Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 31 / 2

Write $a_n=n^{2(1-H)}$ and $b_n=n^{2(1-G)}$. Then $a_n/b_n=n^{2(G-H)}\to \infty$. The large deviation principle on the whole space forces $\inf I=0$; compact sublevel sets and lower semicontinuity make that infimum attained. The unique possible zero is therefore mu, and $I(\mu )=0$.

If a closed set C contains mu, its new infimum rate is zero and the claimed upper bound follows from probabilities being at most one. If it excludes mu, then $\eta ={\inf }_{C}I>0$, possibly infinite: otherwise a sequence of rates tending to zero would have a convergent subsequence in a compact sublevel set, with limit in C and zero rate, contradicting uniqueness. This is the fact that a good rate function is separated from zero away from its minimizer.

Choose any $0<c<\eta$. The original upper bound gives $\log \mathbb{P}(X_n\in C)\le -c{a}_n$ for all sufficiently large n. ThusTogether these cases prove the closed-set upper bound at the slower speed, with rate zero only at mu. Its finite sublevels are the compact singleton $\{\mu \}$, so it is a good rate function.

For an open neighborhood D of mu, its complement is closed and excludes mu. The preceding upper bound is minus infinity, so, for example, eventually $\mathbb{P}(X_n\notin D)\le e^{-b_n}\to 0$. Hence the sequence concentrates in every neighborhood of mu.

If an open E contains mu, its probability tends to one. Its logarithm divided by $b_n\to \infty$ therefore tends to zero, which equals $-{\inf }_E{I}^{\prime }$. If E does not contain mu, that latter value is minus infinity and its lower bound is automatic. ThusWith the closed-set upper bound this proves the slower-speed degeneration of a large deviation principle. This degenerate rate has no strictly positive finite value, so it does not give the sequence a different Hurstiness under the stated definition.

Let $I_X,I_Y$ be the original rates and ${\mu }_X,{\mu }_Y$ their unique zero points. Suppose first that $H<G$. At speed $b_n=n^{2(1-G)}$, the preceding argument gives X the rate zero at ${\mu }_X$ and infinity elsewhere, while Y retains $I_Y$. Independence allows the product large-deviation principle, and addition is continuous, so the contraction principle for large deviations gives the sum rateThis is good, has unique zero ${\mu }_X+{\mu }_Y$, and inherits a strictly positive finite value by translation from $I_Y$. Thus the sum has Hurstiness G. The case $G<H$ is identical with the roles exchanged.

If $G=H$, the same product and contraction results apply to the original rates at their common speed, giving their infimal convolutionIt is good. Every finite fiber infimum is attained: a minimizing sequence lies in a product of compact sublevel sets and has a subsequence converging in the closed addition fiber. Consequently its only zero is ${\mu }_X+{\mu }_Y$. Choose $\hat{x}$ with $0<I_X(\hat{x})<\infty$ and set $\hat{z}=\hat{x}+{\mu }_Y$. Then $\hat{z}$ differs from the zero point and $0<K(\hat{z})\le I_X(\hat{x})<\infty$. The nondegeneracy requirement is therefore satisfied here too. In all cases,The general results used are the independent product large-deviation theorem and the continuous contraction principle. Checking the zero set and a positive finite value is necessary in addition to obtaining an LDP.