# maximum of exponentials

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.

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Presumably the $X_i$ should be assumed independent. Can you quote memorylessness? Compute the distribution of $\epsilon_1$, the minimum. This is straightforward, anyway it has probably already been done in your course. The rest follows by memorylessness and induction. – André Nicolas Apr 17 '13 at 17:53
Yes, sorry I forgot to say that they are independent, so memoryless can be used – daniel Apr 17 '13 at 17:57
What do you mean by compute $\varepsilon_1$, the minimum? – daniel Apr 17 '13 at 18:05
Let random variable $\epsilon_1$ be the minimum of the $X_i$, so it is the lifetime of the first thing to die. The probability that $\epsilon_1 \gt t$ is the probability everybody is alive at time $t$, which is $e^{-n\lambda t}$. So $\epsilon_1$ is exponential parameter $n\lambda$. – André Nicolas Apr 17 '13 at 18:12
It is distribution of the first term in your sum. Now suppose first component to die has just done so. The additional lifetimes of the remaining $n-1$ components have exponential distribution parameter $\lambda$. The next minimum gives you $\epsilon_2$. – André Nicolas Apr 17 '13 at 18:18

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 $exp(1/\lambda)$.

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