I am stuck with the following problem. Let the mean life time of a component be exponentially distributed with mean τ. If we switch on all of them at the same time, how is the waiting time until the first failure distributed? What is its mean? How is the waiting time until the last failure distributed?
Now, I know that the exponential distribution is special in the sense that it does not matter how much time has past, the porbability of failure in the next moment is always the same. However, I don't know if this is asked here and, if yes, how I would go into the problem.
Thank you all for your answers already.
EDIT: I believe I got the first part of the problem:
Let $X_1,X_2 \cdots X_N$ ~ $Exp(\lambda)$, then the cumulative distribution function is given by \begin{align} F_{X_i}(x) = P(X_i \leq x)=1-e^{-\lambda x_i} \end{align}
So if we define $Y = min\{X_1,X_2,\cdots X_N\}$, we can write the cumulative distribution function of Y as:
\begin{align} F_Y(y) &= P(Y\leq y) =\\ &= 1 - P(Y\geq y) =\\ &=1 - P(min\{X_1,X_2,\cdots X_N\} \geq y) =\\ &= 1 - P(X_1 \geq y) P(X_2\geq y) \cdots P(X_N\geq y) =\\ &=1 - e^{-\lambda y}e^{-\lambda y}\cdots e^{-\lambda y} = \\ &= 1 - e^{-n\lambda y} \end{align} which means that $Y$ ~ Exp ($n\lambda$). Still I don't see what the distribution of the maximum should be.
As you've already shown, the minimum has an $\text{Exp}(N\lambda)$ distribution, with mean $\frac{1}{N\lambda}$. The maximum is a little different: $$P(\max X_i\le x)=P(\forall i (X_i\le x))=\prod_iP(X_i\le x)=(1-e^{-\lambda x})^N.$$I don't think this has a nice name, although it does imply $\exp-\lambda \max X_i$ has a Beta distribution, as I'll leave you to show.
Although the moments of $\max X_i$ aren't easy to obtain, there's an elegant result for its cumulants. Define $H_{n,\,m}:=\sum_{k=1}^{n}k^{-m}$ so the $n$th Harmonic number $H_{n}=H_{n,\,1}$. An $\text{Exp}\left(\lambda\right)$ variable has cgf $-\ln\left(1-\frac{t}{\lambda}\right)$, so its cumulants are $\kappa_{m}=\left(m-1\right)!\lambda^{-m}$. To get $N$ failures we wait for the initial one, which takes a period $\sim\text{Exp}(\lambda)$; then we wait for the last of the other decays. Induction therefore gives $\kappa_{m}=\left(m-1\right)!H_{N,\,m}\lambda^{-m}$.