1
$\begingroup$

Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $I\subseteq\mathbb R$
  • $(\mathcal F_t)_{t\in I}$ be a filtration on $(\Omega,\mathcal A)$
  • $(E,\mathcal E)$ be a measurable space
  • $X$ be a $(E,\mathcal E)$-valued $\mathcal F$-Markov process with transition semigroup $\left(\kappa_{s,\:t}:s,t\in I\text{ with }s\le t\right)$

We can show that$^1$ $$\operatorname P\left[X_{t_1}\in B_1,\ldots,X_{t_n}\in B_n\right]\\=\operatorname E\left[1_{\left\{\:X_{t_1}\:\in\:B_1\:\right\}}\operatorname E_{X_{t_1}}\left[1_{\left\{\:X_{t_2}\:\in\:B_2\:\right\}}\operatorname E_{X_{t_2}}\left[\cdots\operatorname E_{X_{t_{n-1}}}\left[1_{\left\{\:X_{t_n}\:\in\:B_n\:\right\}}\right]\right]\right]\right]\tag1$$ almost surely for all $n\in\mathbb N$, $t_1,\ldots,t_n\in I$ with $t_1\le\cdots\le t_n$ and $B_1,\ldots,B_n\in\mathcal E$.

By definition of a Markov process, $$\operatorname P\left[X_t\in B\mid\mathcal F_s\right]=\kappa_{s,\:t}(X_s,B)\tag2$$ almost surely for all $B\in\mathcal E$ and $s,t\in I$ with $s\le t$.

How can we conclude from $(1)$ and $(2)$ that $$\operatorname P\left[X_{t_1}\in B_1,\ldots,X_{t_n}\in B_n\right]\\=\int\operatorname P\left[X_{t_0}\in{\rm d}x_0\right]\int_{B_1}\kappa_{t_0,\:t_1}(x_0,{\rm d}x_1)\cdots\int_{B_n}\kappa_{t_{n-1},\:t_n}(x_{n-1},{\rm d}x_n)\tag3$$ almost surely for all $n\in\mathbb N$, $t_0,\ldots,t_n\in I$ with $t_0\le\cdots\le t_n$ and $B_1,\ldots,B_n\in\mathcal E$?


$^1$ In $(1)$ I'm using the notation $\operatorname E_{\mathcal G}[Y]:=\operatorname E\left[Y\mid\mathcal G\right]$.

$\endgroup$
3
  • $\begingroup$ So where exactly is your problem? Do you know how to prove $(3)$ for $n=2$? (It seems a bit weird that there is a $t_0$ at the right-hand side of $(3)$ whereas there is no $t_0$ on the left-hand side... but I suppose that this is just a typo.) $\endgroup$
    – saz
    Oct 14 '18 at 18:03
  • $\begingroup$ @saz It actually is not a typo. I've seen this equation with $I=[0,\infty)$ and $t_0=0$ and thought it wouldn't make any difference. $\endgroup$
    – 0xbadf00d
    Oct 14 '18 at 18:16
  • $\begingroup$ Ah I see, sorry, my mistake (I mistook the first integral for $\int_{B_0} P(X_{t_0} \in dx_0) $ and not $\int P(X_{t_0} \in dx_0)$.) $\endgroup$
    – saz
    Oct 14 '18 at 18:22
1
$\begingroup$

I will prove $(3)$ only for $n=2$; for larger $n$ you can proceed by iterating the reasoning which I describe below (or, formally, perform a proof by induction).

Using the tower property for conditional expectation, we find from $(2)$ that

$$\begin{align*} \mathbb{E}(1_B(X_t) \mid X_s) = \mathbb{E} \big[ \mathbb{E}(1_B(X_t) \mid \mathcal{F}_s) \mid X_s \big] &\stackrel{(2)}{=} \mathbb{E} \big[ \kappa_{s,t}(X_s,B) \mid X_s \big] \\ &= \kappa_{s,t}(X_s,B) \end{align*}$$

for any $s \leq t$ and $B \in \mathcal{E}$. In your notation, this means that

$$\mathbb{E}_{X_s}(1_B(X_t)) = \kappa_{s,t}(X_s,B). \tag{4}$$

Using a standard monotone class argument, it is not difficult to see that this implies

$$\mathbb{E}_{X_s}(h(X_t)) = \int h(y) \, \kappa_{s,t}(X_s,dy) \tag{5}$$

for any bounded measurable function $h$. Now fix $B_1,B_2 \in \mathcal{E}$ and $t_0 \leq t_1 \leq t_2$. By $(1)$, we have

$$\begin{align*} \mathbb{P}(X_{t_1} \in B_1,X_{t_2} \in B_2) &= \mathbb{P}(X_{t_0} \in E, X_{t_1} \in B_1,X_{t_2} \in B_2) \\ &= \mathbb{E} \big[ 1_{\{X_{t_0} \in E\}} \mathbb{E}_{X_{t_0}} \big[ 1_{\{X_{t_1} \in B_1\}} \mathbb{E}_{X_{t_1}} \big[ 1_{\{X_{t_2} \in B_2\}} \big] \big] \big]. \end{align*}$$

By $(4)$, we get

$$\begin{align*} \mathbb{P}(X_{t_1} \in B_1,X_{t_2} \in B_2) = \mathbb{E} \big[ 1_{\{X_{t_0} \in E\}} \mathbb{E}_{X_{t_0}} \big[ 1_{\{X_{t_1} \in B_1\}} \kappa_{t_1,t_2}(X_{t_1},B_2) \big] \big] \tag{6} \end{align*}$$

Applying $(5)$ for $h(y) := 1_{B_1}(y) \kappa_{t_1,t_2}(y,B_2)$, $s:=t_0$ and $t:=t_1$ shows

$$\mathbb{E}_{X_{t_0}} \big[ 1_{\{X_{t_1} \in B_1\}} \kappa_{t_1,t_2}(X_{t_1},B_2) \big] = \int 1_{B_1}(x_1) \kappa_{t_1,t_2}(y,B_2) \kappa_{t_0,t_1}(X_{t_0},dx_1).$$

Plugging this into $(6)$ we obtain that

$$\begin{align*} \mathbb{P}(X_{t_1} \in B_1,X_{t_2} \in B_2) =\mathbb{E}\left(1_{\{X_{t_0} \in E\}} \int_{B_1} \kappa_{t_1,t_2}(x_1,B_2) \, \kappa_{t_0,t_1}(X_{t_0},dx_1) \right). \end{align*}$$

Writing all the expressions on the right-hand side in terms of integrals, we get

$$\mathbb{P}(X_{t_1} \in B_1,X_{t_2} \in B_2) = \int_E \left( \int_{B_1} \left( \int_{B_2} \kappa_{t_1,t_2}(x_1,dx_2) \right) \kappa_{t_0,t_1}(x_0,dx_1) \right) \, \mathbb{P}(X_{t_0} \in dx_0)$$

which is $(3)$ (for $n=2$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.