= Solution
Because $W$ is continuous and <adapted>, the Borel function $\operatorname{sgn}(W)$ is <predictable>. Its value at zero is $-1$, so its square is exactly one everywhere. Consequently $A$ is a zero-starting <continuous local martingale> with
$$
[A]_t=\int_0^t\operatorname{sgn}(W_s)^2ds=t.
$$
The <Lévy characterization of Brownian motion> therefore makes $A$ a <Brownian motion> in the same <filtration>. Here that characterization is the deterministic-clock case of the conditional-exponential argument in question 2.
Apply the <Itô formula> to $V=W^2$:
$$
dV_t=2W_t\,dW_t+dt.
$$
Since $\sqrt{V_t}=|W_t|$ and $|W_t|\operatorname{sgn}(W_t)=W_t$, including at zero, substitution yields
$$
\boxed{dV_t=2\sqrt{V_t}\,dA_t+dt.}
$$
Thus $W^2$ is a <squared Bessel process> of dimension one.
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