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

Solution 1:

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.