$f(t)-$a characteristic function. Prove that $|f(t)|$ does not have to be a characteristic function based on the example: Discrete random variable
$$ X: \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ 1 \\ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{2}{3} \ \ \ \ \ \ \ \ \ \ \ \frac{1}{3}$$
$f(t)=Ee^{itX}=\frac{2}{3}+\frac{1}{3}e^{it}?$
Well to be frank I do not know how to do such I have tried some one the following properties of characteristicc functions ($f$), but they apply to this random variable it seems, I'll type them out anyway, I might be mistaken.
$$1.\ \ \ f(0)=1; |f(t)|<1;f(-t)=\overline{f(t)} \\ $$
Since the characteristic function is the expectation of a certain function $f(t)$ needs to absolutely converge. I do not know how to prove such. Though I haven't tried these theories on this random variable, mainly because I do not know how to:
Theory A : Function $f(t)$ is a characteristic function iff the following are true:
-$1.)f(0)=1$
-$2.)|f(t)|\leq 1$
-$3.) f(t) \text{ is continuous}$
-$4.) f(t) \text{ is positively definite meaning for every collection of real numbers $t_1,t_2,...,t_n$ and for every collection of complex numbers $z_1,z_2,...,z_n$ applies $\sum_{j,k=1}^{n}f(t_j-t_k)z_j\overline{z_k}\geq 0$}$
Theory B If $f(t)$ is a characteristic function of discrete type, then the distribution function is found in the following manner: $$P\{X=x\}=\lim_{T \to \infty }{\frac{1}{2T}\int_{-T}^{T}e^{itX}f(t)dt}$$ I'm guessing with this one all one would have to prove is that this isn't a properly defined probability.
I will very likely put this up for bounty.
It is straightforward to see that $\vert f(t)\vert=\dfrac{1}{3}\sqrt{5+4\cos t}$. Now consider $$(t_1,t_2,t_3,t_4)=\left(-\dfrac{\pi}{2},0,\dfrac{\pi}{2},\pi\right)$$ Then the matrix $M=\left(\vert f(t_i-t_j)\vert\right)_{1\le i,j\le 4}$ is easily calculated: $$M=\left( \begin{array}{cccc} 1 & \frac{\sqrt{5}}{3} & \frac{1}{3} & \frac{\sqrt{5}}{3} \\ \frac{\sqrt{5}}{3} & 1 & \frac{\sqrt{5}}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{\sqrt{5}}{3} & 1 & \frac{\sqrt{5}}{3} \\ \frac{\sqrt{5}}{3} & \frac{1}{3} & \frac{\sqrt{5}}{3} & 1 \\ \end{array} \right)$$ and $\det M=-\dfrac{16}{18}<0$, so $M$ is not positive. This proves that condition 4 in theorem A is violated, and $\vert f\vert$ is not the characteristic function of a random variable.
Remark. One can also consider $Z=(z_1,z_2,z_3,z_4)=(1,-1,1,-1)$ so that $$ZMZ^T=\sum_{1\le i,j\le4}\vert f(t_i-t_j)\vert z_i\bar z_j=-\dfrac{8}{3(2+\sqrt{5})}<0.$$