Markov property for geometric Brownian motion

Question

I want to prove the Markov-property for the geometric Brownian motion $X$ defined by $$X_t=\exp\left(\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W_t\right)$$ where $(W_t)_{t\geq 0}$ is a Brownian motion. The Markov property for a stochastic process is defined as follows:

A stochastic process $X$ with index set $\mathbb R_{\geq 0}$ and values in $(S,\mathcal{S})$ is called a Markov process, if one can find a transition group $(P_{s,t})_{s\leq t}$, such that $$E(f(X_t)\in A\mid \mathcal{F}_s)=P_{s,t}f(X_s):=\int f(x)P_{s,t}(X_s,dx)$$ holds for all Borel functions $f\geq 0$ and all $A\in\mathcal{S}$. If $S$ is a Borel space, then there is a probability kernel $\mu$ such $=P(X_t\in . \mid X_s)=\mu(X_s,.)$ and we can take $P_{s,t}=\mu(X_s,.)$

For the case $f=id$ we have the following: \begin{align} E[X_t\mid \mathcal{F}_s]&=E\left[\exp\left(\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W_t\right)\mid\mathcal{F}_s\right]\\ &=\exp\left(\left(\mu-\frac{\sigma^2}{2}\right)t\right)E\left[\exp(\sigma W_t)\mid\mathcal{F}_s\right]\\ &=\exp\left(\left(\mu-\frac{\sigma^2}{2}\right)t\right)E\left[\exp(\sigma W_t)\right]\\& =\exp\left(\left(\mu-\frac{\sigma^2}{2}\right)t\right)E\left[\exp\left(\sigma W_t\right)\mid W_s\right], \end{align} if we assume that the filtration $(\mathcal{F}_t)_t$ is the natural filtration of $(W_t)_t$.

How can we extend this to Borel functions $f\geq0$? Is it maybe better to prove the equivalent property $$P(X_t\in A\mid\mathcal{F}_s)=P(X_t\in A\mid X_s)\ \mathrm ?$$

Answer 1

First of all, note that the natural filtration of $(X_t)_{t \geq 0}$ equals the natural filtration of $(W_t)_{t \geq 0}$; this follows directly from the relation

$$X_t = \exp \left( \left[ \mu - \frac{\sigma^2}{2} \right] t + \sigma W_t \right).$$

The Markov property of $(X_t)_{t \geq 0}$ can be proved as follows: Fix some bounded Borel-measurable function $f$ and $s \leq t$. For brevity, set $c := \mu - \frac{\sigma^2}{2}$. Then $$\begin{align*} \mathbb{E}(f(X_t) \mid \mathcal{F}_s) &= \mathbb{E} \bigg( f \left[ e^{ct} e^{\sigma(W_t-W_s)} e^{\sigma W_s} \right] \mid \mathcal{F}_s \bigg). \end{align*}$$

Since $W_t-W_s$ is independent from $\mathcal{F}_s = \sigma(W_r; r \leq s)$ and $W_s$ is $\mathcal{F}_s$-measurable, we get

$$\mathbb{E}(f(X_t) \mid \mathcal{F}_s) = g(e^{ct} e^{\sigma W_s}) \tag{1}$$

where

$$g(y) := \mathbb{E} \left( f(y e^{\sigma (W_t-W_s)}) \right).$$

Since the right-hand side of $(1)$ is $X_s$-measurable, the tower property yields

$$\mathbb{E}(f(X_t) \mid X_s) = \mathbb{E} \bigg[ \mathbb{E}(f(X_t) \mid \mathcal{F}_s) \mid X_s \bigg] \stackrel{(1)}{=} g(e^{ct} e^{\sigma W_s}). \tag{2}$$

Combining $(1)$ and $(2)$ we find

$$\mathbb{E}(f(X_t) \mid X_s) = \mathbb{E}(f(X_t) \mid \mathcal{F}_s).$$

Answer 2

I think i got it on my own, but it would be great to have someone to check it.

I use the disintegration theorem to show that $P(X_t\in A|\mathcal{F}_s)=P(X_t\in A|X_s)$ holds for all $A\in\mathcal{S}$.

Fix two measurable spaces $S$,$T$, a $\sigma$-field $\mathcal{A}\subset\mathcal{F}$, a random element $\xi$ in $S$ s.t. $P[\xi\in.|\mathcal{F}]$ has a regular version $\nu$. We further consider an $\mathcal{F}$-measurable random element $\eta$ in $T$ and a measurable unction $f$ on $S\times T$ with $E[f(\xi,\eta)]<\infty$. Then $$E[f(\xi,\eta)|\mathcal{F}]=\int \nu(ds)f(s,\eta)$$ holds.

We now have for $\nu(.)=P[W_t-W_s\in .|\mathcal{F}_s]$ and a probability kernel $\mu$ satisfying $\mu(X_s,.)=P[W_t-W_s\in .|X_s]$ $$ P(X_t\in A|\mathcal{F}_s) \\=P(exp((\mu-\sigma^2/2)t+\sigma W_t)\in A|\mathcal{F}_s) \\=P(X_s exp((\mu-\sigma^2/2)(t-s)+\sigma (W_t-W_s)))\in A|\mathcal{F}_s) \\=E[1_{X_s exp((\mu-\sigma^2/2)(t-s)+\sigma (W_t-W_s)))\in A}|\mathcal{F}_s] \\=\int \nu(dy) 1_{X_s exp((\mu-\sigma^2/2)(t-s))exp(\sigma y)\in A} \\=\int P[W_t-W_s\in dy|\mathcal{F}_s] 1_{X_s exp((\mu-\sigma^2/2)(t-s))exp(\sigma y) \in A} \\=\int \mu(X_s,dy) 1_{X_s exp((\mu-\sigma^2/2)(t-s))exp(\sigma y) \in A} \\=E[1_{X_s exp((\mu-\sigma^2/2)(t-s))exp(\sigma W_t-W_s) \in A}|X_s] \\=P[X_s exp((\mu-\sigma^2/2)(t-s))exp(\sigma (W_t-W_s)) \in A|X_s] \\=P[X_s exp((\mu-\sigma^2/2)(t-s))exp(\sigma (W_t-W_s)) \in A|X_s] \\=P[X_t \in A|X_s] $$