If $B = (B_t)_{t\geq 0}$ is a standard brownian motion, is $B_t + B_{2t}$ a markov process?
My intuition tells me that it's not a markov process because we don't have independent increments (I think). As far as I know, if we take $t_{k} < t_{k+1} < t_{n} < t_{n+1}$, then nothing assures that $[t_{2(n+1)}, t_{2n}]\cap[t_{2{k+1}}, t_{2k}] = \emptyset$. However, I'm not sure whether this is enough or not. Any help, please?
You are right because the increments are not independent. You can explicitly compute the covariance of the increments. For example for the increments between $t=0$ to $t=1$ and between $t=1$ to $t=2$. Assuming $B_0=0$, we compute: \begin{align} C_t &= B_t + B_{2t} \\ \mathrm{Cov}(C_1-C_0, C_2-C_1) &= \mathrm{Cov}(B_1+B_{0.5}, (B_2+B_1)-(B_1+B_{0.5})) \\ &= \mathrm{Cov}(B_1+B_{0.5}, B_2-B_{0.5}) \\ &= \mathrm{Cov}(B_1,B_2)+\mathrm{Cov}(B_{0.5}, B_2)-\mathrm{Cov}(B_1,B_{0.5})-\mathrm{Cov}(B_{0.5},B_{0.5}) \\ &= \mathrm{Cov}(B_1,B_1)+\mathrm{Cov}(B_{0.5}, B_{0.5})-\mathrm{Cov}(B_{0.5},B_{0.5})-\mathrm{Cov}(B_{0.5},B_{0.5}) \\ &= \mathrm{Cov}(B_1,B_1)-\mathrm{Cov}(B_{0.5}, B_{0.5}) \\ &=1-0.5\neq 0 \end{align} Well, not the most beautiful calculation, but it works. In particular we used the fact that $\mathrm{Cov}(B_a,B_b)=\mathrm{Cov}(B_c,B_c)$ with $c=\min(a,b)$. (This is true because $B$ has independent increments.)