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. 



What is the drift for a convex combination of Girsanov measures?
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 
$F=BF\_1+(1−B)F\_2$. 


