Are characteristic functions always differentiable?

Question

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. — Jose Avilez, May 09 '22 at 18:16
@JoseAvilez ah sure that's because of the properties of the probability measure right? But what if I take $X=x$? — user123234, May 09 '22 at 18:25
@JoseAvilez what if I take $X\sim Exp(1)$ and define $Y=\frac{1}{X}$? Or is there a simpler one? — user123234, May 09 '22 at 18:35
You may wish to take a look at Cauchy random variables as a source for counterexamples in statistics and probability. See my answer below. — Jose Avilez, May 09 '22 at 18:40
@JoseAvilez the problem is that we haven't seen cauchy random variables yet, how are they defined? — user123234, May 09 '22 at 18:41
@JoseAvilez you mean that this are a good class of random variables to find counterexamples? — user123234, May 09 '22 at 18:42
Again, see below for a definition. Yes, they are a good family to find counterexamples. — Jose Avilez, May 09 '22 at 18:49

Jose Avilez · Accepted Answer · 2022-05-09T20:25:12.107

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.

Are characteristic functions always differentiable?

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