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I came across the following statement in a book:

If $\varphi(t)$ is a characteristic function, then $|\varphi(t)|$ is not necessarily a characteristic function.

Here's my argument:

By Bochner’s theorem, we can check $|\varphi(t)|$ satisfies the three conditions:

  • continuous at $t=0$
  • $|\varphi(0)|=1$
  • positive definite \begin{align} \sum_{i,j}\xi_i\bar\xi_j|\varphi(t_i-t_j)| &=\sum_{i,j}\xi_i\bar\xi_j|\varphi(t_i)||\varphi(t_j)| \\ &=\left(\sum_i \xi_i|\varphi(t_i)|\right)\overline{\left(\sum_i \xi_i|\varphi(t_i)|\right)}\\ &=\left|\sum_i \xi_i|\varphi(t_i)|\right|^2\geq 0 \end{align} The equality holds only when $\xi_i=0,i=1,2,\ldots,n$.

What's the problem with this argument?

1024
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    Define characteristic function – zhw. Jul 24 '16 at 02:28
  • @zhw. A characteristic function of a random variable $X$ is $\varphi(t) = \mathbb{E}(e^{itX})$ – 1024 Jul 24 '16 at 02:31
  • Why is continuous at t=0? What if 0 is a critical point of Phi? – Martín Vacas Vignolo Jul 24 '16 at 03:27
  • @vvnitram this is a proposition of characteristic function. Now that phi is continous at 0， so is |phi| – 1024 Jul 24 '16 at 03:31
  • @K.L. why? Im(Phi) is a subset of C, but im(|Phi|) is a subset of R – Martín Vacas Vignolo Jul 24 '16 at 03:35
  • @vvnitram but it seems not to be a problem. By the triangular inequality in complex analysis, we have $||\varphi(t_i)|-|\varphi(t_j)||\leq |\varphi(t_i)-\varphi(t_j)|$. That indicates $\varphi$ continous at 0 => $|\varphi|$ continous at 0 – 1024 Jul 24 '16 at 03:45
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    Could you elaborate why the first "=" holds true? In general, $|\varphi(t_i-t_j)| \neq |\varphi(t_i)| , |\varphi(t_j)|$, right? – saz Jul 24 '16 at 19:25

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