Suppose $\{X_i\}_{i = 1}^n$ is a sequence of i.i.d. random variables, such that $E(X_1) = 0$ and $Var(X_1) = 1$. Does there always exist such $\epsilon > 0$, that $P(|\Sigma_{i = 1}^{n} X_i | \leq \sqrt{n}) \geq \epsilon$?
How did this question arise? There is a well known fact, called Van Zuijlen bound:
Suppose $\{X_i\}_{i = 1}^n$ is a sequence of i.i.d. random variables, such that $P(X_1 = 1) = P(X_1 = -1) = 0.5$. Then $P(|\Sigma_{i = 1}^{n} X_i| \leq \sqrt{n}) \geq 0.5$.
A similar fact is true for normal distribution too:
Suppose $\{X_i\}_{i = 1}^n$ is a sequence of i.i.d. random variables, such that $X_1 \sim N(0, 1)$. Then $P(|\Sigma_{i = 1}^{n} X_i| \leq \sqrt{n}) \geq 0.31$.
However, I have never seen similar statements about other distributions being proved or disproved, despite such question seeming quite natural . . .
By CLT $P( |\sum\limits_{i=1}^{n}X_i| \leq \sqrt n)$ tend to a positive number (namely $P(|Z| \leq 1$ where $Z$ has standard normal distribution). So the question now is if this probability can be $0$ for some $n$. If it is $0$ then $|\sum\limits_{i=1}^{n}X_i| >\sqrt n$ almost surely which implies $E(\sum\limits_{i=1}^{n}X_i)^{2} >n$. But $E(\sum\limits_{i=1}^{n}X_i)^{2} =n$ so we have contradiction.