A Course on Large Deviations with an Introduction to Gibbs Measures.

4.2. Varadhan’s theorem and Bryc’s theorem 31
Ln is a random variable in Y = M1(X ). Let νn be its law. These empirical
measures usually contain more information than just the sample mean Sn/n.
In our case however
Ln = Sn
n δ1
+ 1 − Sn
δ0.
n
Hence, if one defines f : X → Y by f(s) = sδ1+(1−s)δ0, then νn = µn◦f −1 .
The contraction principle allows us to conclude a large deviation principle
for νn. The corresponding rate function is given for α ∈ M1(X ) by
H(α) = Ip(s) for α = sδ1 + (1 − s)δ0 with s ∈ [0, 1].
We will see in Chapter 6 that H(α) is a relative entropy and that the LDP
for the empirical measure holds in general for i.i.d. processes.
4.2. Varadhan’s theorem and Bryc’s theorem
A familiar fact about moment generating functions is that for a bounded
measurable function f : X → R, a probability measure µ, and a sequence
rn → ∞,
Hence
c = µ-ess sup f ≥ 1
rn log
≥ 1 log µ{f > c − ε} + c − ε.
rn
1 lim
n→∞ rn log
e rnf dµ ≥ 1
rn log
e
f>c−ε
rnf dµ
e rnf dµ = µ-ess sup f.
(µ-ess sup f = inf{b : µ(f > b) = 0}.) But what happens when µ is replaced
by a sequence µn? If {µn} satisfies a large deviation principle with
normalization rn, then the rate function I comes into the picture. The result
is known as Varadhan’s theorem. It is a probabilistic analogue of the
well-known Laplace method for asymptotics of integrals illustrated by the
next simple exercise.
Exercise 4.5. (Stirling’s formula) Use induction to show that
n! =
∞
e
0
−x x n dx.
Observe that e −x x n has a unique maximum at x = n. Prove that
lim
n→∞
n!
√ 2πn e −n n n
= 1.
Hint: Show that the main contribution to the integral comes from x ∈
[n − ε, n + ε] and use Taylor’s expansion.