Does the difference between two symmetric r.v.'s also have a symmetric distribution? If I have two different symmetric (with respect to the median) distributions $X$ and $Y$, is the difference $X-Y$  also  a symmetric (with respect to the median) distribution?
 A: This is going to depend on the relationship between $x$ and $y$, here is a counter example where $x$ and $y$ are symmetric, but $x-y$ is not:
$$x=[-4, -2, 0, 2, 4]$$
$$y=[-1, -3, 0, 1, 3]$$
$$x-y = [-3, 1, 0, 1, 1]$$
So here the median of $x-y$ is not the same as the difference in the medians and $x-y$ is not symmetric.  
Edit
This may be clearer in @whuber's notation:
Consider the discrete uniform distribution where $x$ and $y$ are related such that you can only select one of the following pairs:
$$(x,y)=(-4,-1); (-2,-3); (0,0); (2,1); (4,3)$$
If you insist on thinking in a full joint distribution then consider the case where $x$ can take on any of the values $(-4, -2, 0, 2, 4)$ and $y$ can take the values $(-3, -1, 0, 1, 3)$ and the combination can take on any of the 25 pairs.  But the probability of the given pairs above is 16% each and all the other possible pairs have probability of 1% each.  The marginal distribution of $x$ will be discrete uniform which each value having 20% probability and therefore symmetric about the median of 0, the same is true for $y$.  Take a large sample from the joint distribution and look at just $x$ or just $y$ and you will see a uniform marginal distribution (symmetric), but take the difference $x-y$ and the result will not be symmetric. 
A: You'll need to assume independence between X and Y for this to hold in general. The result follows directly since the distribution of $X-Y$ is a convolution of symmetric functions, which is also symmetric.
A: Let $X \sim f(x)$ and $Y \sim g(y)$ be PDFs symmetric about medians $a$ and $b$ respectively. As long as $X$ and $Y$ are independent, the probability distribution of the difference $Z = X - Y$ is the convolution of $X$ and $-Y$, i.e.
$$
p(z) = \int_{-\infty}^\infty f(z + y) g(-y) dy,
$$
where $h(y) = g(-y)$ is simply the PDF over $-Y$ with median $-b.$
Intuitively, we would expect the result to be symmetric about $a - b$ so let's try that.
$$
\begin{split}
p(a - b - z) &= \int_{-\infty}^\infty f(a - b - z + y) g(-y) dy \\
&= \int_{-\infty}^\infty f(a + b + z - y) g(y - 2 b) dy \\
&= \int_{-\infty}^\infty f(a - b + z + v) g(-v) dv \\
&= p(a - b + z).
\end{split}
$$
In the second line I used both the symmetry of $f(x)$ about $a$ and of $g(-y)$ about $-b.$ In the third line, I used the substitution $v = 2 b - y$ in the integral. This proves that $p(z)$ is symmetric about $a - b$ if $f(x)$ is symmetric about $a$ and $g(y)$ is symmetric about $b.$
If $X$ and $Y$ were not independent, and $f$ and $g$ were simply marginal distributions, then we would need to know the joint distribution, $X,Y \sim h(x,y).$ Then, in the integral, we would have to replace $f(z + y) g(-y)$ with $h(z + y, -y).$ However, just because the marginal distributions are symmetric, that does not imply that the joint distribution is symmetric about each of its arguments. So you could not apply similar reasoning.
