# Is $(B_t^2)$ Markov where $(B_t)$ is Brownian motion?

I am pretty sure $(B_{t}^{2})$ not Markov because the squared random walk is not.

Showing the square of a Markov process is or isn't Markov

I guess I can repeat the method since to be Markov it must satisfy the discrete.

Thanks

For every $t\geqslant0$, $(B_{t+s}^2)_{s\geqslant0}$ is distributed as $(X_s)_{s\geqslant0}$, where, for every $s\geqslant0$, $$X_s=B_t^2+2\sqrt{B_t^2}\cdot W_s+W_s^2,$$ where $(W_s)_{s\geqslant0}$ is a Brownian motion independent of $(B_u)_{0\leqslant u\leqslant t}$. Thus, indeed, $(B^2_t)_{t\geqslant0}$ is a Markov process.

The discrete analogue of this result is that, if $(X_n)_{n\geqslant0}$ is a random walk with $\pm1$ steps of equal probabilities $\frac12$, then $(|X_n|)_{n\geqslant0}$ is also Markov.

• Thank you. Did you use that $2B_{t}W_{s}\stackrel{d} {\sim}2\sqrt{B^{2}_{t}}W_{s}$? – TKM Nov 3 '14 at 23:13
• Yes, provided one adds a missing factor 2 on the RHS and one passes to the level of identity in distribution of processes, not only of random variables. – Did Nov 3 '14 at 23:14
• what do you mean by "level of identity in distribution of processes"? – TKM Nov 3 '14 at 23:25
• how can $B_{t}\stackrel{d}{\sim} |B_{t}|$? The latter is not even Gaussian. – TKM Nov 3 '14 at 23:26
• You do not need only the identity in distribution mentioned in your first comment to hold for each fixed $s$, but the identity $$(B_tW_s)_{s\geqslant0}\stackrel{d}{=}(|B_t|\cdot W_s)_{s\geqslant0}.$$ – Did Nov 3 '14 at 23:27

Hint:

Use $$B_{t_n}=sign(B_{t_n})|B_{t_n}|$$ and $$\sigma(B_s ^2)= \sigma(|B_{s}| )$$ and independent increment property of B.M to show

$$P(B_t^2 \leq x | B_{t_1}^2 ,..., B_{t_n}^2)=P(B_t^2 \leq x | B_{t_n}^2)$$ where $$0 and $$x \in \mathbb R$$ .

In the process, you need to show $$sign(B_{t_n})$$ is independent with $$\sigma(|B_{t_1}|,|B_{t_2}|,..., |B_{t_n}| )$$ by calculating marginal distributions.