In class we have seen a maximal inequality about discrete sub martingale. The setting was the following : Consider $T=\{0,1,..,N\}$ and $(X_t)_{t\in T}$ a sub martingale. Then we have for all $\lambda>0$

$$ \lambda\mathbb{P}(\{\max_{t\in T}X_t\geq\lambda\}\leq\mathbb{E}(X_{T}1_{\max_{t\in T}X_t\geq\lambda}) $$

I would like to extend it to continuous martingale. Here is my attempt.

First, after a first attempt I decided to consider a martingale because at some point, for convergence issue (you will tell me if I am wrong) I will need to have an increasing sequence of functions to apply monotone convergence. This leads to the use of absolute value and unfortunately this does not preserve a sub martingale.

Now let $T\in(0,\infty)$. Define the set $D_n = \{\frac{kT}{2^n} : k = 0,…,2^n\}$, clearly it is an increasing sequence of sets that equals $[0,T]$ at the limit.

Next define $Y_n(\omega) = \sup_{d\in D_n}\lvert X_d(\omega)\rvert$, it is an increasing sequence which converges almost surely to $Y(\omega) = \sup_{t\in [0,T]}\lvert X_t(\omega)\rvert$. This implies that

$$ 1_{Y_n\geq\lambda}\to_{n\to\infty}1_{Y\geq\lambda} $$

Applying the maximal inequality for finite time set to $\lvert X_d\rvert$ we get, for all $n$

$$ \lambda\mathbb{P}(\{Y_n\geq\lambda\}\leq\mathbb{E}(\lvert X_{T}\rvert1_{Y_n\geq\lambda}) $$

Then by convergence monotone and the continuity of the probability (which can be used here since $X$ is continuous in $t$) we have

$$ \lambda\mathbb{P}(\{Y\geq\lambda\}\leq\mathbb{E}(\lvert X_{T}\rvert1_{Y\geq\lambda}) $$

I would like to know if it is correct please and if not what can be done to improve this attempt.

Thank you a lot !


1 Answer 1


Your approach seems okay, so long as the submartingale $X$ has paths $t\mapsto X_t(\omega)$ that are right continuous with left limits. Then for each sample point $\omega$, $$ \sup_{t\in[0,T]}X_t(\omega) =\lim_n\sup_{t\in D_n}X_t(\omega), $$ the limit on the right being monotone increasing. The absolute values aren't necessary.

  • $\begingroup$ Thank you for these remarks ! $\endgroup$
    – coboy
    Oct 22, 2023 at 19:09

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