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Using Itô’s Lemma to Seek a Martingale Transform

Article Quant Q&A · Author: AB_IM
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Summary

The document considers a process with Brownian diffusion and a time-dependent drift, then asks how to multiply it by an exponential factor so that the result is a martingale. The proposed approach applies Itô’s product rule to the process and the finite-variation exponential. The cross-variation term vanishes because the exponential has finite variation, leaving a stochastic term and drift terms. Setting the total drift coefficient to zero leads to a proposed choice of the function in the exponential.

Key ideas

  • Itô’s product rule separates the stochastic and drift components of the transformed process.
  • The exponential of an ordinary time integral has finite variation, so its cross-variation with the diffusion term is zero.
  • A martingale candidate must have zero drift under suitable integrability conditions.
  • The proposed drift cancellation depends on dividing by the process value and may fail when that value is zero.
  • The answer does not establish that the resulting local martingale is a true martingale.

Tags

StatisticsRisk management
Full text
# Transformation into Martingale


# 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).

Thanks for your help all.

## Answer by pindropsilence (score 1, accepted)

https://quant.stackexchange.com/a/17304

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}

## Answer by AFK (score 0)

https://quant.stackexchange.com/a/17291

$F=0$ seems like a good choice.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.