Median zero and symmetry of the cdf of the difference of two random variables

Question

In the notation below, let $F_X$ denote the cumulative distribution function (cdf) of a random variable $X$.

Consider the following random variables $\epsilon_0,\epsilon_1,\epsilon_2$. I have two doubts to clarify:

1) Assume that $F_{\epsilon_1-\epsilon_0}(0)=F_{\epsilon_2-\epsilon_0}(0)=\frac{1}{2}$. Does this imply that $F_{\epsilon_1-\epsilon_2}(0)=\frac{1}{2}$?

2) Assume that $F_{\epsilon_1-\epsilon_0}$ and $F_{\epsilon_2-\epsilon_0}$ are symmetric around zero. Does this imply that $F_{\epsilon_1-\epsilon_2}$ is symmetric around zero?

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My thoughts: I think that without further restrictions the answer to both questions is "no". I also think that if $\epsilon_1$ and $\epsilon_2$ are independent and identically distributed then $F_{\epsilon_1-\epsilon_2}$ is symmetric around zero (and, hence, $F_{\epsilon_1-\epsilon_2}(0)=\frac{1}{2}$). I would like your confirmation on this.

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Additional question: can you suggest other sets of sufficient restrictions on $\epsilon_1$, $\epsilon_2$ ensuring median zero and/or symmetry of the cdf of $\epsilon_1-\epsilon_2$?

Answer 1

For counter-examples, try the following combinations where each row has probability $\frac16$:

   e0  e1  e2  e1-e0 e2-e0 e1-e2 
    3   4   2    1    -1     2
    3   4   2    1    -1     2
    3   1   5   -2     2    -4
   -3  -1  -2    2     1     1
   -3  -4  -2   -1     1    -2 
   -3  -4  -5   -1    -2     1

This has

1. $F_{\epsilon_0}(0)=F_{\epsilon_1}(0)=F_{\epsilon_2}(0)=F_{\epsilon_1-\epsilon_0}(0)=F_{\epsilon_2-\epsilon_0}(0)=\frac{1}{2}$ but $F_{\epsilon_1-\epsilon_2}(0)=\frac{1}{3} \not = \frac12$

2. $F_{\epsilon_0}, F_{\epsilon_1}, F_{\epsilon_2}, F_{\epsilon_1-\epsilon_0}$ and $F_{\epsilon_2-\epsilon_0}$ are symmetric around zero, but $F_{\epsilon_1-\epsilon_2}$ is not symmetric about anything

For your additional question, as you say, having $\epsilon_1,\epsilon_2$ independent and identically distributed (though not necessarily symmetric) would be sufficient, as would having $\epsilon_1,\epsilon_2$ independent and symmetric about the same value (but not necessarily identically distributed)