Laplace principle (large deviations theory, ${\lim }_{\theta \to \infty }{\theta }^{-1}\log {\int }_A{e}^{-\theta \varphi (x)}dx=-{\mathrm{essinf}}_A\varphi$)

The exponential integral of a fixed measurable function is controlled, on its logarithmic scale, by the function's essential infimum. Suppose $0<{\int }_A{e}^{-\varphi }dx<\infty$ and $\alpha ={\mathrm{essinf}}_A\varphi$ is finite. For $\theta \ge 1$, the integral is at most $e^{-(\theta -1)\alpha }{\int }_A{e}^{-\varphi }dx$. For every $\epsilon >0$, the set $A_{\epsilon }=\{x\in A:\varphi (x)<\alpha +\epsilon \}$ has positive finite Lebesgue measure, and the integral is at least $|A_{\epsilon }|e^{-\theta (\alpha +\epsilon )}$. Taking logarithms, dividing by $\theta$, and then letting $\epsilon \downarrow 0$ proves the displayed limit. The argument needs no isolated or nondegenerate minimizer. The Laplace principle for probability measures treats families whose distributions also change with the scaling parameter.

A family satisfies this principle with speed $a_n\to \infty$ and rate function $I$ when the displayed variational identity holds for every bounded continuous real function $f$. A large deviation principle implies it by the Varadhan lemma. The lower bound restricts the expectation to a neighborhood of a chosen point; the upper bound partitions the bounded range of $f$ into small intervals and applies the closed-set probability bounds.

For nonnegative $X_n$, the identity also givesFor $\mu >0$, truncate $f(s)=-\mu s$ below at $-\mu R$. The resulting bounded-function identity applies, while the difference of the two exponential expectations is at most $e^{-\mu R{a}_n}$. Choosing $R$ sufficiently large makes this error negligible on the claimed exponential scale. The case $\mu =0$ is immediate. This extension identifies moment-decay rates of a passive scalar from fluctuations of its logarithmic stretching.