Let $X$ be a discrete random variable that takes values in $a+b \mathbb{Z}=\{a+b m \mid m \in\mathbb{Z}\}$ [$a\in\mathbb{R},b>0$]. (It is not assumed that image of $X$ equals $a+b \mathbb{Z}$.) Then show that $\left|\phi_{X}(t)\right|<1$ for $0<t<\frac{2\pi}{b}$.
Now i am getting that $|\phi(\frac{2\pi}{b})|=1$. How i should prove that $|\phi(t)|<t$ for $0<t<\frac{2\pi}{b}$. I tried to do by converting $E[e^{itX}]$ to $E[\cos(tX)]+iE[\sin(tX)]$. Then $|\phi(t)|=E[\cos(tX)]^2+E[\sin(tX)]^2$. But unable to do.
How can i prove it
We have $|\phi(2\pi /b)|=1$ Now$$\phi(t)=E[e^{itX}]=E[\cos(tX)+i\sin(tX)]=E[\cos(tX)]+iE[\sin(tX)]$$Therefore $$|\phi(t)|^2=E[\cos(tX)]^2+E[\sin(tX)]^2$$
Now we know $\cos^2(tX)+\sin^2(tX)=1$ Therefore $$E[\cos^2(tX)+\sin^2(tX)]=E[1]=1\iff E[\cos^2(tX)]+E[\sin^2(tX)]=1$$Now\begin{align} |\phi(2\pi /b)|^2 - |\phi(t)|^2 & = 1 - |\phi(t)|^2\\ & = E[\cos^2(tX)]+E[\sin^2(tX)] - \left[ E[\cos(tX)]^2+E[\sin(tX)]^2 \right]\\ & = E[\cos^2(tX)] - E[\cos(tX)]^2+E[\sin^2(tX)]-E[\sin(tX)]^2 \\ & =\underbrace{Var(\cos(tX))}_{\geq 0}+\underbrace{Var(\sin(tX))}_{\geq 0}\geq 0 \end{align}Therefore $$1 - |\phi(t)|^2\geq 0\iff |\phi(t)|\leq 1$$If $X$ is not constant then the variances becomes strictly greater than 0. In that case we have $|\phi(t)|<1$