Is the reflected Brownian Motion a Markov process

Question

Let $W$ be a Brownian Motion (BM). The reflected BM is defined by $X=|X_0+W|$.

We need to show that this process is a Markov process w.r.t. its natural filtration and we need to compute its transition function.

To obtain our result we need to calculate $P_\nu\{X_t\in B|\mathcal{F}_s^X\}$. For this, I think we need to condition further on $\mathcal{F}_s^W$. But I can not figure out the details.

Can anyone give me some help?

Answer 1

Ok. I will do it more formally. Let $x,y\geq 0$ and $t,s\geq 0$ and $f_t(x,y)$ the density of the normal distribution with mean $x$ and variance $t$. We use that we know that for the Brownian Motion the Chapmann Kolmogorov equation is right.

$$g_{t+s}(x,y)=f_{t+s}(x,y)+f_{t+s}(-x,y)=\int_R f_t(z,y)f_s(x,z) + f_t(z,y)f_s(-x,z)dz\\ =\int_{R+} f_t(z,y)f_s(x,z) + f_t(z,y)f_s(-x,z)dz+\int_{R-} f_t(z,y)f_s(x,z) + f_t(z,y)f_s(-x,z)dz\\ =\int_{R+} f_t(z,y)f_s(x,z) + f_t(z,y)f_s(-x,z) + f_t(-z,y)f_s(x,-z) + f_t(-z,y)f_s(-x,-z) dz\\ =\int_{R+} (f_t(z,y)+ f_t(-z,y))*(f_s(x,z) + f_s(-x,z)) dz\\ =\int_{R+} g_t(z,y)g_s(x,z)dz$$

In the last equation I use the symmetrie of the density of the normal distribution, e.g.

$$f_t(-x,-y)=f_t(x,y)$$ und $$f_t(-x,y)=f_t(x,-y)$$

The equation before this is a substitution.

Answer 2

I would do this. Show $$\begin{align}P_t (x,A) &{:=} \int_A g_t (x,y)dy \\ P_0 (x,A) &{:=} \delta_x (A), \end{align}$$ where $g_t$ density of the folded normal distribution, is a Markov kernel (Chapmann K. Equation, etc.).

Then you know there exists a unique Markov process. That's the reflected Brownian motion.

Have someone any comments? Is that enough?

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Perhaps you have to show this too.

For all continuous and bounded functions $f$, $$E[f(|B(t+s)|)|F_s]=E[f(|B(t+s)|)|B_s].$$

But this should be $$E[f(|B(t+s)|)|F_s]=E[f(|B(s)+B (s+t)-B (s)|)|F_s]= E[f(|B(t+s)|)|B_s].$$

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$$g_{t+s}(x,y)= f_{t+s}(x,y)+ f_{t+s}(x,-y)=\int_R...=\int_{R+} ... +\int_{R-}....=\int_{R+}...= Claim $$