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Consider a twice diferentiable function $F$ on $R$ with bounded first derivative $F'$ and a Brownian motion $W$. Show that $F(W_t)-\frac{1}{2} \int_{0}^{t} F'' (W_s)ds$ is a true martingale.

I tried do show it using this, but I only got confused and did not find any solution.

  1. If $M$ is a local martingale with continues trajectories, then it is a true martingale and $E(M_t^2) < \infty$ for all $t\geq 0$ Or

  2. If $M$ is a local martingale with continues trajectories, then it is a true martingale and $E([M]_t)<\infty$ for all t.

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  • $\begingroup$ I'm not sure where your statements 1 and 2 came from, but I don't think they're true in general. A local martingale with continuous trajectories does not have to be a true martingale. There is a standard example here. $\endgroup$ May 28, 2019 at 21:23

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If $F$ were twice continuously differentiable, one could use Ito's Lemma to obtain the stochastic differential equation for $Y_t = F(W_t)$ as

\begin{equation} \mathrm{d}Y_t = \frac{1}{2}F^{\prime \prime}(W_t) \mathrm{d}t + F^\prime(W_t) \mathrm{d}W_t. \end{equation}

It follows that \begin{equation} F(W_t)-\int_0^t \frac{1}{2}F^{\prime \prime}(W_s) \mathrm{d}s = \int_0^t F^\prime(W_s)\mathrm{d}W_s, \end{equation} which is clearly a martingale as $F^\prime$ is bounded.

If $F^{\prime \prime}$ is not continuous, I am not sure the assertion holds anymore.

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