I want to show that $|B_{t}|$ is a Markov Process where, $B_{t}$ is a Standard Brownian Motion. I have seen the proof here and here. But I don't understand why the method below might fail (or if it's correct).
I have the following Lemma at my disposal: (Ref Rene Schilling Brownian Motion Lemma $A.3 $)
Let $X:(\Omega,\mathcal{F})\to (D,\mathcal{D})$ and $Y:(\Omega,\mathcal{F})\to(A,\mathcal{E})$ be measurable maps such that $X$ is $\mathcal{X}$ measurable and $Y$ is $\mathcal{Y}$ measurable and $\mathcal{X}$ and $\mathcal{Y}$ are independent sub-sigma algebras of $\mathcal{F}$. Then for any $\Phi:D\times E\to \Bbb{R}$ be measurable and bounded , we have that $E(\Phi(X,Y)|\mathcal{X})=E[\Phi(x,Y)]\vert_{x=X}=E(\Phi(X,Y)|X)$ .
So if we use the above Lemma, if $u$ be a Bounded measurable function and if I set $\phi(X,Y)=u(|X+Y|)$ where $X=B_{s}$ and $Y=B_{t+s}-B_{t}$ and we set $\sigma(B_{t}:t\leq s)=\mathcal{X}$ and $\mathcal{Y}=\sigma(B_{t+s}-B_{s})$
Then, by the above lemma, we have that $E(\Phi(X,Y)|\mathcal{X})=E\bigg(u(|B_{t+s}|)\bigg|\mathcal{X}\bigg)=E\bigg(u(|B_{t+s}|)\bigg|B_{s}\bigg)$ which is the criteria for a process $X_{t}$ to be a Markov Process.
Where and how am I wrong in arguing this way? Any help is appreciated.
