In Girsanov theorem, the change of probability measure variable $Z_t = \frac{dQ}{dP}|_{\mathcal{F}_t}$, why does it need to be a martingale with respect to measure $P$ for the change of measure $\frac{dQ}{dP}$ to exist?

I am having trouble understanding this. Anyone familiar with this?


After reading up about Girsanov theorem and martingale theory, I can come up with the following observations. First if we have a filtration $\mathcal{F}_t$ and two probability measures $P$ and $Q$ for which Radon-Nikodym derivative $\frac{dQ}{dP}$ exist, then for each $\mathcal{F}_t$ there exists a Radon-Nikodym derivative $D_t$ with respect to $\mathcal{F}_t$ and $D_t$ is a uniformly integrable martingale with respect to $\mathcal{F}_t$ and $P$.

Now if we have measure $P$ and a martingale $Z_t$ with filtration $\mathcal{F}_t$ we can define set function $Q=Z_t\cdot P$ defined on $\cup \mathcal{F}_t$. It will define a unique probability measure $Q$ on $\sigma(\cup \mathcal{F}_t)$ if $Z_t$ has additional properties, $EZ_t\equiv 1$ being one of them.

Going into more details requires reposting some book on this topic, which is not feasible. I read this one. Chapter VIII is a good read for clarifying things up. Naturally other books can be found.

Note that this is really a comment not an answer, I suggest trying to ask at math.SE, with details what exactly you did not understand, the question in the current format can be answered in many different ways, since there is a lot of things going on.

  • $\begingroup$ @mpiktas: I think you should write that $Q$ is defined on $\sigma(\cup \mathcal{F}_t)$, since $\cup \mathcal{F}_t$ is not necessarily (and indeed, usually is not) a $\sigma$-algebra. $\endgroup$
    – cardinal
    Apr 9 '11 at 13:49
  • $\begingroup$ @cardinal, you are right, although in the book I cited, only union is used. I will have to recheck. $\endgroup$
    – mpiktas
    Apr 11 '11 at 3:24
  • $\begingroup$ @mpiktas: interesting, as that is a well-known and highly regarded reference. You might be interested in this question on math.SE. $\endgroup$
    – cardinal
    Apr 11 '11 at 15:27
  • $\begingroup$ @mpiktas On the contrary, at the very next line after they define the set function, the authors mention that it has to be extended to a probability measure on the generated sigma-algebra. $\endgroup$
    – Did
    Apr 12 '11 at 11:16
  • $\begingroup$ @cardinal See comment above. $\endgroup$
    – Did
    Apr 12 '11 at 11:17

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