# Transformation into Martingale

If $f$ is some function of BV on $\mathbb{R}$ and $dZ_t = f(W_t)dW_t + \mu_t dt$ ($W_t$ is a $1$-dimensional standard Brownian Motion), then what choice of real valued function $F$ makes: \begin{equation} M_t:= Z_t e^{\int_0^tF(Z_t)dt} \end{equation} into a martingale?

I feel that I sould use Ito's product rule to solve this and the fact that the term $e^{\int_0^tF(Z_t)dt}$ must be of B.V. (since it is a Riemman integral), however I'm fuzzy on the details (as I'm completetly new to this type of problem).

As you have guessed correctly, these type of questions can be answered using Ito's Lemma.We have: \begin{equation} d(M_t)= d(Z_t e^{\int_0^tF(Z_u)du})=d(Z_t) e^{\int_0^tF(Z_u)du}+Z_t d(e^{\int_0^tF(Z_u)du})+d(Z_t)d(e^{\int_0^tF(Z_u)du}) \end{equation}

For the first two terms on R.H.S, we have: \begin{equation} d(Z_t) e^{\int_0^tF(Z_u)du} = (f(W_t)dW_t + \mu_t dt) e^{\int_0^tF(Z_u)du} \end{equation}

and \begin{equation} Z_t d(e^{\int_0^tF(Z_u)du}) = Z_t e^{\int_0^tF(Z_u)du}F(Z_t)dt \end{equation}

the third term does not contribute anything.

Now, for martingale condition to hold, equate the coefficient of time dependent term to zero and we get

\begin{equation} F(Z_t) = -\mu_t/Z_t \end{equation}

$F=0$ seems like a good choice.

• Sorry I edited the question, I forgot the drift which made it a bit trivial Apr 7 '15 at 18:58