Let $X:\Omega\times[0,T]\longrightarrow \mathbb{R}$ be a non-negative submartingale with cadlag sample paths, $p>1$ and $C>0$. Then I understand that Doob's Maximal Inequality is given by:
\begin{equation*}
\mathbb{P}\left( \sup_{t\in [0,T]} X_t \ge C\right) \le \frac{\mathbb{E}\left[ X_T\right]}{C}
\end{equation*}
And Doob's $\mathcal{L}^p$ Inequality is given by:
\begin{equation*}
\left\Vert \sup_{t\in [0,T]} X_t\right\Vert_p \le \frac{p}{p-1}\left\Vert X_T\right\Vert_p
\end{equation*}
My questions are:
1)What happens when I take $T\rightarrow\infty$ and could you please provide a careful proof of why it is the case (of whatever happens as we take $T\rightarrow\infty$)?
I would suspect that the following inequalities hold:
\begin{equation*}
\mathbb{P}\left( \sup_{t\in \mathbb{R}^+} X_t \ge C\right) \le \frac{\sup_{T\in \mathbb{R}^+}\mathbb{E}\left[ X_T\right]}{C}
\end{equation*}
And:
\begin{equation*}
\left\Vert \sup_{t\in \mathbb{R}^+} X_t\right\Vert_p \le \frac{p}{p-1}\sup_{t\in \mathbb{R}^+}\left\Vert X_T\right\Vert_p
\end{equation*}
Basically because $\mathbb{E}\left[X_T\right] \le \mathbb{E}\left[ X_{T+1}\right]$ $\forall T\in \mathbb{R}^+$ and the sets event/random variables on the LHS of the inequalities are also increasing. (I think I actually kind of proved this to myself…but I am extremely in-confident about my own abilities and hence am asking for a) a sanity check b) it would be nice to see how someone else would prove this as I may learn a new trick/technique/insight from their proof!)
2) Intuitively why is the assumption of cadlag sample paths so crucial?
3) (This really is more of a sanity check) do these inequalities hold (both for $T\in \mathbb{R}$ and as $T\rightarrow\infty$) if we remove the assumption that $X$ is a non-negative martingale and subsequently write the Doobs Maximal Inequality with absolute values? I.e.
\begin{equation*}
\mathbb{P}\left( \sup_{t\in [0,T]}\left\vert X_t\right\vert \ge C\right) \le \frac{\mathbb{E}\left[\left\vert X_T\right\vert\right]}{C}
\end{equation*}
(and Doobs $\mathcal{L}^p$-Inequality does not change, because the absolute values are automatically already built in)
Thanks in advanced.
Best Answer
Define $\displaystyle Y_n:=\sup_{t\in [0,n]} X_t$. Then $\displaystyle Y_n\uparrow \sup_{t\geqslant 0} X_t$ and the monotone convergence theorem gives that $$\begin{equation*} \left\Vert \sup_{t\in \mathbb{R}^+} X_t\right\Vert_p \leqslant \frac{p}{p-1}\sup_{t\in \mathbb{R}^+}\left\Vert X_t\right\Vert_p. \end{equation*}$$
The assumption of càdlàg paths permits us to go from the discrete case to the continuous one.
Yes, it is true. Actually, a non-negative sub-martingale with right continuous paths is the absolute value of a martingale.