Lets have two measures for the same process, please note different volatilitiesdS(t) = mu^P dt + sigma^P dW^PdS(t) = mu^Q dt + sigma^Q dW^QE^Q[S(1Y) ] = E^P[S(1Y) dQ/dP]where dQ/dP = sigma^P/sigma^Q [exp(-(w-mu^Q)^2/(2*(sigma^Q)^2)]/[exp(-(w-mu^P)^2/(2*(sigma^P)^2)]; ratio of two gaussian Prob densities.thus dQ/dP cannot be written as Radon Nikodym derivative form using exponentials dQ/dP != exp(-lambda^2/2+lambda w ), since there needs to be w^2 termDoes this mean these two measures are not equivalent? If so, I dont think it really fits into the usual definition of equivalent measures ( Pr^Q(A)=0 when Pr^P(A)=0), since these are continuous probability densities, which gives no zero or one probability.

No, these two measures are not equivalent if sigma^P != sigma^Q. It can be seen as a consequence of the Girsanov theorem.

Can you provide a proof?Even though these measures are not equivalent, there is a likelihood function that transforms the expectation in one measure to another one. Thus equivalency is not a requirement for measure change.

the quadratic variation is the same on every path but different in the two different measures so they cannot be equivalent.It's important to distinguish between equivalence of measures on the space of paths which does not hold here, and equivalence of densities on a single time horizon which does hold here. I have some discussion of this point in More Mathematical Finance.

Hi MarkI have your book, which section are you referring to.Thanks

the chapter on the non-commutativity of discretization