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12 votes
2 answers
2k views

Can we do better than Azuma-Hoeffding when the variance is small?

The Azuma-Hoeffding Inequality says that if $X_1,X_2, \ldots$ is a martingale and the differences are bounded by constants, $\|X_i - X_{i-1}\| \le 1$ say, then we should not expect the difference $\|X …
Daron's user avatar
  • 1,955
7 votes
1 answer
398 views

Do i.i.d. sums concentrate any faster than martingales?

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 …
Daron's user avatar
  • 1,955
5 votes
1 answer
163 views

Is there an i.i.d sequence in the unit cube $[-1,1]^d$ with $\mathbb E \left[ \Big \| \sum_{...

There are loads of concentration results for sums of scalar-valued independent sums $X_1,X_2,\ldots, X_N$ with $\mathbb E[X_n]=0$. For example Hoeffding's Inequality says if all $|X_1|\le 1$ then $\ma …
Daron's user avatar
  • 1,955
4 votes
0 answers
138 views

For a martingale $f_0,f_1,\ldots $ how can we bound $P(\frac{1}{n} \|f_n\| \le 1$ for all $ ...

Suppose $f_0,f_1, \ldots$ is a martingale (or i.i.d sequence) in $\mathbb R^d$ with $f_0=0$ and all $\|f_n - f_{n-1}\| \le L$ say. There are many concentration results for the initial segment of the m …
Daron's user avatar
  • 1,955