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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?

minimax
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2 Answers2

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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

Did
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TheBridge
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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 $\frac{1}{2}t^2$.