Does convergence in probability of A_n to 0 result $\lim_{n\to\infty}\mathbb{P}(A_n\geq0)=1$?

Question

Assume $A_1, A_2, \cdots$ is an i.i.d sequence of random variables with finite mean and variance. If $A_n$ converges to zero in probability, i.e., $\forall \gamma>0: \lim\limits_{n\to \infty}\mathbb{P}\left(|A_n|\geq\gamma\right)=0$, is the following true? $$\lim\limits_{n\to \infty}\mathbb{P}\left(A_n\geq0\right)=1$$

I think the answer is no and there should be counter examples. Can you help find that?

Answer 1

EDIT: As others pointed out below, my example doesn't meet the iid condition. Sorry about that oversight - looking for another counterexample or a proof.

How about $A_n=-\frac{1}{n}$? Then for any $\gamma>0$, $|A_n|<\gamma$ eventually, and $A_n<0$ for all $n$.

Answer 2

As others have mentioned, iid means that $\mathbb{P}\left(|A_n|\geq\gamma\right)$ is independent of $n$. Therefore $$ \forall \gamma>0: \lim_{n\to \infty}\mathbb{P}\left(|A_n|\geq\gamma\right)=\mathbb{P}\left(|A_1|\geq\gamma\right) $$ and this is $=0$ if $$ \forall \gamma>0: \mathbb{P}\left(|A_1|\geq\gamma\right)=0. $$ This only works for the degenerate distribution with $\mathbb{P}\left(A_1=0\right)=1$.

In that case, $$ \lim_{n\to \infty}\mathbb{P}\left(A_n\geq0\right)=\mathbb{P}\left(A_1\geq0\right)=1. $$