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I have $E[x(t)^2]\leq A\operatorname{exp}(Bt)+C/Bt$ it's clear that for a finite time less than $T$, x(t)^2 is a "local martingale" because $\lim E[x(t)^2]<\infty$. But one can see that if $t$ tends to plus infinity then the limit of $E[x^2]$ is also infinity, so this is not a martingale, and we can not apply the convergence theorem.

*Can one tell me if my analysis is right, or if there is any way to extend the local martingale to a martingale to be able to apply the martingale convergence theorem.

*Is there any relation between $x(t)^2$ and $x(t)$ in sens of martingales.

fidel
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  • The question is actually kind of confuse. You just have the estimation as hypothesis over X to conclude that $X^2$ is a local or a true martingale ? Because if it's the case you can conclude anything since you need to check the martingale property. – Paul Jun 01 '13 at 16:07

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I begin by answering your last question.

Yes, if $X $ is continuous square integrable martingale then $(X_t^2 -\langle X \rangle_t)$ is a continuous martingale.

You are write about not be abble to conclude that $(X^2_t)_{t\geq 0}$ is a martingale since you can not apply the convergence theorem to conclude that $X^2_ t \in L^1(\Omega, \mathcal F_t, \mathbb P) , \ \forall t \geq 0$ (in a filtred probability space $(\Omega, \mathcal F, \mathcal F_t, \mathbb P)$.

However the fact that the $RHS$ of you estimation goes to infinity when $t$ goes to zero or infinity doesn't mean $X^2$ do so.

Paul
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  • i thing now that even for a finite time x^2 is not a local martingale, because limE[x(t)2]=∞ when t=0. – fidel Jun 01 '13 at 15:52
  • but today i read that to prove the martingale in infinite time t>=0, this means that 0<=t<infinity, is this true, because before i have a thouthg that t>=0 means 0<=t<=infinity, because with the first meaning of T>=0 we will have Kexp(t) is a martingale and local martingale even limE[x(t)^2]=∞ when t tends to infinity – fidel Jun 01 '13 at 16:10
  • $\forall t\geq 0$ – Paul Jun 01 '13 at 16:18
  • ∀t≥0 is 0≤t<∞ or 0≤t≤∞ – fidel Jun 01 '13 at 16:22
  • @fidel: second one – Paul Jun 01 '13 at 16:27
  • see here: http://books.google.fr/books?id=JYzW0uqQxB0C&pg=PA187&dq=martingale+converge+theorem&hl=fr&sa=X&ei=OSGqUYqnMMPKkwX27oHQAg&ved=0CDQQ6AEwAA#v=onepage&q=martingale%20converge%20theorem&f=false – fidel Jun 01 '13 at 16:29
  • book: introduction to stochstic calculus with application, 7.3 martingale convergence – fidel Jun 01 '13 at 16:32
  • note that you have to check if $\sup_{t\geq 0} \mathbb E[|X_t|]< \infty$ – Paul Jun 01 '13 at 16:35
  • note also that this is equivalent to the first condition that M is a martingale on 0≤t<∞, i don't need to prove supt≥0E[|Xt|]<∞ in my case – fidel Jun 01 '13 at 16:40
  • Can you reformulate your question please? I must confess I am lost about what are your asking – Paul Jun 01 '13 at 17:10
  • this is my question, we have E[x(t)^2]≤Aexp(Bt). i would like to know if x(t)^2 is a martingale, and can we apply the martingale converge theorem. – fidel Jun 03 '13 at 01:34