# 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 $$E_Q\left[ F(X) \right] = E_P\left[ F(X) \frac{dQ(x)}{dP(x)} \right]$$ . – wsw Dec 6 '12 at 21:27

• 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