1$ has finite mean. $\endgroup$ May 9 at 18:23
  • $\begingroup$ @JoseAvilez ah sure that's because of the properties of the probability measure right? But what if I take $X=x$? $\endgroup$
    – Wave
    May 9 at 18:25
  • $\begingroup$ Same. All constants have finite mean. $\endgroup$ May 9 at 18:30

1 Answer 1


If $E|X| < \infty$ then the derivative of the characteristic function is given by: $$\Phi_X^\prime (t) = E(iX e^{itX})$$ as the Lebesgue Dominated Convergence theorem would allow us to exchange the order of differentiation and integration. Thus, we must look for a random variable $X$ with infinite mean.

Let $X$ be a Cauchy random variable, so that its pdf is given by $$f_X(x) = \frac{1}{\pi (1 + x^2)} \qquad x \in \mathbb{R}$$ and its characteristic function can be computed to be $$\Phi_X(t) = \exp (-|t|)$$ $\Phi_X(t)$ is readily seen not to be differentiable at $t = 0$, as desired.


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