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We say that two random variables $X$ and $Y$ are said to be exchangeable iff $$\left(X,Y\right)\mathrel{\overset d =} \left(Y,X\right)$$ I want to prove that $X-Y$ is symmetric. I was proceeding in the following logic:

$$P\left(X-Y \leq z\right)=P(X \leq Y+z)=P(Y \leq X+z)=P(Y-X \leq z)$$

Hence, $X-Y\mathrel{\overset d =}Y-X$, and we have symmetry about $0$. But I think that my logic is circular. Can anyone verify this and give me the correct way to proceed?

Source of the problem : Theorem $6$, page $121$, An introduction to Probability and Statistics by Rohatgi, Saleh.

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There is nothing circular about it. But I think you would be more confident if you could see $(X, Y) $ changing to $(Y, X) $ in your calculations. Let $B_z=\{\ (a, b) \in \mathbb{ R}^2\mid a-b\leq z\ \} $. Now we have: $$P(X-Y\leq z) =P(\, (X, Y) \in B_z) =P(\,(Y, X) \in B_z) =P(Y-X\leq z) =P(-(X-Y) \leq z) $$

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