$X$ and $Y$ are independent and follow $U(0,1)$. Show $P(f(X) > Y) = \int_0^1 f(x) dx$

Question

Let $X$ and $Y$ be two independent uniformly distributed r.v. on $[0,1]$, and $f$ is a continuous function from $[0,1]$ to $[0,1]$. Show that $P(f(X) > Y) = \int_0^1 f(x) dx$.

I tried to prove it by change of variable but failed. I can only reach the step $$ P(f(X) > Y) = \int_{\{(x,y): f(x) > y\} }I_{[0,1]}(x) I_{[0,1]}(y) dx dy $$ How can I proceed with the proof? Thanks for any help in advance.

Answer 1

$$ \int_{(x,y):f(x)>y} \mathbf{1}_{[0,1]}(x)\mathbf{1}_{[0,1]}(y)\,\mathrm dx\,\mathrm dy=\int_\mathbb{R}\int_\mathbb{R} \mathbf{1}_{[0,1]}(x)\mathbf{1}_{[0,f(x))}(y)\,\mathrm dy\,\mathrm dx $$

Answer 2

Notice that you can switch the integrals because the assumptions of Fubini-Tonnelli theorem are satisfied.

First integrate with respect to $y$: using the fact that $0\leqslant f(x)\leqslant 1$ will help to simplify it.

Answer 3

$$\displaystyle P(Y < f(X)) = \int_0^1 F_Y(f(x)) f_X(x) dx = \int_0^1 f(x) dx$$

$F_Y(.)$ represents the c.d.f of $Y$. And $f_X(.)$ represents the p.d.f of $X$. Can you fill in the gaps?