Let $X$ be a random variable. $X$ is called lattice distributed if there exist real numbers $a, b$ such that $P(X \in a +b\mathbb{Z})=1$.
Show that $X$ is lattice distributed if there exists $v\neq 0$ such that $\left | \varphi(v) \right |=1$.
Here, $\varphi(t)=E\left[e^{itX}\right]$ denotes the characteristic function of $X$.
I have actually found a solution in Rick Durret's Probability: Theory and Examples, Theorem 3.5.1, but a major part of the proof is missing and I think he's using a theorem about strictly convex functions on a function that is convex, but not strictly convex.
Anyway, there must be an easier solution to this I think.
Best Answer
Since $|\varphi(v)|=1$, there is some $\theta \in \mathbb R$ such that $\varphi(v)=e^{i\theta}$. Then $E(e^{i(vX-\theta)})=1$ ie $E(\cos(vX-\theta))+iE(\sin(vX-\theta))=1$, hence $E(\cos(vX-\theta))=1$ and $E(1-\cos(vX-\theta))=0$
But $1-\cos(vX-\theta)\geq 0$, thus $1-\cos(vX-\theta)=0$ a.s., ie $$P(\cos(vX-\theta)=1)=1$$ For $\omega \in \Omega$, $$\cos(vX(w)-\theta)=1 \iff \exists k \in \mathbb Z, \; X(w) = \frac{2\pi}{v}k + \frac{\theta}{v}$$ thus $1=P(\cos(vX-\theta)=1)=P(X\in \bigcup_{k\in \mathbb Z} \left( \frac{\theta}{v} +k \frac{2\pi}{v}\right))$.