1
$\begingroup$

Consider two Girsanov measures $\mu_1$ and $\mu_2$ corresponding to drifts $F_1(t)$ and $F_2(t)$ respectively. By this, I mean that we have that $B(t)\sim F_1(t)+\tilde B(t)$ where $\tilde B(t)$ is a Brownian motion under $\mu_1$. Similarly for $\mu_2$.

For $\lambda \in [0,1]$ we can consider the probability measure $\mu=\lambda \mu_1+(1-\lambda) \mu_2$. $\mu$ is also a Girsanov measure so it corresponds to a drift $F(t)$. What is $F$ in terms of $F_1,F_2$?

I know if $F_1, F_2, F$ are all deterministic then $$F(t)=E_\mu[B(t)]=\lambda F_1(t)+(1-\lambda)F_2(t)$$.

What about in general?

Even in the case where $F_1,F_2$ are deterministic can we say that $F$ is? This itself is pretty tricky.

$\endgroup$
  • $\begingroup$ It is not clear to me why the convex combination of a Girsanov measure should be a Girsanov measure. Where do you get this from? $\endgroup$ – S.Surace Jun 11 '20 at 15:09
  • $\begingroup$ @S.Surace Because it has a density. $\endgroup$ – user158968 Jun 11 '20 at 15:25
  • $\begingroup$ @S.Surace Any measure that is absolutely continuous wrt Wiener measure is a Girsanov measure and corresponds to a $W^{1,2}$ drift. $\endgroup$ – user158968 Jun 11 '20 at 15:35
  • $\begingroup$ Sure, this makes sense. Unfortunately I don't know an answer to this. The exponential martingales and the sum don't seem to go well together. $\endgroup$ – S.Surace Jun 11 '20 at 16:07
1
$\begingroup$

Just take drift $F_1$ with probability $\lambda$ and drift $F_2$ w.p $(1-\lambda)$. If you want an explicit probabilistic description in terms of the drifts $F_1,F_2$, just enlarge the probability space to support an independent Bernoulli $B$ of parameter $\lambda$ and set the drift $F=BF_1+(1−B)F_2$.

$\endgroup$
  • $\begingroup$ I'm sorry, I'm not sure exactly what you're showing here. $\endgroup$ – user158968 Jun 12 '20 at 16:33
  • $\begingroup$ I thought you wanted a drift $F$ that will generate the measure $\mu$. I constructed one for you. Wasn't it your question? BTW, I somewhat disagree with what you wrote in the "deterministic case". $\endgroup$ – ofer zeitouni Jun 12 '20 at 20:32
  • $\begingroup$ That is what I wanted. Thank you. $\endgroup$ – user158968 Jun 12 '20 at 20:34
  • $\begingroup$ Why do you disagree? $\endgroup$ – user158968 Jun 12 '20 at 20:35
  • $\begingroup$ Because $F(t)$ is the expression I wrote, and not the one you did. If you take $F=\lambda F_1+(1-\lambda) F_2$, where $F_i$ are deterministic functions, the measure you will get is not the convex combination of $\mu_1$ and $\mu_2$. $\endgroup$ – ofer zeitouni Jun 12 '20 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.