Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

1) What is the application of Doob's inequality? Can we use Doob's inequality ($L^1$) to prove the convergence (maybe almost surely) of a martingale?

Doob's inequality: Let $X$ be a submartingale taking on non-negative real values, that is, for all times $s$ and $t$ with $s < t$, $E[X_t\mid\mathcal F_s]\geq X_s$. Then, for any constant $C > 0$ and $T>0$ we have

$$ \mathbf{P} \left[ \sup_{0 \leq t \leq T} X_{t} \geq C \right] \leq \frac{\mathbf{E} \big[ X_{T} \big]}{C}. $$

Doob's inequality ($L^p$): Let $X$ be a martingale,

$$S_{t} = \sup_{0 \leq s \leq t} X_{s},\quad\text{for}\ p > 1$$

$$\| X_{T} \|_{p} \leq \| S_{T} \|_{p} \leq \frac{p}{p-1} \| X_{T} \|_{p}.$$

Maybe it is not clear enough. I know that to prove the $L^p$, $p>1$, convergence of a martingale, we can use Doob's inequality in $L^p$ form. However, it seems in the proof of convergence in $L^1$, that Doob's inequality is not used (as far as I understand, it is not enough - we need uniformly integrability).

There are several formly analogous inequalities in probability theory, for example, Komolgorov's inequality, Doob's inequality, and an analogous Doob's inequality in ergodic theory. They all estimate the probability or expectation of a random variable, $|X_n|>M$ where $M$ is a constant. Why are they useful?

2) What is the application for Doob's decomposition theorem? Is it only formly?

share|improve this question
    
If you think it is not a good question, please tell me how to improve it, thanks –  Ginger Nov 27 '12 at 23:38
    
For the second question, in Durrett's book, it says that doobs decomposition is usefull in reducing questions about submartingales to questions about martingales. As far as I understand, if there is a question about submartingale, we can consider to use decomposition to transfer it to a problem on martingale. –  Ginger Nov 28 '12 at 0:16
1  
It is correct that Doob's inequality doesn't apply for $p=1$. In particular a non-negative martingale is bounded in $L^1$ but it is not necessarily uniformly integrable (and hence Doob's inequality can not be true for $p=1$). The most common technique used in proving a.s. convergence (and therefore also $L^1$ convergence) of a martingale is Doob's upcrossing inequality. –  Stefan Hansen Nov 28 '12 at 8:34
    
What I mean Doob's inequality in L1 is following: Let X be a submartingale taking non-negative real values, That is, for all times s and t with s < t, $\mathbf{E} \big[ X_{t} \big| \mathcal{F}_{s} \big] \geq X_{s}.$ Then, for any constant C > 0, $ \mathbf{P} \left[ \sup_{0 \leq t \leq T} X_{t} \geq C \right] \leq \frac{\mathbf{E} \big[ X_{T} \big]}{C}$. –  Ginger Nov 28 '12 at 9:13
    
Sorry, I thought of the inequality of moments which only holds for $p>1$. Just disregard my comment. Also you could type up these definitions of Doob's inequality and Doob's decomposition theorem in your question. It makes it much easier to read :) –  Stefan Hansen Nov 28 '12 at 9:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.