Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 36 / 1

Write $p=1+\epsilon$ and assume $\lambda >0$. The large-deviation speed is $a_N=N^p$, and the answer isHere and below a large deviation principle at speed $a_N$ means the open-set lower bound $\mathrm{liminf}{a}_N^{-1}\log \mathbb{P}(X_N\in G)\ge -{\inf }_{G}I$ and the closed-set upper bound $\mathrm{limsup}{a}_N^{-1}\log \mathbb{P}(X_N\in F)\le -{\inf }_{F}I$. The displayed rate function is nontrivial and is a good rate function.

For $0<u<v$, the exponential distribution givesShrinking an interval about any $x>0$ supplies the local lower bound $-\lambda x$; neighborhoods of zero have probability tending to one. Every open set intersecting the nonnegative half-line contains one of these neighborhoods, giving the open-set bound. If a closed set $F$ contains zero, its upper bound is just $\mathbb{P}(\cdot )\le 1$. If it meets $[0,\infty )$ but excludes zero, closedness gives $b=\inf (F\cap [0,\infty ))>0$, and $\mathbb{P}(A/a_N\in F)\le e^{-\lambda a_{N}b}$. A closed set disjoint from the support has probability zero. This proves the full large deviation principle, including the boundary at zero.

The fixed exponential tail explains the speed: an excursion of size $x{N}^{1+\epsilon }$ costs $e^{-\lambda x{N}^{1+\epsilon }}$. The full Gärtner–Ellis theorem is unnecessary here; the limiting cumulant-generating function is flat on $\theta <\lambda$ and infinite at $\theta \ge \lambda$, so it fails the theorem's essential smoothness hypothesis.

Assume ${\sigma }^2>0$; otherwise the scaled sum is deterministic and no nontrivial rate function is possible. The sum has normal distribution $B_N\sim N(N\mu ,N{\sigma }^2)$. For $X_N=B_N/N^{1+\epsilon }$, use the faster large-deviation speed $b_N=N^{1+2\epsilon }$. The moment-generating function of a normal distribution yieldsThe limiting cumulant-generating function is finite and differentiable on all of $\mathbb{R}$, hence satisfies the Gärtner–Ellis theorem. Its Legendre-Fenchel transform is obtained by maximizing $\theta x-{\sigma }^2{\theta }^2/2$, whose maximizer is $\theta =x/{\sigma }^2$. ThusThe drift disappears because the scaled mean is $\mu /N^{\epsilon }\to 0$. The rate function is nontrivial and good. The corrected scaled cumulant-generating function used here contains an expectation; the corresponding formula in the paper's reference material omits it.

Use the large-deviation speed $a_N=N^{1+\epsilon }$ from part (a). For every $\eta >0$, the Chernoff bound for a normal distribution gives, for all sufficiently large $N$,Consequentlybecause the exponent has order $N^{1+2\epsilon }$, exceeding $a_N$ by a factor $N^{\epsilon }$. Thus $(A+B_N)/a_N$ and $A/a_N$ satisfy exponential equivalence. This conclusion does not require independence between $A$ and $B_N$, since their difference is exactly $B_N/a_N$.

The exponential equivalence theorem transfers the large deviation principle and its good rate function, givingLarge positive excursions are cheapest through the single exponential random variable; a negative excursion would require the Gaussian random variable sum to move on its faster scale. If $\sigma =0$, the deterministic shift $N\mu /a_N\to 0$ gives the same conclusion directly.