Strong Markov property of Bessel processes

Question

I am thinking about the following:

If $(B_t)_{t \geq 0}$ is a Brownian motion in $\mathbb{R}^3$, how can we show that the Bessel process (of order $3$) $(|B_t|)_{t \geq 0}$ has the strong Markov property? Any hints?

Answer 1

First, we show that $(|B_t|)_{t \geq 0}$ is a Markov process (with respect to the canonical filtration of $(B_t)_{t \geq 0}$):

Let $u \in \mathcal{B}_b$ be Borel-measurable and bounded. Using that $(B_t)_{t \geq 0}$ has independent increments, we obtain

$$\begin{align*} \mathbb{E}(u(|B_{t+s}|) \mid \mathcal{F}_t) &= \mathbb{E}(u(|(B_{t+s}-B_t)+B_t|) \mid \mathcal{F}_t) \\ &= \mathbb{E}u(|B_s+y|) \bigg|_{y=B_t}. \tag{1} \end{align*}$$

If we denote by

$$p_s(x) := \frac{1}{\sqrt{2\pi s}} \exp \left(- \frac{|x|^2}{2s} \right)$$

the heat kernel, then we can rewrite $(1)$ in the following way:

$$\begin{align*} \mathbb{E}(u(|B_{t+s}|) \mid \mathcal{F}_t) &= \int_{\mathbb{R}^3} u(|z+y|) p_s(z) \, dz \bigg|_{y=B_t} \\ &= \int_{\mathbb{R}^3} u(|z|) p_s(z-y) \, dz \bigg|_{y=B_t} \\ &=: h(B_t). \end{align*}$$

The next step is to show that $h$ is rotational invariant, i.e. $h(y) = h(R \cdot y)$ for any rotation $R \in \mathbb{R}^{3 \times 3}$. To this end, we note that by the rotational invariance of the Brownian motion

$$\begin{align*} h(Ry) &= \int_{\mathbb{R}^3} u(|z|) p_s(R^{-1} z-y) \, dz \\ &\stackrel{x := R^{-1}z}{=} \int_{\mathbb{R}^3} u(|x|) p_s(x-y) \, dx \\ &= h(y). \end{align*}$$ Consequently, $h$ is rotationally invariant and therefore there exists a function $H:[0,\infty) \to \mathbb{R}$ such that $h(y) = H(|y|)$ for all $y \in \mathbb{R}^3$. Thus

$$\mathbb{E}(u(|B_{t+s}|) \mid \mathcal{F}_t) = H(|B_t|), \tag{1}$$

i.e. $(|B_t|)_{t \geq 0}$ is a Markov process.

Since the process $(|B_t|)_{t \geq 0}$ has continuous sample paths, this implies that the process is a strong Markov process (see e.g. René Schilling, Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Theorem A.25).

Remark: The proof shows, in particular, that $(|B_t|)_{t \geq 0}$ is a Markov process with respect to the filtration

$$\mathcal{G}_t := \sigma(|B_s|; s \leq t).$$

This follows from the fact that $\mathcal{G}_t \subseteq \mathcal{F}_t$ and the tower property:

$$\begin{align*} \mathbb{E}(u(|B_{t+s}|) \mid \mathcal{G}_t) = \mathbb{E} \bigg[ \mathbb{E}(u(|B_{t+s}|) \mid \mathcal{F}_t) \mid \mathcal{G}_t \bigg] &\stackrel{(2)}{=} \mathbb{E}( H(|B_t|) \mid \mathcal{G}_t) \\ &= H(|B_t|); \end{align*}$$

hence,

$$\mathbb{E}(u(|B_{t+s}|) \mid \mathcal{G}_t)= \mathbb{E}(u(|B_{t+s}|) \mid |B_t|).$$