# 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

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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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