2
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Hmm not quite, this only shows it is a local martingale. $\endgroup$ – Lost1 Jan 2 '14 at 23:55
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.