# change of measure martingale

• Mar 30th 2011, 12:41 AM
Juju
change of measure martingale
Hallo!

There's a given probability space $(\Omega,\mathcal{F},P)$ equipped with a filtration $(\mathcal{G}_t)_{t \in [0,T]}$ and a subfiltration $\mathcal{F}_t \subset \mathcal{G}_t$.

And now I want to do a change of measure

$\frac{dQ}{dP}|_{\mathcal{F}_t}= Z_t$, with

$Z_t=\exp\{-\int\limits_0^t\frac{\mu_t -r}{\sigma}dW_t-\frac{1}{2}\int\limits_0^t(\frac{\mu_t -r}{\sigma})^2dt\}$

where
$\mu_t:=E[\mu|\mathcal{F}_t]$ is an estimator for the unknown parameter $\mu$ which has a normal prior distribution $\mathcal{N}(\mu_0,{\sigma_0}^2)$
and $W$ is a brownian motion with respect to $\mathcal{F}_t$
$r,\sigma >0$ are constants
$t \in [0,T]$

I've also an explicit version of $\mu_t$

$\mu_t=\frac{\sigma^2\mu_0+{\sigma_0}^2(\mu t+\sigma V_t)}{\sigma^2 + {\sigma_0}^2t}$
$V$ is a brownian motion with respect to $\mathcal{G}_t$ and $\mu$ is independent of $V$

How can I now prove, that $Z_t$ is a $\mathcal{F}_t-$martingale?

I only know the Novikov condition,
$E[exp\{\frac{1}{2}\int\limits_0^T(\frac{\mu_t -r}{\sigma})^2dt \}] < \infty$.
But I think this condition isn't fulfilled.

Can anybody help me?