compute $\int _0^1 \cdots \int _0^1F_n(x_1,\cdots, x_n)dx_1\cdots dx_n$

Question

Let $f:\mathbb{R}\to \mathbb{R}$ be a $C^\infty$ strictly increasing function and $F_n(x_1,\cdots, x_n):=n(1-\max_{1\leq k \leq n}{\{f(x_k)\}})$ Then, compute $\int _0^1 \cdots \int _0^1F_n(x_1,\cdots, x_n)dx_1\cdots dx_n$

I tried compute that $\sum \int _{x_{i_1}<\cdots <x_{i_n}}F_n$ but I couldn't.

Answer 1

Convert the integration domain to $0 \leqslant x_1 \leqslant \ldots \leqslant x_n \leqslant 1$ and integrate by $x_1, \ldots, x_{n - 1}$. Given the extra multiplier $n$ in your $F_n$, you should get $$n^2 \int_0^1 x^{n - 1} \big(1 - f(x)\big)\ dx.$$

Answer 2

$\newcommand{\E}{\mathbb E}\newcommand{\PM}{\mathbb P}$A probabilistic approach

Let $U_1,...,U_n \sim \mathcal U[0,1]$ i.i.d. then you are interested in: \begin{align} I_n:=n\E[1-\max\{ f(U_1),...,f(U_n)\}] = n -n\E[\max\{ f(U_1),...,f(U_n)\}] \end{align} One knows that for $f(0)\leq x\leq f(1)$: \begin{align} \PM(\max\{ f(U_1),...,f(U_n)\}\leq x)=\PM(f(U_1)\leq x)^n = \PM(U_1 \leq f^{-1}(x) )^n = \left(f^{-1}(x)\right)^n \end{align} Hence: \begin{align} n\E[\max\{ f(U_1),...,f(U_n)\}] = n^2\int^{f(1)}_{f(0)} x\cdot \left(f^{-1}(x)\right)^{n-1} \cdot \frac{1}{f'(f^{-1}(x))}\,dx \end{align} Set $u=f^{-1}(x)$ so that we get: \begin{align} n\E[\max\{ f(U_1),...,f(U_n)\}] = n^2 \int^1_0 f(u) u^{n-1}\,du \end{align} Hence: \begin{align} I_n = n-n^2 \int^1_0 f(u) u^{n-1}\,du \end{align}

Oh, one last remark, where did we use the fact that $f$ is strictly increasing?