How does $\max\{X_1,\dots, X_n\} = \varepsilon_1 + \dots + \varepsilon_n$ if $X_i$'s are exponential? I am really having difficulties to prove the following: consider $X_1,\dots, X_n$ all exponentially distributed with rate $\lambda$ (i.e. $X_i \sim \exp( 1/\lambda)$). Then argue that we can write
$$\max\{X_1,\dots, X_n\} = \varepsilon_1 + \dots + \varepsilon_n$$
where $\varepsilon_1,\dots, \varepsilon_n$ are independent exponentials with respective rates $n\lambda, (n-1)\lambda, \dots, \lambda$.
The hint I get is: Interprete $X_i$ as the lifetime of component $i$ and $\varepsilon_i$ as the time between $i-1$ and the $i$-th failure.
Thank you very much for any help you could offer :)
 A: As the hint suggests, define the random variable $\epsilon_1$ as $\min(X_1,X_2,\dots,X_n)$. For $2\le i \le n$, let $\epsilon_{i}$ be the waiting time between the $(i-1)$-th failure and the $i$-th failure.
Then $\max(X_1,X_2,\dots,X_n)=\epsilon_1+\epsilon_2+\cdots +\epsilon_n$. Assume that the $X_i$ are independent. (We need this condition.) By the memorylessness of the exponential, the $\epsilon_i$ are independent.   
Note that $\epsilon_1=\min(X_1,X_2,\dots,X_n)$. It is a standard and easily verified fact that $\epsilon_1$ has exponential distribution with parameter $n\lambda$.  
The additional lifetimes $Y_1, \dots, Y_{n-1}$ of the $n-1$ survivors are independent and have exponential distribution with parameter $\lambda$. The random variable $\epsilon_2$ is $\min(Y_1,\dots,Y_{n-1})$. By the same argument as the one above, $\epsilon_2$ has exponential distribution with parameter $(n-1)\lambda$. Now let $Z_1,\dots,Z_{n-2}$ be the additional lifetimes of the $n-2$ survivors. Continue. 
One can prove the result more formally, by induction on $n$. The basic idea does not change. 
Remark: Usually, when one says that a random variable has exponential distribution with rate (parameter) $\lambda$, this means that the density is $\lambda e^{-\lambda t}$ for $t\gt 0$. So towards the beginning of the post, we want something like $exp(\lambda)$, not $exp(1/\lambda)$. 
