Symmetric distribution around zero and median equal to zero Let $X$ be a random variable and suppose that it is symmetrically distributed around $0$. 
Is this equivalent to assume that $X$ has median equal to $0$?
If not, are the two assumptions completely unrelated or one implies the other?
 A: Since $X$ is symmetrically distributed about 0, its distribution satisfies $P(X) dX = P(-X) dX$.  But the probability that a value of $X$ selected from this distribution is positive is
$$
P(X > 0) = \int_0^\infty P(X) dX = \int_0^\infty P(-X) dX = -\int_0^{-\infty} P(\tilde{X}) d\tilde{X} = \int_{-\infty}^0 P(\tilde{X}) d\tilde{X} = P(X < 0).
$$
(In the third step, we have substituted $\tilde{X} = -X$.) Since the probability of getting a result above 0 is the same as getting an answer below 0, the median value of $X$ is 0.  A similar sort of argument can be applied to show that $\bar{X} = 0$ for such a distribution as well.
The reverse statement, however, is false;  it is easy to find a distribution for which the median is 0 but the distribution is not symmetric.  As an example, consider the distribution
$$
P(X)  = \begin{cases} 2 & -5/8 < X < -1/2 \\
1/2  & -1/2 < X < 1/2 \\
1 & 1/2 < X < 3/4
\end{cases}
$$
This corresponds to a probability of 1/4 to obtain a value between -5/8 & -1/2 (with uniform distribution in this range);  a probability of 1/2 to obtain a value between -1/2 & 1/2 (with uniform distribution in this range); and a probability of 1/4 to obtain a value between 1/2 & 3/4  (with uniform distribution in this range).  It should be fairly obvious that the median of this distribution is 0, but it is not symmetric.
A: $X$ is symmetric wrt $0$ if and only if $X$ and $-X$ have the same distribution so that: $$P(X\leq0)=P(-X\leq0)=P(X\geq0)\tag1$$
Also we have: $$P(X\leq0)+P(X\geq0)\geq1\tag2$$ and $(1)$ and $(2)$ together lead to: $$P(X\geq0)>\frac12\wedge P(X\leq0)\geq\frac12$$
Then by definition $0$ is a median of the distribution.
The converse is not true.
Let $Y$ be a random variable such that for every constant $c$ the distributions of $Y-c$ and $c-Y$ are distinct (there are plenty). 
The distribution of $Y$ will have a median $m$ (for every distribution a median exists).
Then $0$ will be a median of the distribution of $Y-m$, but $Y-m$ and $m-Y$ have distinct distribution (i.e. $Y-m$ is not symmetric wrt $0$).
A: By "equvalent" I assume you mean there's an if-and-only-if relation between the two (symmetry about 0 and median 0).
The forward implication is not automatically a given, since it depends on your exact definition of the median (consider a symmetric case where the density is zero on $(-t,t)$), so that $F(x)=\frac12$ in an interval. If you define the median to be the center of the interval then yes, but if for example, you define it on the left then no. Sometimes the median is defined to be the entire interval where $F(x)=\frac12$.
Counterexamples of the reverse implication are trivial -- take an $X$ with an asymmetric distribution and subtract its median (Let $Y=X-\tilde{\mu}_X$). The result has median $0$ and is asymmetric.
