# Why Girsanov theorem doesn't allow you to change volatility?

I see Girsanov/Cameron-Martin as a generalization of change of measure from single random variables to stochastic processes (random functions).

It is simple to change measure from one non-degenerate normal distribution to another normal distribution even if their variances are not equal. The likelihood ratio is well-define. So how come when it comes to diffusion processes, Girsanov can only change the drift but not the volatility?

• Have a look here : math.stackexchange.com/questions/79875/… – TheBridge Sep 4 '19 at 11:29
• So long story short any change in the diffusion term coefficient induces that the laws of the two processes become mutually singular as there are events under each law that are equal to 1 for one and 0 for the other (those events derive from the law of the quadratic variations of each process that can't match when they have different diffusion coefficients). Regards – TheBridge Sep 5 '19 at 7:58

## 1 Answer

John Lennon said it best: "There's nothing you can change that can't be changed..." so invariance of volatility for diffusions is a fact of life that is independent of Girsanov's theorem, or any other theorem. This is because volatility is derived from the predictable compensator of quadratic variation and quadratic variation of diffusion processes is continuous therefore predictable. If you like, quadratic variation (for diffusions or more generally for all continuous semimartingales) is equal to its drift under any probability measure. In "one-period" setting the "process" is no longer continuous and hence its quadratic variation is no longer predictable. Now its predictable compensator can be different under different measures, so we are able to change variance by changing measure.

But also note that the long-run variance of a diffusion will typically change after a change of measure. Consider, for example a geometric Brownian motion such that under the new measure the BM has constant non-zero drift rate $$\mu$$. At a fixed horizon $$T$$ the variance will change from $$\exp(2T)-\exp(T)$$ to $$\exp(2\mu T)(\exp(2T)-\exp(T))$$.