Let $(\Omega,F,P)$ be probability space and $T>0$, let $\lbrace W(t), 0\leqslant t\leqslant T\rbrace $ be SBM and $F^W_{s,t}=\sigma \lbrace W(r)-W(s); s\leqslant r\leqslant t\rbrace \vee N$ with $N$ the class of P-null sets of $ F$. My question is why $F^W_{t,T} \subset F^W_{0,T}$?
2026-03-26 02:54:57.1774493697
inclusion of filtration
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For a fixed $r\in [t,T]$ one has $W(r)-W(t) = W(r)-W(0)- (W(t)-W(0)) $ and the quantities on the RHS are all $F_{0,T}^W$ measurable. To be more precise $$ P(W(r)-W(t)<x) = P(\bigcap_{q\in \mathbb Q \cap [0,x]} \{W(r)-W(0)<q\}\cap \{-(W(t)-W(0))<x-q\}) $$ due to continuity of the Brownian motion. The symmetric difference of the two sets in the above equation is of probability 0, hence they are almost surely equal. The left set is in $F_{t,T}^W$ while the right set is in $F_{0,T}^W$.