マルコフ性のためのおべんきょう

In this article, some equalities are introduced for showing the Markov property of stochastic processes. After then, we demonstrate the Markov property of a geometric Brownian motion.

In what follows, $(\Omega,\mathcal{F},P)$ represents a probability space and the expectation under $P$ is denoted by $E[\cdot]$ .
Proposition. Let $(\Omega_1,\mathcal{B}_1)$ and $(\Omega_2,\mathcal{B}_2)$ be measurable spaces, $f\colon \Omega_1\times\Omega_2\to \mathbb{R}$ be a bounded measurable function, and $\mathcal{G}$ be a sub $\sigma$ -field of $\mathcal{F}$ . If an $\Omega_1$ -valued r.v. $X$ is $\mathcal{G}$ -measurable and an $\Omega_2$ -valued r.v. $Y$ is independent of $\mathcal{G}$ , we have the following:
(1) $E[f(X,Y)|\mathcal{G}]=E[f(X,Y)|X]=E[f(x,Y)]\big|_{x=X}$ , a.s.
(2) $E[f(X,Y)|X=x]=E[f(x,Y)]$ , $P_X$ -a.s.

Proof. Recalling the construction of integration in the general measure theory, we may assume that $f$ is the form $\boldsymbol{1}_\Gamma$ for $\Gamma\in\mathcal{B_1}\otimes \mathcal{B_2}$ . Furthermore, the Dynkin's $\pi\text{-}\lambda$ theorem allows $f(x,y)$ to be the form $\boldsymbol{1}_{A}(x)\boldsymbol{1}_B(y)$ for $A\in\mathcal{B_1}$ , $B\in\mathcal{B_2}$ .
(1) Since $X$ is $\mathcal{G}$ -measurable and $Y$ is independent of $\mathcal{G}$ , we see
\begin{align*}
E[\boldsymbol{1}_A(X)\boldsymbol{1}_B(Y)|\mathcal{G}]
&= \boldsymbol{1}_A(X)E[\boldsymbol{1}_B(Y)] \\
&= E[\boldsymbol{1}_A(x)\boldsymbol{1}_B(Y)]\big|_{x=X}, \quad \text{a.s.}
\end{align*}
As for $E[f(X,Y)|X]$ , it is followed by the same argument since $X$ is $\sigma(X)$ -measurable and $Y$ is independent of $\sigma(X)$ .

(2) Suppose $C\in\mathcal{B}_1$ . Then, by the independence between $X$ and $Y$ , we see that
\begin{align*}
E[\boldsymbol{1}_A(X)\boldsymbol{1}_B(Y);X\in C]
&= E[\boldsymbol{1}_B(Y)]E[\boldsymbol{1}_A(X);X\in C] \\
&= \int_{C}P_X(dx) \boldsymbol{1}_A(x) E[\boldsymbol{1}_B(Y)].
\end{align*}
Thus, we obtain
\begin{align*}
E[\boldsymbol{1}_A(X)\boldsymbol{1}_B(Y)|X=x]
&= E[\boldsymbol{1}_A(x)\boldsymbol{1}_B(Y)], \quad \text{$P_X$-a.s.}
\end{align*}
We now complete the proof. □

Remark 1. We note that how to apply the Dynkin's $\pi\text{-}\lambda$ theorem in the proposition.
It is easy to see that
\begin{align*}
\mathcal{P} := \{A\times B~;~ A\in\mathcal{B}_1, B\in\mathcal{B}_2\}
\end{align*}
is a $\pi$ -system in $\Omega_1\times \Omega_2$ ; that is, $\Omega_1\times \Omega_2\in\mathcal{P}$ and it holds $A \cap B \in \mathcal{P}$ for any $A,B\in\mathcal{P}$ .
Denote by $\mathcal{L}$ the collection of $A\in\mathcal{B_1}\otimes\mathcal{B}_2$ satisfying
\begin{align*}
E[\boldsymbol{1}_A(X,Y)|\mathcal{G}]
&= E[\boldsymbol{1}_A(x,Y)]\big|_{x=X}.
\end{align*}
Then, $\mathcal{L}$ is a $\lambda$ -system; that is, all of the followings are fulfilled:

$\Omega_1\times\Omega_2\in\mathcal{L}$ .
$B\setminus A\in\mathcal{L}$ for any $A\subset B$ , $A,B\in\mathcal{L}$ .
$\bigcup_{n=1}^\infty A_n\in\mathcal{L}$ for any $A_1\subset A_2\subset\cdots$ , $A_1,A_2,\dots\in\mathcal{L}$ .

Then one can apply the Dynkin's theorem and get $\sigma(\mathcal{P})=\mathcal{B}_1\otimes\mathcal{B}_2=\mathcal{L}$ . The remainings are also shown in the same way as above.
Remark 2. The proposition above holds even if the assumption of $f$ is relaxed to that $f(X,Y)$ is integrable.
Remark 3. It can be seen directly that $\mathcal{L}$ is a $\sigma$ -field.

We now show the Markov property of a geometric Browinan motion.
Example. Let $X^x_t = xe^{\alpha t + \beta B_t}$ , where $(B_t)_{t\geq 0}$ is a Brownian motion starting at the origin, and let $f\colon\mathbb{R}\to\mathbb{R}$ be a bounded function. And $(\mathcal{F}_t)_{t\geq 0}$ denotes the Brownian filtration. By (1) of the proposition above, we get, for $s < t$ ,
\begin{align*}
E[f(X^x_t)|\mathcal{F}_s]
&= E[f(xe^{\alpha s + \beta B_s}e^{\alpha (t-s) + \beta (B_t-B_s)})|\mathcal{F}_s] \\
&= E[f(ye^{\alpha (t-s) + \beta B_{t-s}})]\big|_{y=X^x_s} \\
&= E[f(X^y_{t-s})]\big|_{y=X^x_s},
\end{align*}
where, in the second equality, we use the fact that $B_t-B_s=B_{t-s}$ holds in distribution.

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マルコフ性のためのおべんきょう