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Let $(B_t)$ a Brownian motion. Let $f:\mathbb R\to\mathbb R$ such that $f\in\mathcal C^2(\mathbb R)$ and such that $f''$ is bounded. Show that $$\lim_{h\to 0^+}\frac{\mathbb E[f(B_{t+h})\mid B_t=x]-f(x)}{h}=\frac{f''(x)}{2}.$$

In the proof, it's written that: by the Markov property of a Brownian motion, if $B_t^x$ is a brownien motion sucht that $B_0=x$, then

$$\mathbb E[f(B_{t+h})\mid B_t=x]\underset{(1)}{=}\mathbb E[f(B_h)\mid B_0=x]\underset{(2)}{=}\mathbb E[f(B_h^x)]\underset{(3)}{=}\mathbb E[f(x+B_h)].$$

I think that that fact that $f$ is continuous, $f(B_t)$ is also a brownian motion, no ? and by translation of the time, we have $(1)$. But I don't understand neither $(2)$ nor $(3)$.

idm
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    If $f=1$ then $f(B_t)$ is not a Brownian motion. – Patissot Jun 12 '15 at 12:26
  • Ok, then how can I get all theses equality ? – idm Jun 12 '15 at 12:53
  • no body can answer ? – idm Jun 12 '15 at 15:53
  • The first step is in fact simply the Markov property of $(B_t)_t$. So what exactly do you not understand about this equality? – saz Jun 12 '15 at 17:00
  • Why $(f(B_t))$ would have the Markov property ? Because it's not necessarily a Brownian motion... Maybe if a stochastic process $(X_t)$ has the Markov property, then $f(X_t)$ has also the markov property for all function in $\mathcal C^2(\mathbb R)$ but I don't know such a theorem. Could you tell me more ? Thanks, – idm Jun 12 '15 at 17:49

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