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

For the central limit theorem, let $X_1,X_2,\dots$ be independent and identically distributed random variables with expectation $\mu$ and finite positive variance ${\sigma }^2$. ThenThus the centered, normalized sum has convergence in distribution to the standard normal distribution. The positive-variance assumption is needed for this normalization; if ${\sigma }^2=0$, each summand equals $\mu$ almost surely and the centered sum is identically zero.

Here is a characteristic function proof using only the finite second moment. Put $U=(X_1-\mu )/\sigma$, so $\mathbb{E}U=0$ and $\mathbb{E}{U}^2=1$. The elementary integral remainder identity gives $|e^{iz}-1-iz|\le z^2/2$ for real $z$. ConsequentlyBy the dominated convergence theorem, the characteristic function satisfies ${\phi }_U(t)=1-t^2/2+o(t^2)$ as $t\to 0$. For a fixed real $t$, independence therefore givesThe allowed complex-number limit makes this tend to $e^{-t^2/2}$, the characteristic function of the standard normal distribution. It is continuous at zero, so the Lévy continuity theorem proves the asserted convergence in distribution. No third moment or boundedness of the summands was used.

For $t\ge 0$, independence factors the survival function of the minimum:The minimum is nonnegative, and this is exactly the survival function of an exponential distribution. HenceThe parameter here is the rate, rather than the mean. The calculation is the simplest example of competing exponential clocks.

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.