Is it true differentiability of characteristic function at 0 implies existence of the first moment.

Question

The context is the following:

Prove $|\cos(t)|$ is not a characteristic function of a distribution, while $\cos(t)$ is a characteristic function.

I know you can prove it by inverse Fourier transformation and check its positivity. However, I am just wondering if it is possible to prove directly by looking at the differentiability of $|\cos(t)|$.

$|\cos(t)|$ is differentiable at 0 while not at $\pi/2$. I believe if a characteristic function is differentiable at 0, then it should be differentiable everywhere. As if a random variable has finite first moment, then $\phi'(0)$ = $cEX$ ($\phi$ is the characteristic function) for some complex constant $c$. But is the converse true?

If $\phi$ is differentiable at 0, then $EX$ exists.

If this is proved, then $|\cos(t)|$ cannot be a characteristic function, as its derivative at $t$ should be $c\mathbb{E}[X\exp(i\pi/2 X)]$ which is also finite.

In general, I just come up with an example satisfying the following: For $f_n\rightarrow f$ a.s. \begin{align*} \lim_{n\rightarrow\infty}\int f_n \neq \int f \quad s.t. \int |f|=\infty \end{align*} Let $f=1/x$ and $f_n$ be a finite symmetric approximation of $f$, then $\int f_n=0$ while $\int f$ does not exist. I guess the statement might have something to do with the property of $\exp(it)$.

Thanks in advanced!

Answer 1

Actually, the converse is not true! I was surprised to learn this. Instead, here is what is true:

Theorem (Pitman): Let $X$ be a random variable, and let $\phi(t)=\mathbb E[e^{itX}]$. Let $k$ be any odd positive integer. Necessary and sufficient conditions for the existence of $\phi^{(k)}(0)$ are

1. $\lim_{x\to\infty}x^k\,\mathbb P(|X|>x)=0$, and
2. $\lim_{T\to\infty}\mathbb E[X^k\cdot 1_{|X|<T}]$ exists and is finite.

In case these conditions are both true, then $\phi^{(k)}(0)=i^k\lim_{T\to\infty}\mathbb E[X^k\cdot 1_{|X|<T}]$.

Side note: When $k$ is an even integer, $E|X|^k<\infty$ if and only if $\phi^{(k)}(0)$ exists.

In particular, with $k=1$, you get exact conditions for $\phi'(0)$ existing. All that remains is to find a random variable with $E|X|=\infty$ such that both conditions hold. The following example suffices, taken from Durrett, exercise 2.2.4. Let $X$ be the random variable for which $$ P(X=(-1)^k\cdot k)=\frac{C}{k^2\log k},\qquad k\in \{2,3,4,\dots\} $$ where $C$ is a normalizing constant.

Durrett, R. (2019). Probability: Theory and Examples (5th ed., Cambridge Series in Statistical and Probabilistic Mathematics). Cambridge: Cambridge University Press. doi:10.1017/9781108591034

E. J. G. Pitman "On the Derivatives of a Characteristic Function at the Origin," The Annals of Mathematical Statistics, Ann. Math. Statist. 27(4), 1156-1160, (December, 1956)