Since $X_1$ is uniform on $\{-1,1\}$,
$$
\psi(\lambda)
=\log\mathbb E[e^{\lambda X_1}]
=\log\left(\frac{e^\lambda+e^{-\lambda}}2\right)
=\log\cosh\lambda.
$$
For $|x|<1$, the supremum in the <Legendre transform of a cumulant-generating function> is attained where
$$
x=\psi'(\lambda)=\tanh\lambda,
\qquad
\lambda=\frac12\log\frac{1+x}{1-x}.
$$
Substitution gives the <Rademacher large-deviation rate function>
$$
\boxed{\psi^*(x)
=\frac{(1+x)\log(1+x)+(1-x)\log(1-x)}2}
$$
for $|x|\leq1$, with $0\log0=0$; it is $+\infty$ for $|x|>1$.

Solution

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

For a uniform random sign $X\in\{-1,1\}$, $\psi(\lambda)=\log\cosh\lambda$ and
$$
\psi^*(x)
=\frac{(1+x)\log(1+x)+(1-x)\log(1-x)}2
$$
for $|x|\leq1$, with value $+\infty$ outside that interval.


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