most recent 30 from http://mathoverflow.net 2013-05-22T15:30:27Z http://mathoverflow.net/feeds/question/120794 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120794/maximal-inequalities-for-certain-functions-of-a-martingale-difference-sequence gmravi2003 2013-02-04T19:11:37Z 2013-02-06T02:42:26Z

<p>Suppose $\xi_1,\ldots \xi_T$ is a martingale difference sequence. Then,</p> <p>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?</p> <p>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 $\mathbb{P}(\sum_{t=1}^T \xi_t\geq z)$?</p> <p>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?</p>

http://mathoverflow.net/questions/120794/maximal-inequalities-for-certain-functions-of-a-martingale-difference-sequence/120821#120821 Alexander Shamov 2013-02-05T01:49:41Z 2013-02-05T01:49:41Z

<p>I must admit that I understand neither of your questions, or rather what sort of nontrivial answer you want.</p> <p>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.</p> <p>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 <a href="http://en.wikipedia.org/wiki/Azuma_inequality" rel="nofollow">http://en.wikipedia.org/wiki/Azuma_inequality</a>.</p>