I want to prove that an infinitely divisible symmetric $\mathbb Z$-valued random variable satisfies $$ \mathbb P(X \in 2\mathbb Z) > P(X \in 2\mathbb Z+1).$$ I know that $X$ cannot be bounded unless it is constant because of this, in which case it would have to be $0$ because of symmetry. If we call $\mathbb P(X=k) =p_k$ we can say that $p_k\xrightarrow{k\to +\infty} 0$, $p_k\xrightarrow{k\to -\infty} 0$ and that
$$\varphi_X(t) = \sum\limits_{k\in\mathbb Z} \exp(itk)p_k = \sum\limits_{n\in\mathbb N} 2\mathfrak{Re}(\exp(itn))p_n=\sum\limits_{n\in\mathbb N}2\cos(nt)p_n.$$ From which follows that $$1=\varphi_X(0) = \sum\limits_{n\in\mathbb N}2\cos(nt)p_n =\sum\limits_{n\in\mathbb N}2p_n =\sum\limits_{k\in\mathbb Z}p_k \implies p_0 = 0.$$ How can I continue from here? Should I look at the $\psi_n: \psi_n^n = \varphi_X$ we get from infinite divisibility? Multiplying discrete characteristic functions looks tedious.
Thank you!
Characteristic functions are uniformly continuous and it is known that characteristic functions $\varphi_X$ of infinitely divisible random variables $X$ are nowhere $0$. Therefore it follows from the mean value theorem that $\varphi_X > 0$ everywhere or $\varphi_X < 0$ everywhere. But $\varphi_X(0)=1$. It has to be $\varphi_X > 0$.
Because $X$ is $\mathbb Z$-valued. We consider $p_k = \mathbb P(X=k), k\in\mathbb Z$. $$\varphi_X(t)=\sum\limits_{k\in\mathbb Z} \exp(ikt)p_k = \sum\limits_{l\in\mathbb Z} \exp(it(2l))p_k + \sum\limits_{l\in\mathbb Z} \exp(it(2l+1))p_k.$$
If we look at $\varphi_X(\pi) = \sum\limits_{l\in\mathbb Z} \exp(i\pi(2l))p_k + \sum\limits_{l\in\mathbb Z} \exp(i\pi(2l+1))p_k = \sum\limits_{l\in\mathbb Z} p_{2l}-\sum\limits_{l\in\mathbb Z} P_{2l+1} >0$ we see that $$\mathbb P(X\in 2\mathbb Z)>\mathbb P(X\in 2\mathbb Z+1).$$