# Version of Girsanov theorem with changing volatility

Is there a version of Girsanov theorem when the volatility is changing?

For example Girsanov theorem states that Radon Nikodym (RN) derivative for a stochastic equation is used to transform the expectation where the sampling is done in one mesaure to an expectation where sampling is done in another measure.

Let's see an example

$dX_t(w) = f(X_t(w))dt + \sigma(X_t(w))dW_t^P(w)$ in P measure.

In P* measure, drift is $f^{*}(X_t(w))$. We multiply the internals of expectation in P measure with RN derivative to get expectation of X in P* measure

$E^{P^*}[X] = E^P[X \frac{dP^*}{dP}]$

where

$\frac{dP^*}{dP}=e^{-0.5 \int (\frac{ f^{*}(X_s(w)) - f(X_s(w))}{\sigma(X_s(w))})^2ds + \int \frac{ f*(X_s(w)) - f(X_s(w))}{\sigma(X_s(w))} dW_s^P(w)}$

What I am looking for is in P* measure, not only drift but also the volatility changes

$dX_t(w) = f^{*}(X_t(w))dt + \sigma^{*}(X_t(w))dW_t^P(w)$

Then what is $\frac{dP^*}{dP}$?

-
Just one small thing. I think you're to do $E^P[X\frac{dP}{dP^*}]$ to change measure from $P$ to $P^*$. Consider this expression in the form of an integral; $\displaystyle \ \ \int_\Omega X (\frac{dP}{dP^*})dP\frac{dP^*}{dP} = E^P[X\frac{dP}{dP^*}]$. –  Jase Nov 12 '12 at 14:48
Agreed with Jase' comment that $$E_Q\left[ F(X) \right] = E_P\left[ F(X) \frac{dQ(x)}{dP(x)} \right]$$ . –  William S. Wong Dec 6 '12 at 21:27

I don't think Girsanov's formula works when the volatilities are different between the P measure and P* measure. P and P* will be singular with respect to each other.

Please see Prof. Goodman's class notes on page 11 at http://www.math.nyu.edu/faculty/goodman/teaching/StochCalc2012/notes/Week10.pdf .

a probability measure assigns relative likelihood to different trajectories of the Brownian motion. Variance of the Ito process can be recovered from the shape of a single trajectory (quadratic variation), so it does not depend on the relative likelihood of the trajectories, hence, does not depend on the choice of the probability measure.

-
I also remember something about only requiring that your vol be a bounded process with finite quadratic variation in order for the theorem to be valid at this step: $$d\tilde W_t = (\frac{\mu-r} \sigma )dt + dW_t$$ such that $$d\tilde W_t$$ is still a brownian motion –  SpeedBoots Nov 14 '12 at 20:44