All Questions
4 questions
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 $\|...
- 1,955
10
votes
3
answers
4k
views
Extension of the Azuma-Hoeffding inequality (when the differences are bounded with large probability)
Let $(X_i)$ be a super-martingale and suppose their differences are bounded ''with high probability'', that is
$$\mathbb{P}(\exists\,i=1,\dots,n\text{ s.t. }|X_i-X_{i-1}|>c_i) \,\leq\, \epsilon$$
...
- 223
4
votes
0
answers
143
views
For a martingale $f_0,f_1,\ldots $ how can we bound $P(\frac{1}{n} \|f_n\| \le 1$ for all $ n \ge N)$?
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 ...
- 1,955
4
votes
1
answer
1k
views
Does variants of Bernstein and Freedman concentration inequalities exist with NO uniform bound on the range of RV or martingale differences
A classic formulation of the Bernstein inequality (from Wikipedia) is as follow:
Let $X_1, \ldots, X_n$ be independent zero-mean random variables. Suppose that $|X_i|\leq M$ almost surely, for all $i$...
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