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Suppose $\xi_1,\ldots \xi_T$ is a martingale difference sequence. Then,

1) For any $a\in \mathbb{R}^{+}$, can we say something about the sequence $\xi_1^2\mathbb{1}(\xi_1\geq a),\ldots, \xi_T^2\mathbb{1}(\xi_T\geq a)$ ? Is it a (sub/sup) martingale difference sequence?

2) Suppose each $|\xi_t|\leq B_t$ a.s., and $B_1\leq B, B_T\leq B$ a.s. then can we provide an upper bound for $\mathbb{P}(\sum_{t=1}^T \xi_t\geq z)$?

I guess if we can prove that the sequence in (1) is a (sub/sup) martingale difference sequence then one can apply standard maximal inequalities to solve (2). However I am not able to resolve (1), and my intuition says that, for the sequence in (1) one cannot claim any (sub/sup) martingale difference behaviour. However I do not have a formal proof or a counterexample. Also if it turns out that the sequence in (2) is not a (sub/sup) martingale then how do we go about establishing maximal inequalities?

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1 Answer 1

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I must admit that I understand neither of your questions, or rather what sort of nontrivial answer you want.

1) $\xi^2 \mathsf{1} \lbrace \xi \ge a \rbrace \ge 0$, so it is a submartingale difference in an extremely boring way. I don't see what you can possibly want from it.

2) No, we can't bound it in a nontrivial way. As an example, consider independent random variables that equal $B$ with high probability and something very negative with small probability. If $|\xi_t| \le B$ instead, then see http://en.wikipedia.org/wiki/Azuma_inequality.

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  • $\begingroup$ Thanks Alex for your reply. Sorry there was a typo in question (2). It should have read as $|\xi_t| \leq B_t$ a.s., and $B1≤B,B_2\leq B, \ldots B_T\leq B$. What is not clear to me though is 1) Why is the first sequence a submartingale difference sequence. 2) Azuma-Hoeffding (AH) bound applies only to (sup)martingales. If the first sequence is a submartingale, then I do not see how one could apply AH? $\endgroup$
    – gmravi2003
    Feb 6, 2013 at 2:46
  • $\begingroup$ 1) Because it is nonnegative. 2) Right, and $\sum_{t=1}^T \xi_t$ is a martingale, so this inequality does apply to it, with no connection to 1) at all. Probably you should at least check what you wrote against the definitions to eliminate trivialities or nonsense. $\endgroup$ Feb 6, 2013 at 8:46

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