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In the book "Brownian Motion and Stochastic Calculus" by Karatzas/Shreve, they state the following problem (chapter 5, problem 4.4):

A continuous, adapted process $W= \{W_t,\mathcal{F}_t;0\leq t < \infty \}$ is a Brownian motion if and only if \begin{equation} f(W_t) - f(W_0) - \frac{1}{2} \int_0^t f^{\prime\prime}(W_s) \mathrm{d}s,\quad \mathcal{F}_t; \quad 0 \leq t <\infty \end{equation} is a continuous local martingale for every $f\in C^2(\mathbb{R})$.

For the "only if" part, one applies Ito's formula and gets that \begin{equation} f(W_t) - f(W_0) - \frac{1}{2} \int_0^t f^{\prime\prime}(W_s) \mathrm{d}s= \int_0^t f^{\prime}(W_s) \mathrm{d}W_s, \end{equation} which is a continuous local martingale.

But I am struggling with the "if" part. Does anyone have a hint on how to prove that?

Thank you very much!

Best, Luke

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  • $\begingroup$ That's a standard result; look up Lévy's characterization of Brownian motion. $\endgroup$
    – saz
    Jun 13, 2019 at 16:03

1 Answer 1

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If you take $f(x) = x$, you see that $W_t$ is a continuous local martingale.

Then, taking $f(x) = x^2$ gives us that $W_t^2 - W_0^2 - t$ is also a continuous local martingale so that $W_t$ has quadratic variation at time $t$ equal to $t$. Now apply Levy's characterisation of Brownian motion to conclude.

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  • $\begingroup$ Thank you very much! $\endgroup$ Jun 13, 2019 at 16:16

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