Let $U,V$ be two Brownian motions. In class we have seen that if $U$ and $V$ are independent, then $B:=xU+\sqrt{1-x^2}V$ is a Brownian motion for $x\in [-1,1]$. Now I asked myself if the independence is really needed. I think so and I would do the following counterexample otherwise:
Let us assume $U=V$ then they are dependent. And $B_t=xU_t+\sqrt{1-x^2}U_t$. Then $$\begin{align}\operatorname{cov}(B_t,B_s)&=\operatorname{cov}\left(xU_t+\sqrt{1-x^2}U_t,xU_s+\sqrt{1-x^2}U_s\right)\\&=x^2(t\wedge s)+2x\sqrt{1-x^2}(t\wedge s)+(1-x^2)(t\wedge s)\\&=(t\wedge s)+2x\sqrt{1-x^2} (t\wedge s)\\&=(t\wedge s)\left(1+2x\sqrt{1-x^2}\right)\end{align}$$We know that $B$ is a Brownian motion iff $\left(1+2x\sqrt{1-x^2}\right)=1$ or equivalently $2x\sqrt{1-x^2}=0$ which implies $x=0$ or $x=\pm 1$. But if I take $x\in (-1,1)\setminus\{0\}$ then I think that I have found an example which shows that independence is needed.
Is this true or am I wrong?
Recall that Brownian motion must have quadratic variation $\left\langle B \right\rangle_t =t$. Suppose $U$ and $V$ are correlated Brownian motions, so that their quadratic covariation is $\left\langle U, V \right\rangle_t = \rho t$. Let's show that the cases you've identified are indeed the only cases which give you Brownian motion. Here I assume that $(U,V)$ is a $2$-dimensional Brownian motion.
Using the polarisation identities, we get:
\begin{align} \left\langle B \right\rangle_t &= x^2 \left\langle U \right\rangle_t + (1-x^2)\left\langle V \right\rangle_t + 2x\sqrt{1-x^2} \left\langle U,V \right\rangle_t \\ &= t + 2x \sqrt{1-x^2} \rho t \end{align}
This quantity is precisely $t$ when $\rho = 0$ (i.e. under the assumption that $(U,V)$ is 2-dimensional Brownian motion, that $U$ and $V$ are independent), or when $x = \pm 1$, as desired.