I am looking for an example of a continuous distribution function where the first moment does not exist but the characteristic function is differentiable everywhere.

Cauchy distributions do not fulfill this, as their characteristic functions are not differentiable at $0$. Does anybody here have an example?


Below, there is a discrete example because I misread the word continuous.

However, if $X$ has the density $f$ given by $$f(x)=\begin{cases}0&\mbox{ if }|x|\leqslant 2,\\ \frac C{x^2\log |x|}&\mbox{ if }|x|>2,\end{cases}$$ then $X$ is not integrable and the characteristic function is given by $$\varphi(t)=2C\int_2^{+\infty}\frac{\cos(tx)}{x^2\log x}\mathrm{d}x.$$

  • Let us show that $\varphi$ is differentiable at $0$ and $\varphi'(0)=0$. We start from $$\frac{\varphi(t)-\varphi(0)}{2tC}=\int_2^{1/t}\frac{\cos(tx)-1}{x^2\log x}\frac{dx}t+\int_{1/t}^{+\infty}\frac{\cos(tx)-1}{x^2\log x}\frac{dx}t=:A(t)+B(t).$$ Then using $|\cos u-1|\leqslant u^2$ $$|A(t)|\leqslant \int_2^{1/t}\frac{t^2x^2}{x^2\log x}\frac{dx}t=t\int_2^{1/t}\frac{dx}{\log x}.$$ Let $y:=\log x$, then $x=e^y$ and $dx=e^ydy$: $$|A(t)|\leqslant t\int_{\log 2}^{-\log t}\frac{e^y}y\mathrm dy,$$ and splitting the interval at $R$, we get $$|A(t)|\leqslant te^R/\log 2+R^{-1},$$ hence $\lim_{to\to 0}A(t)=0$. Since $$|B(t)|\leqslant \frac 1t\int_{1/t}^\infty\frac{2}{x^2\log(x)}dx,$$ we obtain after the substitution $s=1/x$ that $$|B(t)|\leqslant -\frac 2t\int_0^t\frac{ds}{\log s}.$$ An other transformation $tu=s$ yields
    $$|B(t)|\leqslant -\int_0^1\frac{du}{\log u+\log t},$$ which converges to $0$ as $t$ goes to $0$.
  • A similar argument can be used to show the differentiability at each point.

This example is taken from Stoyanov's book Counter-examples in probability.

Take $X$ such that $$\mathbb P(X=(-1)^kk)=\frac C{k^2\log k},\quad k\geqslant 2.$$ Then $\mathbb E|X|$ is infinite, while the characteristic function
$$\varphi_X(t)=\sum_{k\geqslant 2}\frac C{k^2\log k}e^{it(-1)^kk}$$ is differentiable everywhere: compute $\varphi_X(t+h)-\varphi_X(t)$ approximating $e^{ih(-1)^kk}-1$ by $h(-1)^kk+o(h)$.

| cite | improve this answer | |
  • $\begingroup$ @DavideGiraudo How did you treat the term containing the $\sin$? $\endgroup$ – user66906 May 16 '14 at 21:51
  • $\begingroup$ You mean, in the computation of the characteristic function, don't you? The distribution is symmetric. $\endgroup$ – Davide Giraudo May 16 '14 at 21:54
  • $\begingroup$ @DavideGiraudo but if this integral does not exist on each side, then we are calculating something like a Cauchy principal value not the integral itself, kind of similar to $\int_{-\infty}^{\infty}x dx \neq 0$ $\endgroup$ – user66906 May 17 '14 at 9:55
  • 1
    $\begingroup$ I've edited. Is it clearer? $\endgroup$ – Davide Giraudo May 20 '14 at 9:41
  • 1
    $\begingroup$ Done now. ${}{}$ $\endgroup$ – Davide Giraudo May 21 '14 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy