Let $X_1,X_2,...$ be iid random variable with mean zero. If $X_1$ has second moment then by the CLT we have $P(X_1+X_2+...+X_n\geq 0)\rightarrow \frac{1}{2}$, as $n$ goes to infinity. I am curious about whether does this hold without the second moment assumption? I think this might be a well studied problem. Could anyone provide an answer or reference to me? Many thank!
No, this does not hold without the finite secondmoment assumption, in general.
Consider the Levy $\alpha$stable distributions, which will yield a whole family of examples. Using the parametrization, e.g., on the Wikipedia page, if $X$ is $\alpha$stable, for $1 < \alpha < 2$, then it has characteristic function $$ \varphi(t) = \mathbb E e^{i t X} = \exp(i t \mu  ct^{\alpha} (1  i \beta \mathrm{sgn}(t) \tan(\pi \alpha/2))) ~. $$
For the given range of $\alpha$, we also have that $\mathbb E X = \mu$ and $\mathbb E X^2 = \infty$.
Take $\mu = 0$, $c = \beta = 1$ and $\alpha = 3/2$, for simplicity and concreteness. Then, we get $$ \varphi(t) = \exp( t^{3/2} (1 + i \mathrm{sgn}(t))) ~. $$
Clearly, $\varphi(t) \neq \varphi(t)$, so, in particular, $X$ is not symmetric.
However, observe that if $X_1,\ldots,X_n$ are iid with this distribution and $S_n = X_1 + \cdots + X_n$, then $n^{2/3} S_n$ has the same characteristic function as $X$ since $(\varphi(n^{2/3}t))^n = \varphi(t)$, and, hence the same distribution.
Therefore, $\mathbb P(S_n \geq 0) = \mathbb P(X \geq 0) \neq 1/2$.

$\begingroup$ $X$ being asymmetric is not sufficient for $\mathbf P(X\ge 0)\ne \mathbf P(X\le 0)$, so there needs more work than symmetry to show the latter claim. $\endgroup$ – Hans Dec 3 '13 at 4:08
I suspect not. Let $(p^{(n)})$ be a sequence tending to 0 fast and for each $n$, let $X^{(n)}_j$ be an iid sequence of random variables taking the value $1/p^{(n)}$ with probability $p^{(n)}$ and 0 otherwise (so that $\mathbb EX^{(n)}_j=1$ for each $n$ and $j$).
Now let $Z_j=X^{(1)}_j/3\sum _{n=2}^\infty (1/2)^nX^{(n)}_j$. (This is chosen to make sure that $\mathbb EZ_j=0$.
Now if the $p^{(n)}$ are chosen to go to 0 rapidly enough, I claim that the $Z_j$ fail to have the property you're looking for. For example, let $N^{(n)}=10\uparrow 10\uparrow\ldots\uparrow10$ (i.e. a tower of height $n$) and $p^{(n)}=1/N^{(n)}$. Then for $N^{(n1)}\ll J\ll N^{(n)}$, the chances of having seen a contribution at level $n$ or higher is negligible, whereas it's likely that contributions at the previous level will have equilibriated. So the partial sum should be negative with very high probability if $n$ is odd and positive with very high probability if $n$ is even...
I think this is false without the second moment assumption. Construct $X_n$ of the form $$ X_n = 1+\sum_{k\ge 1} Y_n^k, $$ where $Y_n^k\ge 0$, $E[Y_n^k]=2^{k}$, $P(Y_n^k>0)=\epsilon_k$ and $(Y_n^{k+1}>0)\subset(Y_n^k>0)$. Choose the $\epsilon_k$ inductively as follows: Assume $\epsilon_1,\ldots,\epsilon_k$ have already been chosen, so that $$ X_n^k := 1+\sum_{j=1}^k Y_n^j $$ is already defined. Observe that $X_n^k$ has negative expectation, so that by the law of large numbers you can find an arbitrarily large integer $n_k$ such that $$ P\left(\frac{1}{n_k} \sum_{n=1}^{n_k} X_n^k < 0 \right) > 99/100. $$ Then choose $\epsilon_{k+1}$ so small that $$ P\left( Y_1^{k+1}=0,\ldots,Y_{n_k}^{k+1}=0 \right) > 99/100, $$ then $$ P\left(\frac{1}{n_k} \sum_{n=1}^{n_k} X_n < 0 \right) > 98/100. $$