# What is the drift for a convex combination of Girsanov measures?

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 $$F(t)=E_\mu[B(t)]=\lambda F_1(t)+(1-\lambda)F_2(t)$$.

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

• 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? – S.Surace Jun 11 '20 at 15:09
• @S.Surace Because it has a density. – user158968 Jun 11 '20 at 15:25
• @S.Surace Any measure that is absolutely continuous wrt Wiener measure is a Girsanov measure and corresponds to a $W^{1,2}$ drift. – user158968 Jun 11 '20 at 15:35
• 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. – 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 $$F=BF_1+(1−B)F_2$$.
• 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". – ofer zeitouni Jun 12 '20 at 20:32
• 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 $\mu_2$. – ofer zeitouni Jun 12 '20 at 20:37