I have the following probability : $P( W(t) > 0 \mbox{ and }W(2t) > 0)$
on some textbook it is claimed that this is equal to $P( W(1) > 0 \mbox{ and }W(2) > 0)$ due to the scaling property of the Brownian motion.
As far as I know, the scaling property says that for a real value $c$,
$c*W(t/c²)$ is a Brownian motion.
In that case it seems that it'd work with $c = \sqrt t$, but this would imply that $\sqrt(t)*W_1$ is a Brownian motion , which is wrong.
In short : what is the detailed reason why these 2 probabilities are equal ? Thanks!
As the process $(W(t))_{t\ge 0}$ has the same distribution than $(cW(\frac t{c^2}))_{t\ge 0}$, you get with $c = \sqrt t$: $$ P( W(t) > 0 ;\ \ W(2t) > 0) = P\left( cW(\frac t{c^2}) > 0 ;\ \ cW\left(2 \frac t{c^2}\right) > 0\right)\\ = P\left( W\left(\frac t{c^2}\right) > 0 ;\ \ W\left(2 \frac t{c^2}\right) > 0\right)= P( W(1) > 0 ; W(2 )> 0) $$the second equality because $c>0$, and the third because $\frac t{c^2}=1$.