Let $X$ be a real-valued random variable satisfying
- $\mathsf E X = 0$ and
- $\mathsf P(X\le 0)\ge\frac12$.
In other words, $X$ is a (centralised) random variable whose median is below its mean. For instance, the distribution of wealth is (up to a constant) such a distribution: The median of income/net worth/etc. is always below the mean, since the mean is dragged up by a small number of very wealthy/high-income/etc. individuals.
My question. Suppose that $X_1, X_2, \dots, X_n$ for some $n\in\mathbb N$ are all independent and identically distributed like $X$. Let $S_n\overset{\text{Def.}}=X_1+\dots+X_n$. Is it true that
$$\mathsf P(S_n \le 0)\ge \frac12$$ ? In other words, if we take an "unbiased sample" of $X$ with size $n$, is the median of that sample still below its mean? (Or again in other words: If we take a bunch of "unbiased samples" of $X$ of size $n$, would the average "observation" be higher than the median "observation" ?)
Remark 1. By inductive reasoning, it would be enough to prove that if $Y$ is indepedent of $X$ and identically distributed like $X$, then
$$\mathsf P(X+Y\le 0)\ge \frac12.$$
Remark 2. There is a very easy lower bound:
$$\mathsf P(X+Y\le 0)\ge\mathsf P(X\le 0\cap Y\le 0)=\mathsf P(X\le0)\mathsf P(Y\le 0)\ge\frac14,$$
however, I want to achieve $\frac12$, not $\frac14$.
Remark 3. If $X$ doesn't satisfy $\mathsf EX=0$, then the answer is "no". For instance, if $\mathsf P(X=-1)=\mathsf P(X=2)=\frac12$, then
$$\mathsf P(X\le0)=\frac12,$$
but
$$\mathsf P(X+Y\le 0)=\frac14.$$
How about this modification of the example in Remark 3?
If $\mathsf P(X=-1)=\frac23$, $\mathsf P(X=2)=\frac13$, then
$$\mathsf P(X\le0)=\frac23,$$
but
$$\mathsf P(X+Y\le 0)=\frac49.$$