Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 201 / 3 / a

Table of contents
- Solution a

Solution

Since

X_{1}

is uniform on

{- 1, 1}

,

ψ (λ) = lo g E [e^{λ X_{1}}] = lo g (\frac{e ^{λ} + e ^{- λ}}{2}) = lo g cosh λ .

For

∣ x ∣ < 1

, the supremum in the Legendre transform of a cumulant-generating function is attained where

x = ψ^{'} (λ) = tanh λ, λ = \frac{1}{2} lo g \frac{1 + x}{1 - x} .

Substitution gives the Rademacher large-deviation rate function

ψ^{*} (x) = \frac{( 1 + x ) lo g ( 1 + x ) + ( 1 - x ) lo g ( 1 - x )}{2}

for

∣ x ∣ \leq 1

, with

0 lo g 0 = 0

; it is

+ \infty

for

∣ x ∣ > 1

.

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