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 :)

  • 2
    $\begingroup$ 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. $\endgroup$ – André Nicolas Apr 17 '13 at 17:53
  • $\begingroup$ Yes, sorry I forgot to say that they are independent, so memoryless can be used $\endgroup$ – daniel Apr 17 '13 at 17:57
  • $\begingroup$ What do you mean by compute $\varepsilon_1$, the minimum? $\endgroup$ – daniel Apr 17 '13 at 18:05
  • $\begingroup$ 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 $n\lambda$. $\endgroup$ – André Nicolas Apr 17 '13 at 18:12
  • 1
    $\begingroup$ 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 $\epsilon_2$. $\endgroup$ – 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 $exp(1/\lambda)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.