2
$\begingroup$

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.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

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}

Improve this answer
$\endgroup$
0
$\begingroup$

$F=0$ seems like a good choice.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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