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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 $\frac{dP^*}{dP}$?

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    $\begingroup$ 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; $\displaystyle \ \ \int_\Omega X (\frac{dP}{dP^*})dP\frac{dP^*}{dP} = E^P[X\frac{dP}{dP^*}]$. $\endgroup$
    – Jase
    Nov 12, 2012 at 14:48
  • $\begingroup$ Agreed with Jase' comment that $$E_Q\left[ F(X) \right] = E_P\left[ F(X) \frac{dQ(x)}{dP(x)} \right]$$ . $\endgroup$
    – wsw
    Dec 6, 2012 at 21:27

3 Answers 3

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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 .

Also, from [ http://ocw.mit.edu/courses/sloan-school-of-management/15-450-analytics-of-finance-fall-2010/lecture-notes/MIT15_450F10_lec02.pdf ] page 54:

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.

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The Girsanov theorem applies to any compatible change of measure, including a volatility change. The version you have written above is a simplified version for drift changes only, but if you look in any good stochastic calculus book, you will see that full version just requires that you be able to compute the cross-variation of the two processes.

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    $\begingroup$ Come on @Brian B, I know you probably know the answer by heart (haha). Even if not, please include it in your answer it would help the site with another of your good answers... (for later search engines visibility and for later users browsing the site) $\endgroup$
    – SRKX
    Nov 13, 2012 at 0:01
  • $\begingroup$ Can you provide a reference for your comment? $\endgroup$
    – adam
    Nov 13, 2012 at 18:27
  • $\begingroup$ I actually just checked Karatzas and Shreve without finding the version/result I am thinking of -- did I imagine it? I may withdraw this answer. $\endgroup$
    – Brian B
    Nov 14, 2012 at 4:15
  • $\begingroup$ 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 $\endgroup$
    – SpeedBoots
    Nov 14, 2012 at 20:44
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Pls check Shreve‘s Theorem 9.2.2 Change of risk-neutral measure . Is this what you are looking for?

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