Hoeffding's lemma

In probability theory, Hoeffding's lemma is an inequality that bounds the moment-generating function of any bounded random variable.[1] It is named after the FinnishAmerican mathematical statistician Wassily Hoeffding.

The proof of Hoeffding's lemma uses Taylor's theorem and Jensen's inequality. Hoeffding's lemma is itself used in the proof of McDiarmid's inequality.

Statement of the lemma

Let X be any real-valued random variable such that ${\displaystyle a\leq X\leq b}$ almost surely, i.e. with probability one. Then, for all ${\displaystyle \lambda \in \mathbb {R} }$,

${\displaystyle \mathbb {E} \left[e^{\lambda X}\right]\leq \exp {\Big (}\lambda \mathbb {E} [X]+{\frac {\lambda ^{2}(b-a)^{2}}{8}}{\Big )},}$

or equivalently,

${\displaystyle \mathbb {E} \left[e^{\lambda (X-\mathbb {E} [X])}\right]\leq \exp {\Big (}{\frac {\lambda ^{2}(b-a)^{2}}{8}}{\Big )}.}$

Proof

Without loss of generality, by replacing ${\displaystyle X}$ by ${\displaystyle X-\mathbb {E} [X]}$, we can assume ${\displaystyle \mathbb {E} [X]=0}$, so that ${\displaystyle a\leq 0\leq b}$.

Since ${\displaystyle e^{\lambda x}}$ is a convex function of ${\displaystyle x}$, we have that for all ${\displaystyle x\in [a,b]}$,

${\displaystyle e^{\lambda x}\leq {\frac {b-x}{b-a}}e^{\lambda a}+{\frac {x-a}{b-a}}e^{\lambda b}}$

So,

{\displaystyle {\begin{aligned}\mathbb {E} \left[e^{\lambda X}\right]&\leq {\frac {b-\mathbb {E} [X]}{b-a}}e^{\lambda a}+{\frac {\mathbb {E} [X]-a}{b-a}}e^{\lambda b}\\&={\frac {b}{b-a}}e^{\lambda a}+{\frac {-a}{b-a}}e^{\lambda b}\\&=e^{L(\lambda (b-a))},\end{aligned}}}

where ${\displaystyle L(h)={\frac {ha}{b-a}}+\ln(1+{\frac {a-e^{h}a}{b-a}})}$. By computing derivatives, we can conclude

${\displaystyle L(0)=L'(0)=0}$ and ${\displaystyle L''(h)\leq {\frac {1}{4}}}$ for all ${\displaystyle h}$.

From Taylor's theorem, for some ${\displaystyle 0\leq \theta \leq 1}$

${\displaystyle L(h)=L(0)+hL'(0)+{\frac {1}{2}}h^{2}L''(h\theta )\leq {\frac {1}{8}}h^{2}}$

Hence, ${\displaystyle \mathbb {E} \left[e^{\lambda X}\right]\leq e^{{\frac {1}{8}}\lambda ^{2}(b-a)^{2}}}$.