Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 26 / 1

Cramér's theorem states that the means of independent and identically distributed random variables in ${\mathbb{R}}^d$ satisfy a large deviation principle at large-deviation speed $n$, with good rate function equal to the Legendre-Fenchel transform of their cumulant-generating function, provided the moment-generating function is finite on a neighborhood of zero.

Here the exponential distribution hasFor $x>0$, maximize $\theta x-\Lambda (\theta )$. Its derivative is $x-(\lambda -\theta {)}^{-1}$ and its second derivative is negative, so the maximizing parameter is ${\theta }_x=\lambda -x^{-1}$. Therefore the rate function of an exponential sample mean isFor $x\le 0$, letting $\theta \to -\infty$ in the same supremum gives $+\infty$, including the logarithmic divergence when $x=0$. The moment-generating function is finite near zero, so Cramér's theorem supplies the full large deviation principle.

The rate function is nonnegative, with its unique minimum at $\mu ={\lambda }^{-1}$; $I^{\prime \prime }(x)=x^{-2}>0$. It tends to infinity both as $x\downarrow 0$ and as $x\to \infty$. Its finite sublevel sets are closed bounded intervals contained in $(0,\infty )$, hence compact. Thus it is a good rate function.

For fixed $k$, independence and the product large-deviation principle give the vector $(Y_n(1),\dots ,Y_n(k))$ the good rate function ${\sum }_{j=1}^{k}I(y_j)$. Since the minimum is a continuous map, the contraction principle for large deviations givesIn every vector in this fibre, at least one coordinate equals $m$ and all others are at least $m$. The large-deviation rate of a minimum of independent copies therefore reduces the infimum toIf $0<m<\mu ={\lambda }^{-1}$, the other coordinates can equal $\mu$ and cost zero. If $m\ge \mu$, the rate function is increasing on $[\mu ,\infty )$, so every coordinate costs at least $I(m)$ and equality is achieved when all coordinates equal $m$. For $m\le 0$ the cost is infinite. ThusAt the threshold both expressions are zero. Goodness follows from the contraction principle for large deviations. A small minimum requires only one atypically small mean; a large minimum requires all $k$ means to be atypically large, which explains the two costs.

Varadhan's lemma states that if $U_n$ has a large deviation principle with good rate function $K$ and large-deviation speed $a_n$, then for every bounded continuous real function $h$,For a continuous function unbounded above, the same conclusion holds under the exponential tail extension of Varadhan's lemma, for example when the weighted contribution from $h>M$ is superexponentially negligible as $M\to \infty$.

The scalar clipping map $g(m)=\min (b,\max (a,m))$ is continuous. The rate function under scalar clipping has endpoint costs ${\inf }_{m\le a}J(m)$ and ${\inf }_{m\ge b}J(m)$. Since $a<\mu <b$ and the costs decrease towards $\mu$ from the left and increase to its right, both infima are attained at the corresponding endpoints. Thus the good rate function of $Z_n$ equals $J(z)$ on $[a,b]$ and is infinite outside. The function $\log z$ is bounded and continuous on that compact interval, soOn $[a,\mu ]$, the derivative of $\log z-I(z)$ is $2/z-\lambda >0$, so the maximum there is $\log \mu$. On $[\mu ,b]$, the derivative is $(k+1)/z-k\lambda$, with negative second derivative. Its unconstrained maximizer isIt lies in $[\mu ,b]$ whenever $k\ge (\lambda b-1{)}^{-1}$. The right-hand objective increases initially at $\mu$, so this maximizer dominates the whole left-hand interval. Substitution gives the limit for high moments of a clipped minimum of exponential sample meansThe moment here is the expectation of the $n$th power. If the printed notation were instead read as $(\mathbb{E}{Z}_n{)}^n$, its logarithmic limit would be $\log \mu =-\log \lambda$, since boundedness and convergence in probability give $\mathbb{E}{Z}_n\to \mu$.