# Are characteristic functions always differentiable?

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.

• 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. May 9 at 18:16
• @JoseAvilez So would it work if I consider $X=1$
– Wave
May 9 at 18:20
• No. $X=1$ has finite mean. May 9 at 18:23
• @JoseAvilez ah sure that's because of the properties of the probability measure right? But what if I take $X=x$?
– Wave
May 9 at 18:25
• Same. All constants have finite mean. May 9 at 18:30

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.