Is $W_{t}^{2}$ a Markov process?

Question

$W_{t}$ is a Wiener process. I mean, it seems that it is a Markov process, though I don't see an easy way to prove it. Because, I mean, I don't see how it can be proved that it is a Markov process using definition or the fact that for Markov processes $P(X_{s}|X_{t_{n}}, ... , X_{t_{1}}) = P(X_{s}|X_{t_{n}})$. Any thoughts?

Answer 1

Use properties of the Wiener process (aka Brownian motion) to see that if $0<s<t$ then the conditional distribution of $W_t^2$, given $\mathcal\sigma\{W_u^2: 0\le u\le s\}$, is the same as that of $W_s^2+2W_s\sqrt{t-s}Z+(t-s)Z$, where $Z$ is a standard normal random variable independent of $\mathcal\sigma\{W_u^2: 0\le u\le s\}$. The presence of $W_s={\rm sign}(W_s)\sqrt{W_s^2}$ in the term $2W_s\sqrt{t-s}Z$ means that the conditional distribution of $W^2_t$ given the pre-$s$ past depends not just on $W_s^2$ but also on the sign (positive or negative) of $W_s$. This spoils the Markov property.