Varadhan's lemma

Remarks

The lemma is an infinite-dimensional form of Laplace's method: the integral is dominated by the neighbourhood of the point where $\varphi - I$ is largest. For the lower bound, take $x$ with $I(x) < \infty$ and a small open neighbourhood $G$ of $x$ on which $\varphi > \varphi(x) - \delta$; the large deviation lower bound gives $\mu_n(G) \ge e^{-n(I(x) + \delta)}$ for large $n$. For the upper bound, cover a compact sublevel set of $I$ by finitely many small closed neighbourhoods. Use the large deviation upper bound on each, and the bound on $\varphi$ outside.

The boundedness of $\varphi$ from above can be replaced by the tail condition $$\lim_{M \to \infty} \limsup_{n \to \infty} \frac{1}{n} \log \int_{\set{\varphi \ge M}} e^{n \varphi} \, d\mu_n = -\infty.$$ Some such condition is needed, since otherwise rare but huge values of $e^{n\varphi}$ can dominate the integral.

Example: let $\mu_n$ be the normal law with mean $0$ and variance $1/n$, with $I(x) = x^2/2$, and let $\varphi(x) = -\abs{x - 1}$. The lemma gives the limit $\sup_x (-\abs{x - 1} - x^2/2) = -1/2$, attained at $x = 1$.

Going deeper

For a student

Varadhan's lemma is the main tool to compute free energies and other exponential averages once a large deviation principle is known. In statistical mechanics, the limit is the pressure and the maximisers of $\varphi - I$ are the equilibrium states. Its converse, Bryc's theorem, recovers a large deviation principle from the limits of such exponential integrals.