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 $$P(X_t\in A\mid\mathcal{F}_s)=P(X_t\in A\mid X_s)\ \mathrm ?$$

  • $\begingroup$ If $(X_t)$ is Markov and $Y_t=f_t(X_t)$ where every $f_t$ is injective then $(Y_t)$ is Markov. Your case is when $$f_t(x)=e^{at+\sigma x}\qquad\sigma\ne0$$ The transition semigroup $(Q_{s,t})$ of $(Y_t)$ is such that $$\int f(z)Q_{s,t}(y,dz)=\int f(f_t(x))P_{s,t}(f_s^{-1}(y),dx)$$ $\endgroup$ – Did Oct 15 '16 at 16:24
  • $\begingroup$ @Did That's a nice theorem. Do you have a link to the proof? $\endgroup$ – peer Oct 15 '16 at 16:33
  • $\begingroup$ No. The formula included in my comment is supposed to make you see why the result holds. $\endgroup$ – Did Oct 15 '16 at 16:41

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

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


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

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

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

  • $\begingroup$ See more direct way + semi-group in comment on main. $\endgroup$ – Did Oct 15 '16 at 16:29
  • 1
    $\begingroup$ @Did I'm well aware that there are other (more direct) ways to prove the assertion. The reasoning in my answer is very close to what the OP tried to do .... that's why I decided to prove it this way. $\endgroup$ – saz Oct 16 '16 at 5:57

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

  • $\begingroup$ What is not clear to me is the equality $P[W_t-W_s\in dy|\mathcal{F}_s]=\mu(X_s,dy)$. It follows propably from the independence of $W_t-W_s$ and $\mathcal{F}$ and from the independence of $W_t-W_s$ and $X_s$. But how can we proof the last one? $\endgroup$ – peer Oct 15 '16 at 0:57
  • $\begingroup$ Note that $X_s$ is measurable with respect to $\sigma(W_r; r \leq s)$. Since $\sigma(W_r; r \leq s)$ and $W_t-W_s$ are independent (this is a direct consequence of the independence of the increments of Brownian motion), this implies that $W_t-W_s$ and $X_s$ are independent. $\endgroup$ – saz Oct 15 '16 at 8:31

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