MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the paper

you can find a trajectorial version of Doob's inequality. It is given by:

$$\bar{s}^2_T+4\sum_{k=0}^{T-1}\bar{s_k}(s_{k+1}-s_k)\le 4s^2_T$$

$$ \bar{s}_k=\max (s_1,...,s_k) $$

The proof should be straightforward but I am not able to prove it, please help me with that more or less simple inequality.

share|cite|improve this question
Induction should be the ticket for an inequality like this. Try the base case - T=1. The proof of the base case would then give some insight into how to prove the induction step. – Daniel Spector Nov 27 '12 at 8:09
Is it only possible by doing induction? The paper says it should be possible by rearranging terms. – Leitz Nov 27 '12 at 8:41
up vote 6 down vote accepted

I think induction over $T$ will be hard. But you can prove the inequality by considering the times $n$ at which the maximum $\bar s_n$ increases. For example, let $ 1 = k_1, \ldots, k_r \le T $ be the different times at which $s_k = \bar s_k$ attains a maximum (with respect to all previous times). Then it holds $$ \sum_{l=k_j}^{k_{j+1}-1} \bar s_l (s_{l+1} - s_l) = \sum_{l=k_j}^{k_{j+1}-1} \bar s_{k_j} (s_{l+1} - s_l) = \bar s_{k_j} (\bar s_{k_{j+1}} - \bar s_{k_j} ). $$ Therefore we see that $$ \bar s_T^2 + 4\sum_{n=1}^{T-1} \bar s_n (s_{n+1} - s_n) = \bar s_{k_r}^2 + 4\sum_{j=1}^{r-1} \bar s_{k_j} (\bar s_{k_{j+1}} - \bar s_{k_j} ) + 4\bar s_{k_r} (s_T - \bar s_{k_r}) $$ Since the $\bar s_{k_j}$ are increasing, we estimate the right hand side by $$ \bar s_{k_r}^2 + 2\sum_{j=1}^{r-1} (\bar s_{k_{j+1}} + \bar s_{k_j}) (\bar s_{k_{j+1}} - \bar s_{k_j} ) + 4\bar s_{k_r} (s_T - \bar s_{k_r}) $$ which by the third binomial formula yields $$ \bar s_{k_r}^2 + 2\sum_{j=1}^{r-1} (\bar s_{k_{j+1}}^2 - \bar s_{k_j}^2) + 4\bar s_{k_r} (s_T - \bar s_{k_r}) = \bar s_{k_r}^2 + 2(\bar s_{k_{r}}^2 - \bar s_{1}^2) + 4\bar s_{k_r} (s_T - \bar s_{k_r}). $$ Obviously, this is bounded by $$ -\bar s_{k_r}^2 + 4\bar s_{k_r} s_T = -(\bar s_{k_r} - 2s_T)^2 + 4s_T^2 \le 4s_T^2, $$ which finishes the proof. For comparison, also refer to the proof of Lemma 2.2 in the paper you quoted.

share|cite|improve this answer
@Daniel Marahrens : Very nice proof. Best regards – The Bridge Nov 27 '12 at 20:48
Thank you for the answer, I just recognized it is also possible by using integrals. – Leitz Nov 27 '12 at 21:20
@Daniel Marahrens and Leitz: There is a little typo I think, $k_1$ should start at 0 and not at 1, but this comes from a typo in the question itself, where the definition of $\bar{s}_k$ should include $s_0$. Best regards – The Bridge Nov 28 '12 at 20:16
Thank you for the compliments. I think there is some inconsistency in the question by Leitz, but I simply chose to start the indices at 1 in my answer and I tried to be consistent with it. Probably 0 would have been a better choice as that is the choice in the referenced paper. – Daniel Nov 28 '12 at 23:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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