Assume that $X\_n$ is a sequence of a zero-mean and unit variance random variables (and maybe having density w.r.t. to Lebesgue). Can we conclude that $ P(X\_n \in [0,R\_n]) $ is bounded away from zero eventually, say $$\liminf\_{n \to \infty} P(X\_n \in [0,R\_n]) > 0$$ assuming that $R\_n \to \infty$. Intuitively, $X\_n$ should put some nonvanishing amount of mass on the positive real line because of the zero-mean assumption and the condition of unit variance should prevent that mass from escaping to infinity (?)


Here is a partial argument: We have
\begin{align\*}
P(X\_n \in [0,R\_n]) &= 1 - P(X\_n > R\_n) - P(X\_n < 0) \\
&\ge 1 - P(|X\_n| > R\_n) - P(X\_n < 0) \\
&\ge 1 - \frac{E X\_n^2}{R\_n^2} - P(X\_n < 0) \\
&= 1 - o(1) - P(X\_n < 0)
\end{align\*}
Thus, the problem reduces to arguing that $P(X\_n < 0)$ stays away from 1.



Lower bounding the probability that a zero-mean sequence of random variables stays positive
Here's a proof if a third moment condition is satisfied. 


Suppose that $\mathbb EX=0$, $\mathbb EX^2=1$ and $\mathbb E|X|^3\le K$. 
Then let $Z=|X|$. We use Cauchy-Schwarz: $Z^2=Z^{1/2}Z^{3/2}$, so that $(\mathbb EZ^2)^2\le \mathbb EZ\cdot \mathbb EZ^3$. This gives $\mathbb EZ\ge \frac 1K$. Hence $\mathbb EX\mathbf 1\_{X>0}\ge \frac{1}{2K}$. 


Also $\mathbb EX^2\mathbf 1\_{X>0}\le 1$. Now $\mathbb EX\mathbf 1\_{X>4K}\le (1/4K)\mathbb EX^2\mathbf 1\_{X>4K}\le 1/(4K)$, so that $\mathbb EX\mathbf 1\_{X\in [0,4K]}\ge 1/(4K)$ and $\mathbb P(X\in [0,4K])\ge 1/(16K^2)$. 


