Suppose $X_1,X_2, \ldots, X_N \in \mathbb R^d$ are random variables with each $\|X_n\|_2 \le 1/2$ (this choice of the constant simplifies later formulae).
The simplest concentration inequality I know only applies in the case $d=1$ and only when $X_1,X_2, \ldots, X_N$ are i.i.d. The Hoeffding Lemma gives for each $\epsilon >0$ the bound
$$P(|X_1 + \ldots + X_N| \ge \epsilon) \le \exp\left (-\frac{2\epsilon^2}{N} \right).\tag{1}$$
On the other end of the spectrum are results that work under the weaker assumption that $X_1,X_2, \ldots, X_N$ is a martingale, and work for any $d \in \mathbb N$, or indeed for infinite dimensional Banach spaces provided some variant of the parallelogram is satisfied. For example [Theorem 3.5 of this paper of Pinelis][1]Theorem 3.5 of this paper of Pinelis leads to the following variant of the Azuma-Hoeffding inequality.
$$P(\|X_1 + \ldots + X_n\|_2 \ge \epsilon \text{ for some }n\le N) \le \exp\left (-\frac{\epsilon^2}{2N} \right).\tag{2}$$
The exponent is the same as the scalar Azuma Hoeffding. Notice the $2$ is now downstairs rather than upstairs like before.
If we are only dealing with i.i.d scalars and only interested in the final element, we should use $(1)$ because it gives a better bound. If we are dealing with either vectors, martingales, or want a uniform inequality we better use $(2)$ instead.
My problem is between the two extremes. I am dealing with a sequence of i.i.d vectors in $\mathbb R^d$ and I am interested in a uniform bound. I wonder does there exist an appropriate middle-ground between these two results? Perhaps combining the $-2\epsilon^2/N$ of the first with the uniform nature of the second, at the expense of only applying to i.i.d sequences as opposed to martingales. [1]: https://projecteuclid.org/download/pdf_1/euclid.aop/1176988477
