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In our class we have proven Doob's inequality for discrete martingale, i.e. Let $(M_n)_{n \in \mathbb{N}}$ a martingale, then

$$ \| \max_{0\le k\le n} M_k\|_p \le C_p \|X_n\|_p$$

for $p\in (0,\infty)$. How can I use this to prove the same result for a continuous-time martingale which is right continuous?

EDIT:

After the hint of GEdgar, I did the following: Let $A=[0,T] \cap \mathbb{Q}$ a dense subset of $[0,T]$. For each $n$ let $A_n$ be a finite set, such that $A_n \subset A_{n+1}$ and $A=\cup_{n\in \mathbb{N}} A_n$. I'm quite sure, that I can choose such sets, but could someone give a concrete example here? Now we have: $$\sup_{t\in A_n} M_t\le \sup_{0\le t \le T} M_t\, \forall n,$$ hence $$\sup_n\sup_{t\in A_n} M_t\le \sup_{0\le t \le T} M_t.$$ On the other hand, let $(q_m)$ be a decreasing sequence of rationals in $[0,T]$ such that $M_t=\lim_m M_{q_m}$. Let $B_m$ be the set in $(A_n)$ with $q_m\in A_n$ then for all $m\ge1$ $$M_{q_m}\le \sup_{t\in B_m}\le\sup_n\sup_{t\in A_n}M_t.$$ Hence $M_t=\lim_m M_{q_m}\le\sup_n\sup_{t\in A_n}M_t$. Therefore we have shown: $$\sup_{0\le t\le T}M_t=\sup_n\sup_{t\in A_n}M_t.$$

Clearly $N_n:=\sup_{t\in A_n}M_t$ is increasing in $n$, measurable and converges to $\sup_{0\le t\le T}M_t$. Applying discrete Doob to $A_n$ leads to $E[N_n^p]\le c(p)E[M_{\max A_n}^+]$. The left hand side converges to $E[\sup_{0\le t\le T}]$ by Monotone Convergence. Why does the RHS converge to the right thing? Beside of my two questions, is my proof correct?

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3  
Use right-continuity to show the sup is approximated by a finite max? –  GEdgar Jun 3 '12 at 12:50
    
@ GEdgar: How can I approximate the sup by finite max? And how should I handle the $L^p$ norm? Some more details would be appreciated. –  user20869 Jun 3 '12 at 16:22
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If $M(t)$ is right-continuous, then $$ \sup_{0\le t < T}M(t) = \sup_{n \in \mathbb N} \max_{0\le k < T 2^n} M(k/2^n) $$ –  GEdgar Jun 3 '12 at 16:49
    
@ GEdgar: I added a proof of your hint as well as of the whole argument. There are still some small questions around. It would be appreciated, if someone could answer them. –  user20869 Jun 4 '12 at 6:59
    
Now: if $M(t)$ is a martingale, then $|M(t)|^p$ is a submartingale. Conclude $\|M(t)\|_p$ increases with $t$. –  GEdgar Jun 4 '12 at 13:36

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