Blumenthal's zero–one law

on ${\displaystyle [0,\infty )}$ starting from deterministic point has also deterministic initial movement.

Statement

Suppose that ${\displaystyle X=(X_{t}:t\geq 0)}$ is an adapted

${\displaystyle (\Omega ,{\mathcal {F}},\{{\mathcal {F}}_{t}\}_{t\geq 0},\mathbb {P} )}$ such that ${\displaystyle X_{0}}$ is constant with probability one. Let ${\displaystyle {\mathcal {F}}_{t}^{X}:=\sigma (X_{s};s\leq t),{\mathcal {F}}_{t^{+}}^{X}:=\bigcap _{s>t}{\mathcal {F}}_{s}^{X}}$. Then any event in the

sigma algebra

${\displaystyle \Lambda \in {\mathcal {F}}_{0+}^{X}}$ has either ${\displaystyle \mathbb {P} (\Lambda )=0}$ or ${\displaystyle \mathbb {P} (\Lambda )=1.}$

Generalization

Suppose that ${\displaystyle X=(X_{t}:t\geq 0)}$ is an adapted stochastic process on a probability space ${\displaystyle (\Omega ,{\mathcal {F}},\{{\mathcal {F}}_{t}\}_{t\geq 0},\mathbb {P} )}$ such that ${\displaystyle X_{0}}$ is constant with probability one. If ${\displaystyle X}$ has Markov property with respect to the filtration ${\displaystyle \{{\mathcal {F}}_{t^{+}}\}_{t\geq 0}}$ then any event ${\displaystyle \Lambda \in {\mathcal {F}}_{0+}^{X}}$ has either ${\displaystyle \mathbb {P} (\Lambda )=0}$ or ${\displaystyle \mathbb {P} (\Lambda )=1.}$ Note that every

${\displaystyle (\Omega ,{\mathcal {F}},\{{\mathcal {F}}_{t}\}_{t\geq 0},\mathbb {P} )}$ has

strong Markov property

with respect to the filtration ${\displaystyle \{{\mathcal {F}}_{t^{+}}\}_{t\geq 0}}$.

References