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let $$(X_1, X_2, \dots, X_n)$$ be the order statistics of $n$ i.i.d. uniform random variables. that satisfies the following condition. $$0 < x_1 < x_2 < \dots < x_n < 1$$ now consider a continuous function $f : [0,1] \rightarrow \mathbb{R}$ we define the random variable $R$ to be: $$ R = \sum_{i = 0}^{n - 1}f(X_{i+1}) \times (X_{i+1} - X_i) \hspace{0.5cm}X_0 = 0 $$ I need to prove the expected value of R equals: $$ E[R] = \int_{0}^{1}f(t)(1-(1-t)^n)dt $$ What I've tried so far is that I tried to find the pdf of each $X_i$ using the following formula: $$ f_{X_k}(x_k) = \frac{n!}{(k-1)!(n-k)!}F_X^{k-1}(x)[1-F_X(x)]^{n-k}f_X(x) $$ where $f_{X}$ is the pdf of our Uniform random variable(not to be mistaken with the f function described above) and F is the CDF of the aforementioned random variable. Using the said pdf and CDF and the linearity of expected value seems to get me nowhere, and I'm stuck.

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Note that for $0\le i\le n-1$, $$ \mathsf{E}f(X_{i+1})(X_{i+1}-X_i)=\frac{1}{1+i}\mathsf{E}f(X_{i+1})X_{i+1} $$ Thus, \begin{align} \mathsf{E}R&= \sum_{i=0}^{n-1}\frac{1}{1+i}\mathsf{E}f(X_{i+1})X_{i+1} \\ &= \sum_{i=0}^{n-1}\frac{1}{1+i}\int_0^1 f(x)xf_{X_{i+1}}(x)\, dx \\ &=\int_0^1 f(x)\sum_{i=1}^n\binom{n}{i}x^{i}(1-x)^{n-i}\, dx \\ &=\int_0^1 f(x)(1-(1-x)^n)\, dx \end{align} because $\sum_{i=0}^n\binom{n}{i}x^{i}(1-x)^{n-i}=1$.


One can show the first equality by noticing that $(X_i,X_{i+1})\overset{d}{=}\big(U_{i}^{1/i}\cdots U_n^{1/n},U_{i+1}^{1/(i+1)}\cdots U_n^{1/n}\big)$, $1\le i\le n-1$, where $U_1,\ldots,U_n$ are i.i.d. $U[0,1]$ random variables and computing the expectation. That is, letting $Z\equiv U_{i+1}^{1/(i+1)}\cdots U_n^{1/n}$, \begin{align} \mathsf{E}f(X_{i+1})(X_{i+1}-X_i)&=\mathsf{E}f(Z)Z\big(1-U_i^{1/i}\big) \\ &=\mathsf{E}[f(Z)Z]\times \mathsf{E}\big[1-U_i^{1/i}\big] \\ &=\mathsf{E}[f(Z)Z]\times \frac{1}{1+i}. \end{align}

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    $\begingroup$ I don't understand how the first equality was derived, and how it was used to calculate the expected value of R. (I would really appreciate it if you could add parentheses around E so I could know the scope of it) $\endgroup$ Apr 11, 2022 at 21:03
  • $\begingroup$ Thank you so much for clarifying your answer, but I still don't see how we can write the kth order statistic as a product of nth roots of uniform distributions. I know that the kth order statistic sampled from an uniform distribution is a beta distribution itself, but I can't use that to proof what you have stated. I would appreciate it if you could prove that part, or cite some resources where I could read about it. $\endgroup$ Apr 11, 2022 at 22:20
  • $\begingroup$ @nothatcreative5 I learnt this trick from here (Chapter4, p.43). $\endgroup$
    – user140541
    Apr 11, 2022 at 22:33

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