Vtyjkyuk

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 $F_epsilon_1-epsilon_0(0)=F_epsilon_2-epsilon_0(0)=frac12$.
Does this imply that $F_epsilon_1-epsilon_2(0)=frac12$?

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?

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, $F_epsilon_1-epsilon_2(0)=frac12$). I would like your confirmation on this.

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 $epsilon_1-epsilon_2$?

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. $F_epsilon_0(0)=F_epsilon_1(0)=F_epsilon_2(0)=F_epsilon_1-epsilon_0(0)=F_epsilon_2-epsilon_0(0)=frac12$ but $F_epsilon_1-epsilon_2(0)=frac13 not = frac12$




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)