Blumenthal's zero–one law

Suppose that ${\displaystyle X=(X_{t}:t\geq 0)}$ is a stochastic process on a probability space ${\displaystyle (\Omega ,{\mathcal {F}},\mathbb {P} )}$ with ${\displaystyle \Omega =C(\mathbb {R} ^{+},\mathbb {R} )}$, natural filtration ${\displaystyle \{{\mathcal {F}}_{t}\}_{t\geq 0}}$ and canonical identification ${\displaystyle X_{t}(\omega )=\omega (t)}$. If ${\displaystyle X}$ has the strong Markov property then any event in the germ sigma algebra ${\displaystyle \Lambda \in {\mathcal {F}}_{0+}=\bigcap _{s>0}{\mathcal {F}}_{s}}$ has either ${\displaystyle \mathbb {P} (\Lambda )=0}$ or ${\displaystyle \mathbb {P} (\Lambda )=1.}$[2]

The above statement means roughly speaking that the initial datum of the process is deterministic. Indeed, the law ${\displaystyle \rho _{0}:=\mathbb {P} _{X_{0}}}$of the random variable ${\displaystyle X_{0}}$satisfies ${\displaystyle \rho _{0}(\Lambda )=\mathbb {P} (X_{0}\in \Lambda )\in \{0,1\}}$ for all ${\displaystyle {\mathcal {F}}_{0+}}$-measurable event ${\displaystyle \Lambda }$. Such a probability measure is necessarily of the form ${\displaystyle \rho _{0}=\delta _{x}}$for some particular point ${\displaystyle x\in \mathbb {R} }$, and as a consequence ${\displaystyle X_{0}(\omega )=\omega (0)=x}$ for ${\displaystyle \mathbb {P} }$-almost all paths ${\displaystyle \omega \in \Omega }$. In other words, the process starts almost surely from the completely deterministic initial position ${\displaystyle X_{0}=x}$.