Let $f$ be the characteristic function of a real-valued random variable $X$. It is known that if $f$ has a $k$-th order derivative (for some even $k$) then $\mu$ has a finite $k$-th order moment. Is there any reference or discussion the case when $k$ is odd? For example, if we only know that $f$ has its first order derivative, do we know that $\mu$ has finite first moment?
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2 Answers
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A good reference regarding this type of results is the book by Eugène Lukacs "Characteristic Function". For example, chapter 2.3 "Characteristic functions and moments" provides results in this direction. Theorems 2.3.1, 2.3.2 and 2.3.3 might be what you are looking for.
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3$\begingroup$ To complement my post and to counter the down vote: in the same chapter there is an example due to Zygmund of a characteristic function with a continuous first derivative but for which the associated law has an infinite first moment... $\endgroup$ Commented Jun 13, 2020 at 15:36
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$\begingroup$ Thanks. I saw the example. But it was provided without proof that it is differentiable. Did Zygmund prove it? Can you give a reference to the proof of differentiability? $\endgroup$– trisctCommented Jun 13, 2020 at 15:56
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1$\begingroup$ I think it was stated by Zygmund in this paper projecteuclid.org/euclid.aoms/1177730443 but without proof. For a proof, you should have a look at his fundamental work on trigonometric series. $\endgroup$ Commented Jun 13, 2020 at 22:11
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Zygmund's example is the discrete random variable $X$ where $\mathbb{P}(X=\pm n) = \frac{C}{n^2\log(n)}$ for integer $n \geq 2$. $C$ is a unique constant that makes this a probability distribution. You can calculate that $X$ does not have a finite first moment because $$\mathbb{E}(|X|)= \sum_{n=2}^\infty \frac{2C}{n \log(n)} = +\infty$$. It diverges because $\int_2^\infty \frac{1}{x\log(x)}dx$ diverges.
The characteristic function is $$\varphi(t) = C\sum_{n=2}^\infty \frac{\cos(nt)}{n^2 \log(n)}.$$
$\varphi(t)$ is differentiable at zero and the derivative is $\varphi'(0)=0$. Zygmund's argument does not rely on any fancy theory of trigonometric series. You can just check the differentiability directly.
Specifically, look at the quotients $$\frac{\varphi(h) - \varphi(0)}{h} = C\sum_{n=2}^\infty \frac{\cos(nh) - 1}{hn^2\log(n)}$$. Split this into two pieces for a $M \in \mathbb{N}$ to be chosen soon. $$=C\sum_{n=2}^M \frac{\cos(nh) - 1}{hn^2\log(n)} + C\sum_{n=M+1}^\infty \frac{\cos(nh) - 1}{hn^2\log(n)}$$.
For the first sum, where $nh$ is small, we use the Remainder Theorem estimate $$|\cos(nh) -1|\leq \frac{n^2h^2}{2}$$ and for the second sum, where $nh$ is big, we use the estimate $$|\cos(nh) -1|\leq 2.$$
This all leads to the estimate, where $K$ is some constant independent of $h$ whose value changes line to line, $$\left|\frac{\varphi(h) - \varphi(0)}{h}\right| \leq K \sum_{n=2}^M \frac{h}{\log(n)} + K\sum_{n=M+1}^\infty \frac{1}{n^2\log(n) h}$$.
The first sum is trivially less than $KMh$. For the second sum, $$\sum_{n=M+1}^\infty \frac{1}{n^2 \log(n) h} \leq \frac{1}{h\log(M)} \sum_{n=M+1}^\infty \frac{1}{n^2} \leq \frac{K}{hM\log(M)}.$$
These calculations prove that $$\left|\frac{\varphi(h) - \varphi(0)}{h}\right| \leq KMh + \frac{1}{M \log(M) h}$$.
Given $h$, choose integer $M=M(h)$ so that $\frac{1}{M \sqrt{\log(M)}}\approx h$. Then both terms go to $0$ as $h \to 0$, proving that $\varphi(t)$ is differentiable at $t=0$.
