Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 101 / 1 / c

The exponential distribution has no atoms, so all sample values are distinct almost surely, and $M_i$ is the $i$th order statistic. Set $M_0=0$ and $D_i=M_i-M_{i-1}$. Part (b) gives $D_1\sim \mathrm{Exp}(n\lambda )$.

To justify induction and the required independence, let $J$ be the label attaining the minimum. Conditional on its time and label, the remaining clocks are fresh independent rate-$\lambda$ exponential random variables, by the memoryless property. This conditional claim can be checked directly: for a specified label $j$, first ringing time $u$, and residual lifetimes $y_k>0$ for $k\ne j$, the joint probability density isThus the first minimum, its uniform winning label and the surviving residual clocks are independent. Repeat this argument after each minimum. At the $i$th step there are $n-i+1$ clocks, so the independent exponential order-statistic spacings satisfySince $Y_k/k$ has rate $k\lambda$ whenever $Y_k\sim \mathrm{Exp}(\lambda )$, summing the gaps provesIn particular the maximum is $M^{\ast }=M_n$. By variance additivity for independent random variables and the Laplace transform of an exponential distribution, its requested quantities areThe product follows from independence of the gaps, not merely from their marginal probability distributions.