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I'm trying to show that for a standard Brownian motion and some twice continuously differentiable function $f$ that $f(B_t)$ is a local martingale iff $f'' = 0$.

Applying Ito's formula gives, and since $\langle B_s\rangle = s$: $$f(B_t) = f(B_0) + \int_0^t f'(B_s) dB_s + \frac{1}{2} \int_0^t f''(B_s) ds$$

Ito's formula tells us that $f(B_t)$ is a semimartingale, so has unique decomposition $$f(B_0) + M_t + A_t$$ where $M_t$ is a local martingale and $A_t$ is a process of finite variation. Presumably we use uniqueness of the decomposition to match up the terms of these expressions and the result follows, so my only questions are why

  1. $\int_0^t f'(B_s) dB_s$ is a local martingale?
  2. $\int_0^t f''(B_s) ds$ is a process of finite variation?

Sorry for the long preamble, thank you for any help!

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  • 2
    $\begingroup$ Why not take a look at the literature? The (local) martingale property of the stochastic integral with respect to Brownian motion is discussed in (almost) any book on this topic. $\endgroup$ – saz Jun 10 at 16:35
  • $\begingroup$ I think I understand why the integral w.r.t brownian motion is a martingale, and it doesn't seem like the inclusion of $f'(B_s)$ makes much difference. The integral in (2) seems like a less standard result though, what exactly is the integral of $f''(B_s)$ w.r.t. $s$ defined as? $\endgroup$ – watt Jun 10 at 17:41

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