• $(\Omega,\mathcal A,\operatorname P)$ 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 $\operatorname E_{\mathcal G}[Y]:=\operatorname E\left[Y\mid\mathcal G\right]$.

  • $\begingroup$ 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

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 $n=2$).


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