Bernstein tail bound for independent bounded sums

If $m=0$, then $S_m=V=0$, so the zero-variance clause holds. Suppose $m\ge 1$, fix one centered variable $Z$ with $\mathrm{\mid }Z\mathrm{\mid }\le b$, and let ${\sigma }^2=\mathbb{E}[Z^2]$. For every integer $k\ge 2$, the power functions are Borel and $x\mapsto e^{\lambda x}$ is continuous by [F9], so their compositions with $Z$ are measurable by [F10]; the almost-sure bound gives integrability of each power. Thus $\mathrm{\mid }\mathbb{E}[Z^k]\mathrm{\mid }\le \mathbb{E}[\mathrm{\mid }Z{\mathrm{\mid }}^k]\le b^{k-2}{\sigma }^2$. For fixed real $\lambda$, the exponential series converges uniformly on $[-b,b]$: its tail is bounded by ${\sum }_{k>N}(\mathrm{\mid }\lambda \mathrm{\mid }b{)}^k\mathrm{/}k!\to 0$. Since $\mathrm{\mid }Z\mathrm{\mid }\le b$ almost surely, this uniform convergence and finite linearity [F2] permit termwise expectation in [F3], and the centered first-order term vanishes.

For $k\ge 2$, $k!\ge 2\cdot 3^{k-2}$, by induction from $2!=2$ and the fact that each next factor is at least $3$. If $\mathrm{\mid }\lambda \mathrm{\mid }b<3$, [F3] and step 1.1 therefore give $\mathbb{E}{e}^{\lambda Z}\le 1+{\sigma }^2{\sum }_{k\ge 2}\frac{\mathrm{\mid }\lambda {\mathrm{\mid }}^k{b}^{k-2}}{k!}\le 1+\frac{{\lambda }^2{\sigma }^2}{2(1-\mathrm{\mid }\lambda \mathrm{\mid }b\mathrm{/}3)}\le \exp \text{\hspace{0.17em}⁣}\left(\frac{{\lambda }^2{\sigma }^2}{2(1-\mathrm{\mid }\lambda \mathrm{\mid }b\mathrm{/}3)}\right),$ where the last inequality follows from $1+x\le e^x$ for $x\ge 0$, which follows termwise from [F3].

For $0<\lambda <3\mathrm{/}b$, repeated use of [F4] gives $e^{\lambda S_m}={\prod }_{i=1}^m{e}^{\lambda Z_i}$. Each $Z_i$ is measurable, scalar multiplication preserves measurability by [F10], and $x\mapsto e^x$ is Borel by [F9], so the factors are measurable by composition. They are nonnegative functions of the independent variables, so [F5] and step 2.1 yield $\mathbb{E}{e}^{\lambda S_m}={\prod }_{i=1}^m\mathbb{E}{e}^{\lambda Z_i}\le \exp \text{\hspace{0.17em}⁣}\left(\frac{{\lambda }^{2}V}{2(1-\lambda b\mathrm{/}3)}\right)\mathrm{.}$

If $V>0$, take $\lambda =t\mathrm{/}(V+bt\mathrm{/}3)$; then $0<\lambda b\mathrm{/}3=bt\mathrm{/}(3V+bt)<1$. The finite sum $S_m$ and its threshold event are measurable by [F10]. Since [F9] makes $e^x$ increasing, on $\{S_m\ge t\}$ one has $e^{\lambda S_m}\ge e^{\lambda t}$. Markov [F6] and step 3.1 give $\mathbb{P}(S_m\ge t)\le \exp \text{\hspace{0.17em}⁣}(\frac{{\lambda }^{2}V}{2(1-\lambda b\mathrm{/}3)}-\lambda t)=\exp \text{\hspace{0.17em}⁣}(-\frac{t^2}{2(V+bt\mathrm{/}3)})\mathrm{.}$

Applying step 4.1 to $-Z_1,\dots ,-Z_m$ gives the lower-tail bound, since this family remains independent, centered, bounded by $b$, and has the same total second moment $V$. If $V=0$, each nonnegative $Z_i^2$ has expectation zero, so [F7] makes every $Z_i=0$ almost surely; their finite sum is then zero almost surely.