I am trying to prove Blumenthal's zero-one law using Kolmogorov's zero-one law. I use that $B_t$ Brownian $\iff$ $tB_{1/t}$ Brownian. Can I change the intersection and union as follows?

Start of my proof

If $(B_t)_{t\geq 0}$ is a Brownian motion then $(B_t^*)_{t\geq 0}:=(tB_\frac{1}{t})_{t\geq 0}$ is a Brownian motion. We have $$\mathcal{F}_{0+}^B=\cap_{t>0}\mathcal{F}_t^B=\cap_{t>0}\sigma(B_s:s\leq t) .$$ For any event in this $\sigma$-algebra there must also exist an event in $\mathcal{F}_{0+}^{B^*}$ with the same probability. But \begin{align*}\mathcal{F}_{0+}^{B^*}&=\cap_{t>0}\mathcal{F}_t^{B^*}=\cap_{t>0}\sigma(B_s^*:s\leq t)=\cap_{t>0}\sigma(sB_\frac{1}{s}:s\leq t)=\cap_{t>0}\sigma(B_s:s>t)\\ &= \cap_{t\in\mathbb{N}}\sigma(B_s:s\geq t) = \cap_{t\in\mathbb{N}}\cup_{n\in\mathbb{N}} \sigma(\sigma(B_t),\sigma(B_{t+k2^{-n}}-B_{t+(k-1)2^{-n}}:k\in\mathbb{N}))\\ &= \cap_{t\in\mathbb{N}}\cup_{n\in\mathbb{N}} \sigma(B_{t+k2^{-n}}-B_{t+(k-1)2^{-n}}:k\in\mathbb{N})\\ &=\cup_{n\in\mathbb{N}}\cap_{t\in\mathbb{N}} \sigma(B_{t+k2^{-n}}-B_{t+(k-1)2^{-n}}:k\in\mathbb{N}) \end{align*} By application of Komogorov's zero-one law we see that all events in the latter set must have either probability one or zero. QED.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.