Proof $P(X<Y)=1/2$ when $X$ and $Y$ i.i.d continuous (symmetry)

Question

I'd like to prove with the integral that $P(X < Y)=1/2$ when $X$ and $Y$ i.i.d continuous random variable (they are symmetry therefore).

I tried with convolution calling $Z=X-Y$ and $$P(Z<0)=\int\int f_X(z+y)f_Y(y)dydx$$ but then I wasn't able to reach a result.

Answer 1

$P(X<Y)=P(Y<X)$ because $(X,Y)$ and $(Y,X)$ are identically distributed. Also $P(X=Y)=0$ because the common distribution is continuous. Hence $P(X<Y)=P(Y<X)=\frac 1 2 (P(X<Y)+P(Y<X))=\frac 1 2$.

Note: $P(X=Y)=\int P(X=y) dF_Y(y)=0$ because $P(X=y)=0$ for each $y$.

$P(X<Y)=\int I_{x<y} dF_X(x)dF_Y(y)=\int I_{x<y}dF_Y(y) dF_X(x)=P(Y<X)$ where I have used Fubini's Theorem to interchange the integrals.