= Solution
<Hoeffding lemma> states that if $a\leq X\leq b$ almost surely, then for every real $\lambda$,
$$
\log\mathbb E e^{\lambda(X-\mathbb EX)}
\leq\frac{\lambda^2(b-a)^2}{8}.
$$
Convexity of $e^{\lambda x}$ bounds it on $[a,b]$ by the secant joining its endpoint values. Taking expectations reduces the centered moment-generating function to that of a two-point variable on $\{a,b\}$ having the same mean. After rescaling to $[0,1]$, its logarithm is
$$
-\lambda q+\log(1-q+qe^\lambda),
$$
where $q$ is its mean. Twice differentiating in $\lambda$ shows that the second derivative is a Bernoulli variance and hence at most $1/4$. The value and first derivative vanish at zero, so Taylor's theorem gives at most $\lambda^2/8$. Rescaling proves the claim.
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