Why $\epsilon (X – X’)$ and $(X – X’)$ have exactly the same distribution

probability

Let $$\epsilon \in \{-1, 1\}$$ be a Rademacher random variable,
$$X$$ be a random variable and $$X'$$ be an independent copy of $$X$$.
Why is $$\epsilon (X – X')$$ and $$(X – X')$$ have the same distribution?

Given that $$X-X'$$ is symmetric around $$0$$, this is intuitively correct, but how to prove it?

You probably need to add the condition that $$\epsilon$$ and $$X$$ are independent, otherwise the conclusion may not hold. For example, if $$X, X' \text{ i.i.d. } \sim N(0, 1)$$, $$\epsilon = 2I_{\{X - X' > ~0\}} - 1$$. Since $$X - X' \sim N(0, 2)$$, on one hand, $$P(X - X' \leq 1) = \Phi(1/\sqrt{2})$$. On the other hand, \begin{align*} P(\epsilon(X - X') \leq 1) &= P(X - X' \leq 1, X - X' > 0) + P(X - X' \geq -1, X - X' \leq 0) \\ &= P(0 < X - X' \leq 1) + P(-1 \leq X - X' \leq 0) \\ &= \Phi(1/\sqrt{2}) - \frac{1}{2} + \frac{1}{2} - \Phi(-1/\sqrt{2}) \\ &= 2\Phi(1/\sqrt{2}) - 1 \neq \Phi(1/\sqrt{2}). \end{align*}
When $$\epsilon$$ and $$X$$ are assumed independent, the conclusion indeed holds: as for any $$x \in \mathbb{R}$$, \begin{align*} P(\epsilon(X - X') \leq x) &= P(X - X' \leq x, \epsilon = 1) + P(X - X' \geq -x, \epsilon = -1) \\ &= P(X - X' \leq x)P(\epsilon = 1) + P(X - X' \geq -x)P(\epsilon = -1) \\ &= \frac{1}{2}P(X - X' \leq x) + \frac{1}{2}P(X - X' \geq -x) \\ &= P(X - X' \leq x). \end{align*} The last equality holds because if $$F$$ is the common distribution function of $$X$$ and $$X'$$, then \begin{align*} & P(X - X' \leq x) = \int_{\mathbb{R}}P(X - t \leq x)~\mathrm dF(t) = \int_{\mathbb{R}}F(x + t)~\mathrm dF(t), \\ & P(X - X' \geq -x) = \int_{\mathbb{R}}P(t - X' \geq -x)~\mathrm dF(t) = \int_{\mathbb{R}}F(x + t)~\mathrm dF(t). \\ \end{align*}