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?


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.

  • $\begingroup$ 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 '13 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. $\endgroup$ – Alexander Shamov Feb 6 '13 at 8:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.