Is $W_{2t}-W_t$ a brownian motion?

Question

Is $W_{2t}-W_t$ a brownian motion?

$(W_t)_{t\geq 0}$ is a brownian motion, I have to show that $X_t:=W_{2t}-W_t$ is a brownian motion as well.

$$W_{2t}= 1/\sqrt{2} W_t$$ (by scaling property)

then I can write $$1/\sqrt{2} W_t-W_t$$ and its distribution is Normal, with mean $0$ and variance $\frac{3-2\sqrt{2}}{2}$.

I would conclude it is a brownian motion.

Answer 1

For each fixed $t$, the random variable $X_t=W_{2t}-W_t$ is centered normal with variance $t$ hence indeed distributed like $W_t$.

But the process $(X_t)$ is not a Brownian motion. To see this, note that $X_2=W_4-W_2$ and $X_1=W_2-W_1$ are the increments of a Brownian motion on some disjoint time intervals hence they are independent while $W_2$ and $W_1$ are not (for example, $E(X_2X_1)=0$ and $E(W_2W_1)=1$).

Answer 2

$X_\frac{1}{2} = W_1 - W_{\frac{1}{2}}$ and $X_1 - X_{\frac{1}{2}} = (W_2 - W_1)-(W_1 -W_\frac{1}{2})$. They are not independent, while a Brownian motion has independent increments.

Remark that we can only say $W_{2t} = \sqrt{2}W_t$ in distribution, they are not equal almost surely.

Answer 3

You could also check the fact that $X_t$ is not a martingale and hence cannot be a Brownian motion. In particular, $E[X_t|F_s] = 0 \neq X_s$ for $s\leq t$.