I was going through some notes on concentration inequalities when I noticed that there are two commonly-cited forms of McDiarmid's inequality. Long story short: I know how to prove the weaker one from Azuma's inequality. I also know how to prove the stronger one *not* directly from Azuma. But is there a way to prove the strong one from Azuma's inequality?

In what follows I will assume for simplicity that all differences are bounded by 1.

Azuma's inequality:Let $S_n = \sum_{i=1}^n X_i$ and $S_0 = 0$. Let the filtration $\{\mathcal{F}_n\}$ be the usual one defined by $\mathcal{F}_n = \sigma(X_1, \ldots, X_n)$. If $\{S_n\}_{n \geq 0}$ is a martingale and $|X_n| \leq 1$ for all $n$, then for $\lambda>0$, $$P(S_n \geq \lambda) \leq \exp \left(-\frac{\lambda^2}{2n} \right)$$

Now here is the "weak" version of McDiarmid's which is an obvious application of Azuma's inequality:

McDiarmid's inequality:Suppose $\xi_1, \ldots, \xi_n$ are independent. Let $Z = f(\xi_1, \ldots, \xi_n)$. Also, assume that: $$ \Big|\; f(\xi_1, \ldots, \xi_k, \ldots, \xi_n) - f(\xi_1, \ldots, \xi_k', \ldots, \xi_n) \;\Big| \leq 1$$ for two realizations of the same random variable $\xi_k$ and $\xi_k'$. Then for $\lambda>0$, $$ P(Z-EZ \geq \lambda) \leq \exp \left(-\frac{\lambda^2}{2n} \right) $$

Now the strong version of McDiarmid's inequality has the same setup except that the "2" in the denominator is moved up to the numerator, so the bound is: $$P(Z-EZ \geq \lambda) \leq \exp \left(-\frac{2\lambda^2}{n} \right)$$ Clearly this bound is sharper than the one in the version presented above. I know a proof for this bound but it does not explicitly use Azuma's inequality above.

**The question:** Is it possible to adapt Azuma's inequality to prove this second bound. More specifically, is there a more general form of Azuma's inequality that can be applied to the case of McDiarmid's inequality to yield the tighter bound?

**Edit:** After pondering it over, I now think that it is **not** possible. Just intuitively, the bounded martingale difference condition in Azuma's inequality seems somehow "weaker" than the condition in McDiarmid's inequality, which is a Lipschitz-like condition on the function $f$. So one would expect that using Azuma's inequality would not give the sharpest bound.

**Edit 2:** To shed some more light on this, the proof of the "weak" McDiarmid's using Azuma's inequality goes like this. Define $S_n = E(Z \;|\; \xi_1, \ldots, \xi_m) - EZ$. Then $\{S_m\}_{m \geq 1}$ is a martingale. Also, for $m=n$, $S_m = S_n = Z - EZ$.

To prove that the martingale differences are bounded, let $\xi_1', \ldots, \xi_n'$ be independent copies of the $\xi_1, \ldots, \xi_n$. Then we have:
$$ E\big[f(\xi_1, \ldots, \xi_m, \ldots, \xi_n) \;\big|\; \xi_1, \ldots, \xi_{m-1}\big] = E\big[f(\xi_1, \ldots, \xi_m', \ldots, \xi_n) \;\big|\; \xi_1, \ldots, \xi_m\big] $$
Thus we have:
$$\big|S_m - S_{m-1}\big| \leq E\big[|\, f(\xi_1, \ldots, \xi_m, \ldots, \xi_n) - f(\xi_1, \ldots, \xi_m', \ldots, \xi_n)\,| \;\big|\; \xi_1, \ldots, \xi_m\big]$$
So to satisfy the bounded martingale difference in Azuma's inequality, we only need the **expectation** of the difference of $f(\cdot) - f(\cdot)$ (conditional on $\xi_1, \ldots, \xi_m$) to be bounded, whereas the assumption in McDiarmid's is slightly stronger. However, the bounded martingale difference must also hold for all $m$, and in particular for the case of conditioning on $\xi_1, \ldots, \xi_n$. In that case it seems to reduce to assumption. So... it does seem like Azuma's inequality (in this form) is not strong enough. Thoughts?

Here is a resource presenting the "standard" proof of the strong version of McDiarmid's inequality: http://empslocal.ex.ac.uk/people/staff/yy267/McDiarmid.pdf