Is the sum of two Brownian motions always a martingale, even if the two are possibly correlated?
I'm not sure, because on the internet I couldn't find the specific answer. I get correlation plays a role, but not sure how.
EDIT: Thanks to anyone who will answer. However, I sometimes find the lack of sympathy on this platform a bit unsettling. I thought this was a place where any student could ask a question, as simple as it might be, but I often get arrogant answers and of no help whatsoever.
Also, I'd rather not share the whole questions, for reasons you may infer by yourselves. I just know I have two brownian motions which might or might not be correlated, and I'm asked to state if their sum is always martingale or if they necessarily have to be independent/uncorrelated. I hope this clarifies.
I interpret the question to mean: If $\{B(t)\}$ and $\{W(t)\}$ are two Brownian motions defined on the same probability space, is $\{B(t)+W(t)\}$ necessarily a Martingale even if $B$ and $W$ are dependent?
If you assume that $B$ and $W$ are both adapted and satisfy the Martingale property with respect to the same filtration $({\mathcal F}_t)$ i.e., $$\forall s>t, \quad E[B(s)| {\mathcal F}_t]=B(t) \,,$$ $$\forall s>t, \quad E[W(s)| {\mathcal F}_t]=W(t) \,,$$ then the answer is positive from the definitions.
Without some assumption on the dependence of $B$ and $W$, the answer is negative. Let $B$ be a standard Brownian motion, and recall that $E[B(1) | B(2)]= B(2)/2.$
Define a second Brownian motion by $W(t)=B(t+1)-B(1)$ and let $Z(t)=B(t)+W(t)$, so $Z(1)=B(2)$. Then $$E[Z(2)-Z(1)|Z(1)]=E[B(3)-B(1) | B(2)]= B(2)-B(2)/2=B(2)/2 \,,$$ so $Z$ is not a Martingale.