For context, consider a sum $X$ of $n$ independent random variables in $[0,1]$.  For this situation, some Chernoff bounds bound the probability of a deviation with additive error, while others consider multiplicative error.  E.g., letting $\mu = E[X]$, one standard Chernoff bound bounds the probability of deviating by an additive error:

$$\Pr[X \ge \mu + \epsilon n] ~\le~ \exp(-\epsilon^2 n).$$

Another gives a bound on the probability of deviating by a _multiplicative_ error:

$$\Pr[X \ge (1+\epsilon) \mu] ~\le~ \exp(-\epsilon^2 \mu / 2).$$

This variant is independent of $n$, and stronger when $\mu \ll n$.

Now, [McDiarmid's inequality](https://en.wikipedia.org/wiki/Doob_martingale#McDiarmid.27s_inequality) as it is usually stated gives a bound on the first kind of deviation (additive error).  

>  Is there a form of McDiarmid's inequality that gives a multiplicative-error bound, of the form above? (Assuming here we are using it to bound, say, a function $f(x_1,..,x_n)$ that is always non-negative, and changes by at most 1 when any $x_i$ is changed.)

In case the answer to that is no, here is the specific application I have in mind.  Does a multiplicative-error bound hold for this application?

Fix $n$ arbitrary values $x_1, x_2, ..., x_n$ in $[0,1]$, and an integer $k$.  Obtain $k$-set $S$ by drawing $k$ times randomly _without replacement_ from $\{1,2,..,n\}$.  Define r.v. $X = \sum_{i \in S} x_i$.

Following the hint [here](https://mathoverflow.net/questions/120163/concentration-bounds-for-sums-of-random-variables-of-permutations), we can use McDiarmid's inequality to give a bound such as

$$\Pr[X \ge \mu + \epsilon k] ~\le~ \exp(- \epsilon^2 k / 3),$$

where $\mu = E[X]$.  

> In this case, does it also hold that, say,

> $$\Pr[X \ge \mu (1+\epsilon)] ~\le~ \exp(- \epsilon^2 \mu / c)$$

> for some constant $c > 0$ (independent of $k$ and $\mu$, e.g. $c=3$)?

It seems plausible that, at least in this specific case, one might be able to prove a multiplicative-error bound using some carefully designed super-martingale.

A related but different question was asked [here](https://mathoverflow.net/questions/119584/does-multiplicative-version-of-azumas-inequality-hold).