Let $\{X_n\}$ be a sequence of independent random variables. For every n we have $E[X_n] = 0$ and $E[|X_n|] = 1$. Then $P(\liminf(X_n) < 0) > 0$.
The exercise consists on deciding whether that statement is true or false, giving a proof or finding a counterexample. I think it is true, but I'm struggling to find the proof. I have tried stating that $P(\liminf(X_n) < 0) = 0$ and finding a contradiction with the hypothesis, but I can't find a relation between $P(\liminf(X_n) < 0)$ and $E[X_n]$ or $E[|X_n|]$. The statement could also be wrong, but I can't find a counterexample either.
Any help would be appreciated, thank you beforehand!
The statement is false. Let $X_n$'s be independent random variables such that $X_n$ takes the value $-\frac {n^{2}} 2$ with probability $\frac 1 {n^{2}}$ and the value $\frac 1 {2(1-\frac 1 {n^{2}})}$ with probability $1-\frac 1 {n^{2}}$. Then $EX_n=0$, $E|X_n|=1$ and $\sum P(X_n=-\frac {n^{2}} 2) <\infty$. By Borel - Cantelli Lemma it follows that $X_n =-\frac {n^{2}} 2$ infinitely often with probability $0$. Hence there is probability $1$ that $X_n=\frac 1 {2(1-\frac 1 {n^{2}})}$ for all $n$ sufficiently large which makes $\lim \inf X_n \geq 0$ almost surely.