Symmetric Distribution of Random Variable

Question

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.

Answer 1

Prove $\phi_{X-Y}(t) \in \mathbb{R}$

Answer 2

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

Answer 3

$$ (X,Y) = \begin{cases} (a,a) & \text{with probability }p, \\ (a,b) & \text{with probability }q, \\ (b,a) & \text{with probability }r, \\ (b,b) & \text{with probability }s. \end{cases} $$ $$ p+q=\Pr(X=a)=\Pr(Y=a)=p+r\text{; therefore }q=r. $$ 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.)