Get ready Part 12

The joint probability distribution function of $x_i$ and $x_{i+1}$ (for $1 \le i \le n-1$ ) is $p_i(s,t) \; = \; \frac{n!}{(i-1)!(n-1-i)!}s^{i-1}(1-t)^{n-1-i} \hspace{2cm} 0 \le s < t \le 1$ which implies that the probability distribution function of $x_i$ (for $1 \le i \le n$ ) is $q_i(s) \; = \; \frac{n!}{(i-1)!(n-i)!}s^{i-1}(1-s)^{n-i} \hspace{2cm} 0 \le s \le 1$ If $1 \le i \le n-1$ then $\begin{aligned} E[x_if(x_{i+1})] & = \; \iint_{0 \le s < t \le 1}s f(t) p_i(s,t)\,ds\,dt \; = \; \frac{n!}{(i+1)(i-1)!(n-i-1)!}\int_0^1 t^{i+1}(1-t)^{n-i-1}f(t)\,dt \\ & = \; i\binom{n}{i+1}\int_0^1 t^{i+1}(1-t)^{n-i-1}f(t)\,dt \end{aligned}$ while $E[x_if(x_{i+1})] = 0$ if $i = 0,n$ . On the other hand, if $1 \le i \le n$ then $E[x_if(x_i)] \; = \; \int_0^1 tf(t)q_i(t)\,dt \; = \; i\binom{n}{i}\int_0^1 t^i(1-t)^{n-i}f(t)\,dt$ while $E[x_{n+1}f(x_{n+1})] = 0$ . Thus $\begin{aligned} E\left[\sum_{i=0}^n (x_{i+1}-x_i)f(x_{i+1})\right] & = \; \sum_{i=1}^n E[x_if(x_i)] - \sum_{i=0}^n E[x_if(x_{i+1})] \\ & = \; \sum_{i=1}^n i\binom{n}{i}\int_0^1 t^i(1-t)^{n-i}f(t)\,dt - \sum_{i=1}^{n-1}i\binom{n}{i+1}\int_0^1 t^{i+1}(1-t)^{n-i-1}f(t)\,dt \\ & = \; \sum_{i=1}^n i\binom{n}{i}\int_0^1 t^i(1-t)^{n-i}f(t)\,dt - \sum_{i=2}^{n}(i-1)\binom{n}{i}\int_0^1 t^{i}(1-t)^{n-i}f(t)\,dt \\ & = \; \sum_{i=1}^n \binom{n}{i}\int_0^1 t^i(1-t)^{n-i}f(t)\,dt \; = \; \int_0^1\big[1 - (1-t)^n\big]f(t)\,dt \end{aligned}$ so that $P(t) = 1 - (1-t)^n$ is a polynomial of degree $n$ with $0 \le P(t) \le 1$ for all $0 \le t \le 1$ .

It is an elegant consequence of this that $\lim_{n \to \infty}\,E\left[\sum_{i=0}^n (x_{i+1}-x_i)f(x_{i+1})\right] \; = \; \int_0^1 f(t)\,dt$

This is simply the definition of a weight function, since the problem has us consider a distribution.