# Trajectorial version of Doob's $L^2$ inequality

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.

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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

## 1 Answer

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.

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@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