Doob's inequality for martingale "convolution"

Question

Let $(X_t, t \in \mathbb{N})$ be a martingale, and let $a \leq b \leq T \in \mathbb{N}$ be constants. Is there something like Doob's inequality for $\mathbb{E} \sup_{a \leq t \leq b} X_t(X_T-X_t)$, i.e. is it possible to bound this supremum by something involving just a deterministic variance? I am hoping for one of the following inequalities to be true: $$ \mathbb{E} \sup_{a \leq t \leq b} \left(X_t(X_T-X_t)\right)^2 \leq C\mathbb{E} \left(X_b(X_T-X_a)\right)^2 $$ or maybe even $$ \mathbb{E} \sup_{a \leq t \leq b} \left(X_t)(X_T-X_t)\right)^2 \leq C\mathbb{E} X_b^2 \mathbb{E}(X_T-X_a)^2 $$ for some deterministic constant $C$. In my application, $X_t$ is the sum process of iid mean-0 random variables; perhaps that makes life easier?

Answer 1

Assume that $X_t$ have independent and centered increments, but not necessarily identically distributed. Let $D_i=X_i-X_{i-1}$ for $i\geqslant a+1$ and $D_a=X_a$. Let $ S_t=X_t\left(X_T-X_t\right). $ Then $$S_t=\left(\sum_{i=a}^tD_i\right)\left(\sum_{j=t+1}^TD_j\right)$$ and \begin{align} S_{t+1}-S_t= \left(\sum_{i=a}^{t+1}D_i\right)\left(\sum_{j=t+2}^TD_j\right)-\left(\sum_{i=a}^tD_i\right)\left(\sum_{j=t+1}^TD_j\right)\\ = \left(\sum_{i=a}^{t}D_i\right)\left(\sum_{j=t+2}^TD_j\right)+ D_{t+1}\left(\sum_{j=t+2}^TD_j\right)-\left(\sum_{i=a}^{t}D_i\right)\left(\sum_{j=t+2}^TD_j\right)-\left(\sum_{i=a}^{t}D_i\right) D_{t+1} \\ = D_{t+1}\left(\sum_{j=t+2}^TD_j\right) -\left(\sum_{i=a}^{t}D_i\right) D_{t+1} \end{align} hence letting $$ D'_t= -\left(\sum_{i=a}^{t}D_i\right) D_{t+1};\quad D''_t:= D_{t+1}\left(\sum_{j=t+2}^TD_j\right), $$ the following equality holds $$ S_u= S_a+\sum_{t=a}^{u-1}D'_t+\sum_{t=a}^{u-1}D''_t. $$ Then $$ \mathbb E\left[\max_{a\leqslant t\leqslant b}S_t^2\right] \leqslant 4\mathbb E\left[D_a^2\left(\sum_{j=a+1}^TD_j\right)^2\right] +4\mathbb E\left[\max_{a\leqslant u\leqslant b}\left(\sum_{t=a}^{u-1}D'_t\right)^2\right] +4\mathbb E\left[\max_{a\leqslant u\leqslant b}\left(\sum_{t=a}^{u-1}D''_t\right)^2\right]. $$ Using independence, the first term can be simplified. For the second one, we have maximum of a martingale, and for the third of a reversed martingale. At the end, we get $$ \mathbb{E}\left[\max_{a \leqslant t \leqslant b} \left(X_t(X_T-X_t)\right)^2 \right]\leq C\mathbb{E} X_a^2 \mathbb{E}(X_T-X_b)^2. $$