I need to prove that if $W$ is a Brownian motion then $W'(t)=tW(1/t), t >0, W'(0)=0$ is a Brownian motion.
It is continuous and by using the law of large numbers for Brownian motion it is continuous in 0 aswell. I am able to prove stationary increments, and that the increments are normally distributed with variance of the length of the increment. My difficulty is independent increments.
Let $0<s_1<s_2<\ldots s_n$.
I need to prove that $W'(s_n)-W'(s_{n-1}),W'(s_{n-1})-W'(s_{n-2}),\ldots, W'(s_1)$, are mutually independent. This means I need to prove that:
$s_nW(1/s_n)-s_{n-1}W(1/s_{n-1}), s_{n-1}W(1/s_{n-1})-s_{n-2}W(1/s_{n-2}), \ldots,s_1W(1/s_1)$ are mutually independent.
Since we have that $1/s_n<1/s_{n-1}<\ldots <1/s_1$. I have from the independence of Brownian increments that: $W(1/s_1)-W(1/s_2), W(1/s_2)-W(1/s_3),\ldots, W(1/s_{n-1})-W(1/s_n), W(1/s_n)$ are mutually independent.
I have tried calculating the characteristic functions of the increments, but I don't get it to work. I was only able to do this for two points. Is there any other way to do it? Do you have any idea on how to show the independent increments?
Best Answer
Hints:
Show that $(W_t')_{t \geq 0}$ is a Gaussian process. To this end, show that any vector $Y := (W_{t_1}',\ldots,W_{t_n}')$ can be written in the form $$Y = A \cdot X$$ for some (deterministic) matrix $A$ and $X := (W_{1/t_1},\ldots,W_{1/t_n})$.
Recall the following statement: