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.

  • $\begingroup$ 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

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 $F=BF_1+(1−B)F_2$.

  • $\begingroup$ 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

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