To formulate my question, I present the relevant definitions from page 24 of these notes.
Denote by $p_t(x, B)$ a Markov transition function, where $t \geq 0$, $x \in \mathbb{R}^d$, $B \in \mathcal{B}( \mathbb{R}^d)$. This means that $B \mapsto p_s (x,B)$ is a distribution on $\mathbb{R}^d$; $(s, x) \mapsto p_s (x, B)$ is Borel measurable; the Chapman-Kolmogorov equations hold, i.e.
$$
p_{s+t} ( x, B ) = \int_{\mathbb{R}^d} p_t (y, B) p_s ( x, dy ) \quad s,t \geq 0.
$$
Now let $P^x$ be a probability measure on the measurable space $(\Omega, \mathcal{F})$ for each $x \in \mathbb{R}^d$. Consider an $\mathbb{R}^d$-valued stochastic process $(X_t)_{t \geq 0}$ adapted to the filtration $( \mathcal{F}_t)_{t \geq 0}$. Let us call $(X_t)_{t \geq 0}$ a Markov process if
- $P^x ( X_0 = x ) = 1$;
- $p_t ( x, B ) := P^x ( X_t \in B)$ is a Markov transition function;
- $P^x ( X_t \in B | \mathcal{F}_s) = p_{t -s} ( X_s, B )$ $P^x$-a.s. Taking into account the previous point, this means that $P^x ( X_t \in B | \mathcal{F}_s) = P^{X_s} ( X_{t-s} \in B)$ $P^x$-a.s., $s \leq t$.
Now assume that $0 \leq t_1 < t_2 <t_3$ and $B_1, B_2, B_3 \in \mathcal{B}( \mathbb{R}^d)$. I am trying to show that
\begin{align} &P^x ( X_{t_1} \in B_1, X_{t_2} \in B_2, X_{t_3} \in B_3) \\ &= \int_{B_1} \left( \int_{B_2} \left( \int_{B_3} p_{t_3-t_2} (x_2, d x_3) \right) p_{t_2 -t_1} ( x_1, dx_2)\right) p_{t_1} ( x, d x_1)
\end{align}
My attempt:
\begin{align*}
&P^x ( X_{t_1} \in B_1, X_{t_2} \in B_2, X_{t_3} \in B_3) = E^x \left[ 1_{ \{ B_1 \times B_2 \times B_3 \} } ( X_{t_1}, X_{t_2}, X_{t_1} ) \right] \\ &= E^x [ 1_{ B_1 } (X_{t_1}) \cdot 1_{ B_2 } (X_{t_2}) \cdot 1_{ B_3 } (X_{t_3}) ] = E^x [ 1_{ B_1 } (X_{t_1}) \cdot E^x [ 1_{ B_2 } (X_{t_2}) | \mathcal{F}_{t_1}] \cdot E^x [ 1_{ B_2 } (X_{t_3}) | \mathcal{F}_{t_2}] ] \\
&= E^x [ 1_{ B_1 } (X_{t_1}) \cdot P^x [ X_{t_2} \in B_2 | \mathcal{F}_{t_1}] \cdot P^x [ X_{t_3} \in B_3 | \mathcal{F}_{t_2}] ] \\
&= E^x [ 1_{ B_1 } (X_{t_1}) \cdot p_{t_2-t_1} ( X_{t_2}, B_2) \cdot p_{t_3-t_2} ( X_{t_3}, B_3) ] = \ldots
\end{align*}
How can one proceed from here? I think the Chapman-Kolmogorov equations should somehow come into play.
Best Answer
I think your idea is good. Using the properties of conditional expectation, it could go like this:
\begin{align*} &P^x ( X_{t_1} \in B_1, X_{t_2} \in B_2, X_{t_3} \in B_3) = E^x \left[ 1_{B_1}(X_{t_1}) \cdot 1_{B_2}(X_{t_2}) \cdot 1_{B_3}(X_{t_3}) \right] \\ &= E^x \left[ 1_{B_1}(X_{t_1}) \cdot E^x \left[ 1_{B_2}(X_{t_2}) \cdot 1_{B_3}(X_{t_3}) | \mathcal{F}_{t_1} \right] \right] \\ &= E^x \left[ 1_{B_1}(X_{t_1}) \cdot E^x \left[ 1_{B_2}(X_{t_2}) \cdot E^x \left[ 1_{B_3}(X_{t_3}) | \mathcal{F}_{t_2} \right] | \mathcal{F}_{t_1} \right] \right] \\ & = E^x \left[ 1_{B_1}(X_{t_1}) \cdot E^x \left[ 1_{B_2}(X_{t_2}) \cdot E^{X_{t_2}} \left[ 1_{B_3}(X_{t_3-t_2}) \right] | \mathcal{F}_{t_1} \right] \right] \\ & = E^x \left[ 1_{B_1}(X_{t_1}) \cdot E^x \left[ 1_{B_2}(X_{t_2}) \cdot \left( \int_{B_3} p_{t_3-t_2} (X_{t_2}, d x_3) \right) | \mathcal{F}_{t_1} \right] \right] \\ & = E^x \left[ 1_{B_1}(X_{t_1}) \cdot E^{X_{t_1}} \left[ 1_{B_2}(X_{t_2-t_1}) \cdot \left( \int_{B_3} p_{t_3-t_2} (X_{t_2-t_1}, d x_3) \right) \right] \right] \\ & = E^x \left[ 1_{B_1}(X_{t_1}) \cdot \left( \int_{B_2} \left( \int_{B_3} p_{t_3-t_2} (x_2, d x_3) \right) p_{t_2-t_1} (X_{t_1}, d x_2) \right) \right] \\ &= \int_{B_1} \left( \int_{B_2} \left( \int_{B_3} p_{t_3-t_2} (x_2, d x_3) \right) p_{t_2-t_1} (x_1, d x_2) \right) p_{t_1} (x, d x_1) \end{align*}
Chapman-Kolmogorv equations are an equivalent writing, meaning that:
\begin{align*} & P^x \left( X_{t_1} \in B_1, X_{t_2} \in B_2, X_{t_3} \in B_3 \right) \\ &= \int_{B_1} \left( \int_{B_2} \left( \int_{B_3} p_{t_3-t_2} (x_2, d x_3) \right) p_{t_2-t_1} (x_1, d x_2) \right) p_{t_1} (x, d x_1) \\ & = \int_{B_1 \times B_2 \times B_3} p_{t_1} (x, d x_1) p_{t_2-t_1} (x_1, d x_2) p_{t_3-t_2} (x_2, d x_3) \end{align*}
Hope it helps!
Remark: we could do the same in dimension n, using recurrence.