TITLE: probability of IID sum being positive
QUESTION [7 upvotes]: 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!

REPLY [9 votes]: No, this does not hold without the finite second-moment 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, $\mathbb P(S_n \geq 0) = \mathbb P(X \geq 0) \neq 1/2$.