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Let $W_t$ be standard Brownian motion. It is well known that $W_t^2-t$ is a martingale. One way to show this is by applying Ito's lemma to calculate that $d(W_t^2-t)/dt = 2W_t dW_t$, which has no drift. Therefore $W_t^2-t$ is a martingale. I am a novice in stochastic process so I want to ask which theorems one use in this proof?

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From Itô's lemma you have :

$Y_t=W_t^2 - t= \int_0^t W_sdW_s$

So what you have here is that $Y_t$ is a local martingale. To prove that it is indeed a martingale it suffices to show that :

$\forall t>0, E[\langle Y\rangle_t]<\infty$

as you can check in lemma 3 which is not too hard I think.

Best regards

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  • $\begingroup$ Added \langle and \rangle (hope you don't mind) and upvoted (hope you don't mind either...). $\endgroup$ – Did Oct 15 '13 at 9:49
  • $\begingroup$ @ Did : It's ok as long as it is not a mathematical mistake in my post :-). Regards $\endgroup$ – TheBridge Oct 15 '13 at 10:00
  • $\begingroup$ why is $Y_t$ a local martingale? $\endgroup$ – mastro Jul 14 '15 at 17:05
  • $\begingroup$ @ mastro : Because : $\forall t>0, E[\langle Y\rangle_t]<\infty$ which is enough to show that $Y$ is a martingale (once an easy calculation is done). Best regards. $\endgroup$ – TheBridge Jul 14 '15 at 20:33
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We can use Corollary 3.2.6 in page 33 of Oksendal's book, where the definition of $\mathcal{V}(0,T)$ is in page 25. So to prove $\int_0^tW_sdW_s$ is a martingale, we can simply verify that $E(\int^t_0W_s^2ds)<\infty$, which is true since this integral is $\frac{1}{2}t^2$.

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