Show that $M(t)=e^{\frac{t}{2}}\sin(B(t))$ is a martingale by using Ito s formula.

$B(t)$ is brownian-motion.

i must show $s \le t $ then $E(M(t)|\mathcal F_{s})=M(s)$ but i dont know how use ito formula
thanks for help

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    $\begingroup$ How would you start? What are you having difficulty with? Taking (partial) derivatives? Recalling Ito's formula? Something else? $\endgroup$ – cardinal Jan 1 '14 at 17:39

Your exact question is answered on Quant Stackexchange here. Essentially, if you can express your stochastic process $M(t)$ as $M(t) = \int A(t) dB(t)$ where $B(t)$ is a martingale, then $M(t)$ is also a martingale. Since Brownian motion is a martingale, it suffices to express $M(t)$ as $\int A(t) dB(t)$ where $A(t)$ is something and $B(t)$ is Brownian motion. In the notation of stochastic calculus, you have to show $$dM(t) = A(t) dB(t)$$ for some $A(t)$.

Ito's formula allows you to calculate $dM(t)$ because $M(t) = f(t, B(t))$ where $f(x,y) = e^{x/2}\sin(y)$ and $B(t)$ is Brownian motion. Ito's formula gives you

$$dM(t) = \left(\frac{\partial f}{\partial x} + \frac{1}{2} \frac{\partial^2 f}{\partial y^2}\right) dt + \frac{\partial f}{\partial y}(t, B(t)) dB(t)$$

It doesn't matter what $\frac{\partial f}{\partial y}$ is. You just need to show that

$$\frac{\partial f}{\partial x} + \frac{1}{2} \frac{\partial^2 f}{\partial y^2} = 0$$

and it follows that $M(t) = (something) \times dB(t)$ is a martingale.

Edit: @Lost1 points out in the comments that this only proves that $M(t)$ is a local martingale. It also needs to satisfy some grwoth conditions in order to be a martingale. According to the Wikipedia article on "local martingale" it is sufficient that for every $\varepsilon > 0$ and every $t$ there exits a constant $C(\varepsilon, t)$ such that $|f(s,x)| \le Ce^{\varepsilon x^2}$ for all $x \in \mathbb{R}$ and all $0 \le s \le t$. In this case, $|f(s,x)| \le e^{s/2} \le e^{t/2}e^0$ seems to be sufficient. Feel free to edit if you don't think this is a good approach.

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  • $\begingroup$ Hmm not quite, this only shows it is a local martingale. $\endgroup$ – Lost1 Jan 2 '14 at 23:55
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    $\begingroup$ And this guy posted a lot of stochastic analysis questions on here and math.se without trying them. Many of them were closed. $\endgroup$ – Lost1 Jan 2 '14 at 23:59
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    $\begingroup$ I am happy with the solution now. In fact, you can add $\cos(B_t)e^{t/2}$ to $i\sin(B_t)e^{t/2}$ to get something which you can show to be a complex martingale using only characteristic functions of normal distribution. So real and imaginary part must be martingales. @pualambagher please! At least look at the formula tell people how you are stuck. As it stands, none of the questions you asked meets the standard required of the forum. I will help you if you will give more details on what you tried! $\endgroup$ – Lost1 Jan 3 '14 at 11:45

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