$X_1, \ldots , X_n$, $n \ge 4$ are independent random variables with exponential distribution: $f\left(x\right) = \mathrm{e}^{-x}, \ x\ge 0$. We define $$R= \max \left( X_1, \ldots , X_n\right) – \min \left( X_1, \ldots , X_n\right)$$
Calculate $\mathbb{E}R$.
So I know that: $$\mathbb{E}R =\mathbb{E}\left( \max \left( X_1, \ldots , X_n\right) \right)- \mathbb{E}\left(\min \left( X_1, \ldots , X_n\right)\right)$$
And I can calculate
$$\mathbb{E}\left(\min \left( X_1, \ldots , X_n\right)\right) = \int\limits_{0}^{\infty}\left(1-F_{min}\left(x\right)\right) \mathrm{dx}=\int\limits_{0}^{\infty}\left(\mathrm{e}^{-nx}\mathrm{dx} \right) = \frac{1}{n}$$.
The problem is to calculate:
$$\mathbb{E}\left(\max \left( X_1, \ldots , X_n\right)\right) = \int\limits_{0}^{\infty}x \cdot n\cdot \mathrm{e}^{-x}\left( 1-\mathrm{e}^{-x}\right)^{n-1} \mathrm{dx} = \ldots$$
I don't know how to calculate the above integral.
Best Answer
You can go for calculating another integral:
$$\begin{aligned}\mathbb{E}\max\left(X_{1},\dots,X_{n}\right) & =\int_{0}^{\infty}P\left(\max\left(X_{1},\dots,X_{n}\right)>x\right)dx\\ & =\int_{0}^{\infty}1-P\left(\max\left(X_{1},\dots,X_{n}\right)\leq x\right)dx\\ & =\int_{0}^{\infty}1-\left(1-e^{-x}\right)^{n}dx\\ & =\int_{0}^{\infty}\sum_{k=1}^{n}\binom{n}{k}\left(-1\right)^{k-1}e^{-kx}dx\\ & =\sum_{k=1}^{n}\binom{n}{k}\left(-1\right)^{k-1}\int_{0}^{\infty}e^{-kx}dx\\ & =\sum_{k=1}^{n}\binom{n}{k}\left(-1\right)^{k-1}\left[-\frac{e^{-kx}}{k}\right]_{0}^{\infty}\\ & =\sum_{k=1}^{n}\binom{n}{k}\left(-1\right)^{k-1}\frac{1}{k} \end{aligned} $$
There might be a closed form for it, but I haven't found it yet.
Edit:
According to the comment of @RScrlli the outcome can be proved to equal harmonic number: $$H_n=\sum_{k=1}^n\frac1{k}$$
This makes me suspect that there is a way to find it as the expectation of:$$X_{(n)}=X_{(1)}+(X_{(2)}-X_{(1)})+\cdots+(X_{(n)}-X_{(n-1)})$$