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The exponent printed in the original PDF is too strong under the stated increment bound. Take $M_0=0$, let $M_1$ equal $1$ or $-1$ with equal probabilities, and set $M_k=M_1$ for $k\ge 1$. This is a martingale with $c_1=1$ and $c_k=0$ for $k>1$. For $n=1$ and $x=1$,Thus the printed inequality is false. The intended two-sided Azuma-Hoeffding inequality isHere the $c_k$ are deterministic increment bounds. We prove this corrected assertion from conditional exponential moments.

Let $D_k=M_k-M_{k-1}$. The martingale property gives $\mathbb{E}[D_k\mid {\mathcal{F}}_{k-1}]=0$. For $c_k>0$, convexity of the exponential function gives the secant bound on $[-c_k,c_k]$,Taking conditional expectation removes the terms containing $D_k$, soThe last inequality follows term by term from $(2j)!\ge 2^{j}j!$ in the power series for $\cosh z$ and $e^{z^2/2}$. If $c_k=0$ the increment is zero and the same bound holds directly. This is the bounded-increment case of the Hoeffding lemma.

Repeated conditioning, without any independence assumption, now givesWrite $C={\sum }_{k=1}^n{c}_k^2$. If $C>0$, the Markov inequality gives for $\theta >0$,Minimizing at $\theta =x/C$ gives $e^{-x^2/(2C)}$. Apply the same argument to the martingale $-M$ for the lower tail, and use the union bound to obtain the boxed result. If $C=0$, all increments vanish almost surely, so the deviation probability is zero for every $x>0$. This handles the degenerate case without dividing by zero and proves the correct bound in full.