My question is if two continuous Markov processes $(Y(t))$ and $(X(t))$ have the same transition kernel is it then true that they have the same finite-dimensional distribution?
More precisely following the notation from p. 49 and onwards in Brownian Motion by Mörters and Peres we say that $p:[0,\infty)\times \mathbb{R}\times \mathscr{B}\rightarrow \mathbb{R}$, where $\mathscr{B}$ denote the Borel subsets of $\mathbb{R}$, is a Markov transition kernel if
- $p(\cdot,\cdot ,A)$ is measurable
- $p(t,x,\cdot)$ is a Borel probability measure.
- For all $x\in \mathbb{R}$, $t,s>0$ and $A\in \mathscr{B}$
$$p(t+s,x,A) = \int_{\mathbb{R}}p(t,y,A)p(s,x,\mathrm{d}y).$$
We say that $X:[0,\infty)\times \Omega\rightarrow \mathbb{R}$ is a Markov process if there exists a Markov transition kernel $p$ and a filtration $\mathcal{F}(s)$ on $\Omega$ such that
$$\mathbb{P}(X(t)\in A\mid \mathcal{F}(s)) = \mathbb{E}(1_{X(t)\in A}\mid \mathcal{F}(s)) = p(t-s,X(s),A).$$
My question regards when two Markov processes are the same (in distribution). Suppose that $(Y(t))$ and $(X(t))$ are two Markov processes such that $t\mapsto (X(t,\omega), Y(t,\omega))$ is continuous for each $\omega$. Let $p_X$ and $p_Y$ be the Markov transition kernels of $X$ and $Y$ respectively. If $p_X = p_Y$
does it then hold that $X$ and $Y$ have the same finite-dimensional distributions?
To show this we need to show that for each Borel set $A\in \mathbb{R}^n$
$$\mathbb{P}((X(t_1),\dotsc,X(t_n))\in A) = \mathbb{P}((Y(t_1),\dotsc,Y(t_n))\in A).$$
I only see how this follows in the case where $n=1$, otherwise I don't see how to do this, could someone help me with this?
Best Answer
Provided $X$ and $Y$ have the same initial distributions, then Yes: the finite dimensional distributions of $X$ (for example) can be expressed in terms of integrals involving its initial distribution and the transition kernel.
Taking $n=2$, if $X$ has initial distribution $\mu$ then for $t_1<t_2$, $$ \Bbb P[(X(t_1),X(t_2))\in A] =\int\int\int 1_A(y,z)\mu(dx) p(t_1,x,dy)p(t_2-t_1,y,dz). $$ (Each of the integrals is over $\Bbb R$.)