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It is shown here, without using stochastic calculus, that if $W_t$ is a standard Brownian motion, then $$ f(W_t)-\frac{1}{2}\int_0^t f''(W_s)ds $$ is a martingale, where $f\in C^2$ and compactly supported. According to Problem 4.4 in Chapter 5 of Karatzas and Shreve's Brownian Motion and Stochastic Calculus, the converse is also true. Namely, if $W$ is a continuous adapted process with $W_0=0$ such that $$ f(W_t)-\frac{1}{2}\int_0^t f''(W_s)ds\tag{1} $$ is a martingale whenever $f\in C^2$ and compactly supported, then $W$ is a standard Brownian motion. This is essentially part of the proof of Lévy's characterization of Brownian motion.

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A martingale is characterized by its characteristic function.

  • apply the property with $f(x) = x$ proves that $W$ is a martingale
  • fix $u$ and apply with $f(x) = \exp(iux)$; define $g(u, t) = \mathrm E(\exp(iuW_t)); M(t) = f(W_t) - \frac 12\int_0^t f''(W_s) ds$ and you get $$ 1 = M(0) = \mathrm EM(t) = g(u,t) + \frac 12 u^2\int_o^t g(u, s) ds $$

whose solution is $$ g(u, t) = \exp\left(-\frac 12 u^2 t\right) $$

From this, you get that $W$ is a brownian motion.

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    $\begingroup$ you didn't show independence of increments of BM. $\endgroup$ – Calculon Apr 25 '15 at 0:21
  • $\begingroup$ The second characterisation is easy: take $f(x) = x^2$. $\endgroup$ – mookid Apr 25 '15 at 0:57
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    $\begingroup$ @mookid Of course I will, but could you please complete your chosen method of proving this, by showing that the increments are stationary and independent? $\endgroup$ – theLowerCrust Apr 25 '15 at 1:05
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    $\begingroup$ @theLowerCrust I think that you can find quite a complete proof here: imada.sdu.dk/~njn/MM24/levy.pdf $\endgroup$ – mookid Apr 25 '15 at 1:10
  • $\begingroup$ @mookid The functions $f(x) := x$ and $f(x) := x^2$ don't have a compact support. And in the question it is only assumed that the martingale property holds for compactly supported functions $f$. Moreover, your answer doesn't show that the process is a Brownian motion; the only thing you prove is that $W_t \sim N(0,t)$. $\endgroup$ – saz Apr 25 '15 at 18:08
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Step 1:

The assumption holds for all $f \in C_b^2$, i.e. functions which are twice differentiable and have bounded continuous derivatives.

To this end, pick a function $\chi \in C^2_b$ such that $0 \leq \chi \leq 1$, $\chi(x) = 1$ for $x \in (-1,1)$ and $\chi(x)=0$ for all $|x|>2$. Then the function

$$f_n(x) := f(x) \cdot \chi \left( \frac{x}{n} \right), \qquad x \in \mathbb{R}, \tag{1}$$

is twice differentiable and has compact support. By assumption,

$$\mathbb{E} \left( f_n(W_t) - \frac{1}{2} \int_0^t f_n''(W_r) \, dr \mid \mathcal{F}_s \right) = f_n(W_s) - \frac{1}{2} \int_0^s f_n''(W_r) \, dr$$

for all $s \leq t$. Definition $(1)$ implies that $$\|f_n\|_{\infty}+ \|f_n'\|_{\infty} + \|f_n''\|_{\infty} \leq C$$ for some constant $C>0$ (which does not depend on $n$). Therefore, it follows from the continuity of $f,f''$ and the (conditional) dominated convergence theorem that

$$\mathbb{E} \left( f(W_t) - \frac{1}{2} \int_0^t f''(W_r) \, dr \mid \mathcal{F}_s \right) = f(W_s) - \frac{1}{2} \int_0^s f''(W_r) \, dr.$$

Step 2:

$(W_t)_{t \geq 0}$ is a Brownian motion.

The claim is proved in this answer (note that, by the first step, the assumptions in the linked question are satisfied). Don't hesitate to ask if you don't get along with it (notation is a bit different there).

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  • $\begingroup$ Thanks a lot for your answer and comment. mookid actually provided a link which showed how to complete the argument and obtain independence. So it remained to approximate $f$ as you did, which I was actually fine with. But I definitely appreciate your careful answer, which I'm sure will also help others on this site. $\endgroup$ – theLowerCrust Apr 25 '15 at 19:13

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