Difference of i.i.d. is symmetric

Question

I want to prove that if $X, Y$ are independent and identically distributed, then $X-Y$ has a symmetric distribution.

I started with characteristic functions, knowing that if $X-Y$ and $Y-X$ would have the same characteristic function, then they would have the same distribution.

Hence I got: $h_{X-Y}(t) = h_X(t) \overline{h_Y(t)}$ and $h_{Y-X}(t) =\overline{h_X(t)}h_Y(t)$. However, if either of them would have complex and not real values, then those two functions wouldn't necessarily be equal. How do I know that they in fact are the same?

Answer 1

You're essentially there: if $h_X(t)$ and $h_Y(t)$ are the characteristic functions of $X$ and $Y$, then $h_X(t)=h_Y(t)$ since $X$ and $Y$ have the same distribution. Therefore $$ h_{X-Y}(t)=h_X(t)\overline{h_Y(t)}=|h_X(t)|^2=\overline{h_X(t)}h_Y(t)=h_{Y-X}(t) $$

Answer 2

Why not use exchangeability? $X$ and $Y$ are iid, therefore exchangeable, so the joint distribution of $(X,Y)$ is the same as the joint distribution of $(Y,X)$. Consequently, the same function $g(a,b) = a-b$ applied to each joint distribution must also result in the same distribution, hence the distribution of $X - Y$ is symmetric about $0$: $$\Pr[X - Y \le t] = \Pr[Y - X \le t].$$