Suppose $X_1, X_2, ... \sim X$ are i.i.d. random variables on $\mathbb{Z}$. Then the sequence $\{P(\sum_{i=1}^{d(X)n} X_i = 0)^{\frac{1}{d(X)n}}\}_{n=1}^\infty$ converges to some constant $\rho(X) \in [0;1]$ almost surely by Kingman’s subadditive ergodic theorem. Here $d(X) = \min\{d \in \mathbb{N}|P(\sum_{i=1}^{d} X_i = 0)>0\}$.
For some random variables $X$ the value $\rho(X)$ can easily be calculated. For example:
If $P(X \geq 0) = 1$ or $P(X \leq 0) = 1$, then $\rho(X) = P(X = 0)$ trivially.
If $$X = \begin{cases} -a & \quad \text{ with probability } p \\ b & \quad \text{ with probability } 1-p \end{cases}$$ where $a, b \in \mathbb{N}$, $p \in (0;1)$, then $\rho(X)=\frac{p^{1-x}(1-p)^{x}}{x^x(1-x)^{1-x}}$, where $x = \frac{a}{a+b}$. That is due to the fact, that $P(\sum_{i=1}^{(a+b)n} X_i = 0) = C^a_{a+b}p^ac^b$.
I want to determine, whether the inequality $|E[X]| + \rho(X) \geq 1$ holds for all random variables $X$ on $\mathbb{Z}$.
So far I only managed to "prove" it in three trivial cases:
If $P(X \geq 0) = 1$. Then, by Markov inequality $1 - \rho(X) = 1-P(X=0) = P(X \geq 1) \leq E[X] = |E[X]|$.
If $P(X \leq 0) = 1$. Then, by Markov inequality $1 - \rho(X) = 1-P(X=0) = P(-X \geq 1) \leq -E[X] = |E[X]|$.
If $|E[X]| \geq 1$. That is because $\rho(X) \in [0;1]$
However, I do not know how to prove this inequality in general case. Neither have I found any counterexamples.
Related question on upper bound for $\rho(X)$: $\rho(X)E[X]^2 \leq Var[X]$?