Proving (a part of) Hoeffding's lemma Hoeffding lemma goes like this: *Let $X$ be a scalar variable taking values in an interval $[a,b]$. Then for any $t>0$ $$\mathbb{E} e^{tX}\leq e^{t\mathbb{E} (X)}(1+O(t^2\mathbb{V}(X)\exp(O(t(b-a)))).$$In particular $$\mathbb{E} e^{tX}\leq e^{t\mathbb{E} (X)}\exp(O(t^2(b-a)^2).$$(This version is taken from T. Taos book on Random matrices - it's on pp. 61 Lemma 2.1.2 from the draft from his website)
My question is: How can one get the second inequality from the first ? Tao gives as a hint $ $$\mathbb{V}(X)\leq(b-a)^2$.    
 A: Let me get rid of the $O(\cdot)$-notation: Tao's proof shows
$$\mathbb{E}e^{tX} \leq e^{t \mathbb{E}X} \bigg( 1+ \frac{t^2}{2} \mathbb{V}X \cdot e^{t (b-a)} \bigg).$$
We consider two cases separately:


*

*$t \leq \frac{1}{b-a}$: For $t \leq \frac{1}{b-a}$, we have $e^{t (b-a)}\leq e^1 \leq 4$, hence $$\mathbb{E}e^{tX} \leq e^{t \mathbb{E}X} \bigg( 1+ 2t^2 \mathbb{V}X \bigg).$$ As $1+x \leq e^x$ for any $x \geq 0$, we conclude $$\mathbb{E}e^{tX} \leq e^{t \mathbb{E}X} \exp(2 t^2 \mathbb{V}X).$$ Finally, $\mathbb{V}X \leq (b-a)^2$ implies $$\mathbb{E}e^{tX} \leq e^{t \mathbb{E}X} \exp(2 t^2 (b-a)^2).$$

*$t \geq \frac{1}{b-a}$: Obviously, $1 \leq e^{t(b-a)}$. Consequently, we find $$\mathbb{E}e^{tX} \leq e^{t \mathbb{E}X} e^{t(b-a)} \left(1+ \frac{t^2}{2} \mathbb{V}X \right).$$ Using $\mathbb{V}X \leq (b-a)^2$ and $1+\frac{x^2}{2} \leq e^x$, $x \geq 0$, we arrive at $$\mathbb{E}e^{tX} \leq e^{t \mathbb{E}X}e^{2t (b-a)}.$$ As $t (b-a) \geq 1$, the monotonicity of $x \mapsto e^x$ entails $$\mathbb{E}e^{tX} \leq e^{t \mathbb{E}X}e^{2t (b-a) \cdot 1} \leq e^{t \mathbb{E}X}e^{2t^2 (b-a)^2}.$$

