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).

Thanks for your help all.


2 Answers 2


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.

  • 1
    $\begingroup$ Sorry I edited the question, I forgot the drift which made it a bit trivial $\endgroup$
    – ABIM
    Commented Apr 7, 2015 at 18:58

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