Symmetric Distribution of Random Variable Prove: Let $X$ and $Y$ be random variables with the same distribution. If $X$ and $Y$ take only two values​​, then $X - Y$ are symmetrically distributed around zero. 
Note: 1 - You can use characteristic function; 
      2 - Being symmetric means $F_{X}(x)=F_{-X}(x)$.
  3 - Nothing is talked about X and Y are independent!

  4 - Exercise 6, letter b, page 257 of the book Probability of Barry James.

 A: Without loss of generality, assume that $X$ and $Y$ are Bernoulli random variables with the same parameter $p = P\{X=1\} = P\{Y=1\}$.  Thus,
$X-Y$ takes on values $-1, 0, 1$.  Then, we have that
$$\begin{align}
P\{X=1\} = p &= P\{X=1, Y=1\} + P\{X=1, Y=0\}\tag{1}\\
P\{Y=1\} = p &= P\{X=1, Y=1\} + P\{X=0, Y=1\}\tag{2}.
\end{align}$$
From $(1)$ and $(2)$ we see that 

$P\{X=1, Y=0\} = P\{X-Y = 1\}$
  equals $P\{X=0, Y=1\} = P\{X-Y=-1\}$.

A: $$
(X,Y) = \begin{cases}
(a,a) & \text{with probability }p, \\
(a,b) & \text{with probability }q, \\
(b,a) & \text{with probability }r, \\
(b,b) & \text{with probability }s.
\end{cases}
$$
$$
p+q=\Pr(X=a)=\Pr(Y=a)=p+r\text{; therefore }q=r.
$$
Therefore
$$
X-Y =\begin{cases}
0 & \text{with probability }p+s, \\
a-b & \text{with probability }q, \\
b-a & \text{with probability }r.
\end{cases}
$$
Now use the fact that $q=r$.
(Off hand I don't know a simpler way that uses characteristic functions, so I wouldn't use those here.)
A: Prove $\phi_{X-Y}(t) \in \mathbb{R}$
