Connection between max and min of symmetric random variables I bumped into a problem and I wonder if the following statement is true:
Let $X_1, X_2, ..., X_n$ be symmetric random variables (possibly dependent) centered around zero (i.e. $\mathbb{E}(X_i) = 0$ for $i\in\{1,2,...,n\}$). Then
\begin{equation}
\mathbb{P}\left(max_{i\in\{1,2,...,n\}} X_i < 0\right) \stackrel{?}{=} \mathbb{P}\left(min_{i\in\{1,2,...,n\}} X_i > 0\right).
\end{equation}
In case of independence the proof can be done by using set operations, superlevel and sublevel sets: $\mathbb{P}\left(max_{i\in\{1,2,...,n\}} X_i < 0\right) = \mathbb{P}\left(\bigcap_{i\in\{1,2,...,n\}} \{X_i < 0\}\right)  = \prod_{i\in\{1,2,...,n\}}\mathbb{P}\left(X_i < 0\right) = \prod_{i\in\{1,2,...,n\}}\mathbb{P}\left(X_i > 0\right) = \mathbb{P}\left(\bigcap_{i\in\{1,2,...,n\}} \{X_i > 0\}\right) = \mathbb{P}\left(min_{i\in\{1,2,...,n\}} X_i > 0\right).$
Can this argument be modified, in order to prove the general case stated above?
 A: The claim isn't even true for a single variable. Let the sample space be $\{0, 1\}$ with $P(0) = 1/3$, $P(1) = 2/3$ and let $X_1(0) = -2$, $X_1(1) = 1$. Then $\mathbb{E}[X_1] = 0$ but $P(X_1 > 0) \neq P(X_1 < 0)$.
Edit: because I slightly misread the question, I originally gave a counter example which didn't actually work. I have removed it and replaced it.
Let's consider a stronger version of "symmetric"; that for each $i$, the random variable $X_i$ hasn't the same probability distribution as $-X_i$ does. This actually still isn't enough. For consider random variables $X_1, X_2, X_3$ defined as follows:
There is a 1/3 chance that $X_1 = X_2 = X_3 = 1$. For the other $2/3$ of the time, independently assign each $X_i$ to $0$ or $-1$ with equal probability.
Formally, the sample space is $\{(1,1,1)\} \cup \{0, -1\}^3$ with probability distribution $P(x) = 1/3$ if $x = (1,1,1)$ and $1/12$ otherwise. The variables $X_1, X_2, X_3$ are the projection functions.
Clearly, each $X_i$ has a $1/3$ chance of being each of $1$, $0$, and $-1$ and is therefore symmetric under this stronger definition. The probability of the maximum value being negative is only $\frac{2}{3} (\frac{1}{2})^3 = \frac{1}{12}$ since this only occurs when $X_1 = X_2 = X_3 = -1$, while the probability of the minimum being positive is $\frac{1}{3}$.
