How to find the quadratic variation of $Y_t=B_{2t}$ for a Brownian motion $(B_t)_{t \geq 0}$?
I know that $\langle B \rangle_t=t$, but I have difficult to find what's $\langle Y \rangle_t$ when $Y_t:=B_{2t}$ and $B_t$ is Brownian Motion. The quadratic variation definition I use is : Quadratic variation of $g$ is $$\langle g,g \rangle_t=\lim_{((t_i-t_{i-1})\to 0)}\sum_{i=1}^n|g(t_i)-g(t_{i-1})|^2$$
Hint: Use that $$W_t := \frac{Y_t}{\sqrt{2}}$$ is a Brownian motion and the fact that $$\langle c\cdot W,c \cdot W \rangle_t = c^2 \cdot \langle W,W \rangle_t$$ for any constant $c \in \mathbb{R}$.