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Let $(B_{t},\mathcal{F}_{t})$ be a standard 1-d Brownian motion on some $(\Omega, \mathcal{F}, \mathbb{P}) \ $, and let's assume $\mathcal{F}_{t}$ is the augmentation of $\mathcal{F}_{t}^{W}$.
Let $T>0$ and define $Z_{T}=\exp\{W_{T}-\frac{1}{2}T\} \ $, then $\frac{dP^{T}}{dP}=Z_{T}$ defines a probability measure $P^{T}$ on $\mathcal{F}_{T}$, which is equivalent (i.e. mutually absolutely continuous) w.r.t. $P$ on $F_{T}$.
If we assume $A \in \mathcal{F}_{T}, \ B \in \mathcal{F}_{T}$ are two independent events under $P$, then do we have that $A$ and $B$ are independent under $P^{T}$?

Intuitively I think the answer should be no, but I can't think of a counterexample. Could anyone help on this? Thanks a lot!

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up vote 3 down vote accepted

Try $A=[W_T\gt0]$ and $B=[|W_T|\lt x]$ with $x\gt0$, then $A$ and $B$ are independent under $\mathrm P$ because the distribution of $W_T$ under $\mathrm P$ is symmetric. Independence under $\mathrm P^T$ would mean that $$ \mathrm E(\mathrm e^{W_T}\,;\,0\lt W_T\lt x)\cdot\mathrm e^{T/2}=\mathrm E(\mathrm e^{W_T}\,;\,|W_T|\lt x)\cdot\mathrm E(\mathrm e^{W_T}\,;\,W_T\gt0). $$ If this identity holds for every $x\gt0$, differentiating it with respect to $x$ yields $$ 2\mathrm e^{x}\cdot\mathrm e^{T/2}=(\mathrm e^{x}+\mathrm e^{-x})\cdot\mathrm E(\mathrm e^{W_T}\,;\,W_T\gt0), $$ which cannot hold for every $x$.

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Thanks! This is ingenious. – user7762 Aug 20 '12 at 21:00

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