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This question already has an answer here:

Let ${W_t}$ be 1 dim Brownian motion and $X_t:=\exp(t/2)\cos W_t$ $t\in[0,T]$.

Show that $X_t$ is martingale.

My try is below.

I understood $df(t,W_t)=-\exp(t/2)\sin xdW_t$ , but I don't know why it become $X_t=1-\int_0^t \sin X_sdW_s$.

$f(0,X_0)=e^0 \sin X_0=1$? $X_0=?$

Please tell me.

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marked as duplicate by Nate Eldredge, user61527, user940, qwr, Davide Giraudo Jun 7 '14 at 20:00

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It seems to me that you applied Itô's formula without understanding properly how to apply it. The equality

$$d f(t,W_t) = - \exp \left( \frac{t}{2} \right) \sin x \, dW_t$$

is not correct; instead it should read

$$d f(t,W_t) = - \exp \left( \frac{t}{2} \right) \sin \color{red}{W_t} \, dW_t.$$

This might look like a minor mistake to you, but, in fact, it is crucial. Then, it follows from the very definition of the stochastic exponential that

$$X_t - X_0 = f(t,W_t)-f(0,W_0) = \int_0^t \, df(s,W_s) =- \int_0^t \exp \left( - \frac{s}{2} \right) \sin W_s \, dW_s.$$

As $|\exp(-s/2) \cdot \sin W_s| \leq 1$, this shows that $(X_t)_{t \geq 0}$ is a martingale since it is a stochastic integral (of a properly integrable function). Moreover, by definition of $X_t$,

$$X_0 = \exp(0) \cdot \cos 0 = 1 \cdot 1 = 1.$$

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  • $\begingroup$ $X_0=e^0\cos W_0=1*1=0$ where $W_0=0$ ($\because$ Brownian motion's definition) right? $\endgroup$ – KMHR Jun 7 '14 at 14:03
  • $\begingroup$ @KMHR Exactly... $\endgroup$ – saz Jun 7 '14 at 14:11

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