# Upper bound for the $\sup$ of a martingale defined as a stochastic integral of a general continuous martingale

Consider a probability filtred space $(\Omega, \mathcal F, \mathcal F_ t, \mathbb P)$ and a continuous $\mathcal F _t$-martingal starting from $0$, $M = (M_t)_{t \geq 0}$, such that $\left \langle M \right \rangle_\infty \leq 1$ $\mathbb P$-ps. Now, we define by recurence $\forall n \in \mathbb{N}$ $$I^{(o)}_t \equiv 1, \ I^{(n+1)}_t \int _0 ^t I^{(n)}_s d M_s \ , \ t \geq 0$$

The question: How to show that $I^{(n)}$ is a $\mathcal F _t$-martingal with $I^{(n)}_t$ in $L^2(\Omega, \mathcal F, \mathbb P)$ and that we have

$$\mathbb{E} \left \{ \sup_ {t \geq 0} \left | I^{(n)}_t \right | ^2 \right \} \leq 4^n$$

Element of answer: for the last inequality we can maybe start using Burkholder-Davis-Gundy inequality, then we'll have :

\begin{align} \mathbb{E} \left \{ \sup_ {0 \leq t \leq T} \left | I^{(n)}_t \right | ^2 \right \} & \overset{BDG}{\leq} C\mathbb{E} \left \{ \left \langle I^{(n)} \right \rangle_T \right \} \\&= C\mathbb{E} \left \{ \int_0 ^T \left |I^{(n-1)}_s\right | ^2 d\left \langle M\right \rangle _s \right \} \\ & \overset{Fubini}{=} C \int_0 ^T \mathbb{E} \left \{\left |I^{(n-1)}_s\right | ^2 \right \}d\left \langle M\right \rangle _s \\ & \overset{Ito's isometry}{=} C \int_0 ^T \mathbb{E} \left \{\int _0 ^s\left |I^{(n-2)}_s\right | ^2 d\left \langle M\right \rangle _{s_1}\right \}d\left \langle M\right \rangle _s \\ &=(...) \\& = C \int_0 ^T \int_0 ^s \int_0 ^{s_1} ... \int_0 ^{s_n-2} 1 \ d\left \langle M\right \rangle _{s_{n-2}} ... d\left \langle M\right \rangle _{s_2} d\left \langle M\right \rangle _{s_1} d\left \langle M\right \rangle _{s}\end{align}

EDIT

Did sugestion: Could you please, check if I understood ?

You should notice that all the steps where the Fubini's theorem was applied are wrong!

\begin{align} \mathbb{E} \left \{ \sup_ {0 \leq t \leq T} \left | I^{(n)}_t \right | ^2 \right \} & \overset{BDG}{\leq} 4\mathbb{E} \left \{ \left \langle I^{(n)} \right \rangle_T \right \} \\&=4\mathbb{E} \left \{ \int_0 ^T \left |I^{(n-1)}_s\right | ^2 d\left \langle M\right \rangle _s \right \} \\ & \overset{Fubini}{=} 4 \int_0 ^T \mathbb{E} \left \{\left |I^{(n-1)}_s\right | ^2 \right \}d\left \langle M\right \rangle _s \\ & \overset{}{\leq} 4 \int_0 ^T \mathbb{E} \left \{\sup_{0\leq u \leq s}\left |I^{(n-1)}_u\right | ^2 \right \}d\left \langle M\right \rangle _s \\ & \overset{BDG}{=} 4 ^2 \int_0 ^T \mathbb{E} \left \{\int _0 ^s\left |I^{(n-2)}_s\right | ^2 d\left \langle M\right \rangle _{s_1}\right \}d\left \langle M\right \rangle _s \\ &=(...) \\& \leq 4^n \int_0 ^T \int_0 ^s \int_0 ^{s_1} ... \int_0 ^{s_n-2} 1 \ d\left \langle M\right \rangle _{s_{n-2}} ... d\left \langle M\right \rangle _{s_2} d\left \langle M\right \rangle _{s_1} d\left \langle M\right \rangle _{s}\\ &\leq 4^n\end{align} because $\left \langle M \right \rangle = (\left \langle M \right \rangle_t)_{t\geq 0}$ is an increasing process with $\left \langle M \right \rangle_\infty \leq 1$

This is direct if one recalls that $C=4$ and if one replaces $\left |I^{(n-1)}_s\right | ^2$ by $\sup\limits_{s\geqslant 0}\left |I^{(n-1)}_s\right | ^2$ as soon as possible (and note that the localization by $t\leqslant T$ is not needed).

• Can you detail the calculation please?
– Paul
Commented Dec 31, 2012 at 1:13
• It IS detailed (just read).
– Did
Commented Dec 31, 2012 at 1:20
• Could you please, check at the end of the question if I understood ? I still do not understand how to complete the proof neither why it is not necessary to localizate by $t\leq T$.
– Paul
Commented Dec 31, 2012 at 1:38
• Edit part, second line of displayed equations: $C=4$. Third line and every subsequent line: this is not Fubini, watch out for random variables vs constants. Everywhere: please follow closely my hint.
– Did
Commented Dec 31, 2012 at 9:05
• OP: Why not trying to follow my suggestions? One of the first things to do is to correct the identity in the third line (the one topped by the mention Fubini), which is invalid because $d\langle M\rangle_s$ is random, hence it must stay inside the integral.
– Did
Commented Jan 1, 2013 at 15:43