# How to find quadratic variation of $Y_t=B_{2t}$ for a Brownian motion $(B_t)_{t \geq 0}$

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 $$\langle g,g \rangle_t=\lim_{((t_i-t_{i-1})\to 0)}\sum_{i=1}^n|g(t_i)-g(t_{i-1})|^2$$

• It would be helpful to know which definition of "quadratic variation" you are using. Moreover, you should make your question self-contained, i.e. explain the notation you are using (what is $B_t$?) – saz Dec 4 '16 at 12:47
• Sorry I didn't state the question clear, $B_t$ is Brownian Motion. And I'm studying stochastic calculus. The quadratic variation definition of the note book is :quadratic variation of g is $[g,g]_t=\sum_{i=1}^n|g(t_i)-g(t_{i-1})|^2$ as the max of ($t_i-t_{i-1}$) is very small to zero – Lexi Dec 4 '16 at 13:00
• No, that's not a correct definition of the quadratic variation. Please check it again. – saz Dec 4 '16 at 13:02
• as the max of ($t_i-t_{i-1}$) is very small to zero – Lexi Dec 4 '16 at 13:10
• Limit in which sense? Pointwise, in probability, in $L^2$....? – saz Dec 4 '16 at 13:11

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 $c \in \mathbb{R}$.
• Is it necessary to appeal to the similarity properties of Brownian motion? If $Z_t=X_{a(t)}$ for some nondecreasing nonnegative deterministic continuous function $a$ then $\langle Z\rangle_t=\langle X\rangle_{a(t)}-\langle X\rangle_{a(0)}$, no? – Did Dec 4 '16 at 21:10