$\rho(X)E[X]^2 \leq Var[X]$?

Question

Suppose $X_1, X_2, \ldots \sim X$ 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 $d(X) = \min\{d \in \mathbb{N}\mid P(\sum_{i=1}^d X_i = 0)>0\}$.

For some random variables $X$ the value $\rho(X)$ can easily be calculated. For example:

1. If $P(X \geq 0) = 1$ or $P(X \leq 0) = 1$, then $\rho(X) = P(X = 0)$ trivially.

2. 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 $P(\sum_{i=1}^{(a+b)n} X_i = 0) = C^a_{a+b}p^b(1-p)^a$.

I want to determine, whether the inequality $ \rho(X)E[X]^2 \leq Var[X]$ holds for all random variables $X$ on $\mathbb{Z}$.

So far I only managed to "prove" it in four trivial cases:

1. If $P(X \geq 0) = 1$. Then, by Chebyshev inequality $\rho(X) = P(X=0) \leq P(|X - E[X]| \geq E[X]-\epsilon) \leq \frac{Var[X]}{(E[X]-\epsilon)^2}$, for all $\epsilon > 0$. Now we can get our inequality by taking the limit $\epsilon \to 0$

2. If $P(X \leq 0) = 1$. Then, by Chebyshev inequality $\rho(X) = P(X=0) \leq P(|X - E[X]| \geq -E[X]-\epsilon) \leq \frac{Var[X]}{(E[X]+\epsilon)^2}$, for all $\epsilon > 0$. Now we can get our inequality by taking the limit $\epsilon \to 0$

3. If $E[X]^2 \leq Var[X]$. That is because $\rho(X) \in [0;1]$

4. If $E[X] = 0$, then the statement is trivial due to non negativity of variance.

However, I do not know how to prove this inequality in general case. Neither have I found any counterexamples.

Related question on lower bound for $\rho(X)$: $|𝐸[𝑋]|+𝜌(𝑋)≥1$?

Answer 1

False for the second example you gave with $a=2,b=1,p=0.9$ (unless I made a mistake).