$\mathbb P(X<Y)=\mathbb P(X>Y)$ for i.i.d. rvs: formal conceptual proof

Question

Let $X,Y$ be i.i.d. random variables. Intuitively it's obvious (in fact I'd never thought about proving it until some minutes ago) that $$\mathbb P(X<Y)=\mathbb P(X>Y)$$ And this can be proved formally by computing the probability via Fubini's theorem (independence is invoked to construct a product measure, as usual): \begin{align} \mathbb P(X>Y)&=\int\mathbf 1_{X>Y}~d\mathbb P\\ &=\iint\mathbf 1_{x>y}~dF(x)dF(y)\\ &=\int (1-F(y))dF(y)\\ &=\int (1-F(x))dF(x)\\ &=\mathbb P(Y>X) \end{align} But the proof is not satisfying for me since it involves too many tools and is based on non-direct computation. I have the feeling that a nice proof of this "simple" fact should be almost conceptual and computation-free. But unfortunately such a proof found no way into my mind. Is there a way that is more elegant, i.e., conceptual, to attack the problem? Am I missing something obvious?

Answer 1

Since $X$ and $Y$ are i.i.d. one has $$ P_{(X,Y)}=P_X\otimes P_Y=P_{(Y,X)}. $$ In particular, if $A=\{(u,v)\in\mathbb{R}^2\mid u<v\}$, then $$ P(X<Y)=P_{(X,Y)}(A)=P_{(Y,X)}(A)=P(Y<X). $$