# Why is $\operatorname{sign} B_t$ a predictable process?

Does anybody know why $\operatorname{sign} {B_t}$ is a predictable process if $B_t$ is a Brownian motion and sign denotes the signum function with the convention that $\operatorname{sign} (0) := -1$ ?

Thanks for your help! Regards, Si

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Being predictable is just being measureable with respect to a certain sigma-field. If $\mathcal{P}$ denotes the predictable sigma-field on $\mathbb{R}_+\times \Omega$ then a process $(X_t)_{t\geq 0}$ is said to predictable if $$(t,\omega)\mapsto X_t(\omega)$$ is $(\mathcal{P},\mathcal{B}(\mathbb{R}))$-measureable. Since $(B_t)_{t\geq 0}$ is continuous and adapted, it is indeed predictable. Now the process $(\text{sign}(B_t))_{t\geq 0}$ is just a composition and using the fact that $x\mapsto \text{sign}(x)$ is $(\mathcal{B}(\mathbb{R}),\mathcal{B}(\mathbb{R}))$ measureable we get that $$(t,\omega)\mapsto \text{sign}(X_t(\omega))$$ is is $(\mathcal{P},\mathcal{B}(\mathbb{R}))$-measureable and hence $(\text{sign}(B_t))_{t\geq 0}$ is predictable.
What is the notation $(t,\omega)\mapsto X_t(\omega)$? –  Jase Nov 9 '12 at 4:26
@Jase: A stochastic process $(X_t)_{t\geq 0}$ can be viewed in several ways. 1) For every $t\in \mathbb{R}_+$ we have a random variable $\omega\mapsto X_t(\omega)$ (it is a random variable by definition of $(X_t)_{t\geq 0}$ being a stochastic process). 2) for every $\omega\in\Omega$ we have a sample path $t\mapsto X_t(\omega)$, or we can 3) look at it as simultaneously as a mapping $$\mathbb{R}_+\times \Omega\ni (t,\omega)\mapsto X_t(\omega).$$ When talking about joint measurability or progressive measurability, it is this last notion that is used. –  Stefan Hansen Nov 9 '12 at 7:12