Problem:
Show $(B_t )^2$ i.e. square of a Brownian motion is a Markov process.
To do this, I want to show $$P( B_t ^2 | B_{t_1} ^2 , ... , B_{t_n} ^2 )= P( B_t ^2 | B_{t_n} ^2 ) $$ where $0<t_1<...<t_n<t$.
My attempt:
$ B_t ^2 = (B_t -B_{t_n})^2 +2 (B_t -B_{t_n})B_{t_n} + B_{t_n} ^2$ and $(B_t -B_{t_n})$ is independent with $ \sigma (B_{t_1} ^2 , ... , B_{t_n} ^2) $ while $B_{t_n} ^2$ is measurable w.r.t. $ \sigma (B_{t_1} ^2 , ... , B_{t_n} ^2)$. However, $ B_{t_n}$ is not measurable w.r.t. $\sigma (B_{t_1} ^2 , ... , B_{t_n} ^2)$ so I cannot apply the standard lemma which states that I can integrate the "independent part".
I have also tried to calculate the joint p.d.f of $B_{t_1} ^2 , ... , B_{t_n} ^2 , B_t ^2$ by change of variables from the marginal distribution of Brownian motion, but it is not very easy.
Any hint is appreciated.
