Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 77 / 3 / Solution

A standard form of Varadhan's integral theorem is as follows. Let ${\mu }_L$ be probability measures on a Polish space $E$ satisfying a large deviation principle with speed $L\to \infty$ and good rate function $I$. If $F:E\to \mathbb{R}$ is continuous and bounded above, thenThe usual bounded-continuous version follows immediately. If $F$ is unbounded above, the same conclusion holds under the additional exponential tail conditionThis is the exponential tail extension of Varadhan's lemma. A convenient sufficient hypothesis isAn integrability hypothesis is essential for general unbounded functions; the ordinary large deviation principle alone does not control arbitrarily large exponential weights.

For example, let $X^L=L$ with probability $e^{-L^2}$ and $X^L=0$ otherwise. It satisfies the large deviation principle at speed $L$ with rate zero at zero and infinite elsewhere. But for $F(x)=x^2$, $\mathbb{E}{e}^{LF(X^L)}\ge e^{L^3-L^2}$, whose logarithmic rate diverges, while $\sup (F-I)=0$. This shows why the tail hypothesis cannot simply be omitted.

For the lower bound, fix $x$ with $I(x)<\infty$ and $\epsilon >0$. By continuity choose an open set $G$ containing $x$ on which $F(y)\ge F(x)-\epsilon$. ThenApply the open-set lower bound of the large deviation principle:Let $\epsilon \downarrow 0$ and take the supremum over $x$ to obtain the required lower bound. This argument also applies when $F$ is not bounded above.

For the upper bound, suppose $F\le M$ and set $S={\sup }_E(F-I)$, which is finite. Choose a large $K$ and partition $[-K,M]$ into finitely many closed intervals $[a_j,a_j+\epsilon ]$ of length at most $\epsilon$. The preimagesare closed sets. The overlapping endpoints do not matter for an upper estimate. Split off the lower tail and bound each remaining exponential weight:For a finite sum of nonnegative terms the upper logarithmic rate is at most the largest of their upper logarithmic rates. Applying the closed-set upper bound givesFor $x\in C_j$, $a_j+\epsilon \le F(x)+\epsilon$, hence every inner term is at most $S+\epsilon$. Empty preimages contribute no term. Let $K\to \infty$ and then $\epsilon \downarrow 0$ to get the upper bound $S$. Together with the lower bound this proves the bounded-above theorem.

For the unbounded extension, put $F_M=\min (F,M)$. This is continuous and bounded above, so the proved result applies, withThe full integral is at most the integral with $F_M$ plus the integral on $\{F>M\}$. The upper logarithmic rate is therefore at most the larger of $S_M$ and the tail rate. The tail hypothesis and then $M\to \infty$ give the upper bound $S$; the earlier local argument gives the lower bound. If $S=+\infty$, that lower argument already yields the asserted infinite limit.

Finally, the sufficient exponential-moment condition really implies the required tail estimate: on $F>M$,After integration the upper logarithmic rate is bounded by a finite constant minus $(\gamma -1)M$, which tends to $-\infty$. This completes the statement and proof of Varadhan's lemma, including the hypotheses needed for the unbounded version.