Solved – show $M(t)=e^{frac{t}{2}}sin(B(t))$ is a martingale by using Ito s formula

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

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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