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Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$
  • $E$ be a metric space
  • $(X_t)_{t\ge0}$ be an $E$-valued right-continuous time-homogeneous Markov process on $(\Omega,\mathcal A,\operatorname P)$
  • $\kappa_t$ be a regular version of the conditional probability of $X_t$ given $X_0$, i.e. $\kappa_t$ is a Markov kernel on $(E,\mathcal B(E))$ with $$\operatorname P\left[X_{s+t}\in B\mid\mathcal F_s\right]=\operatorname P\left[X_t\in B\mid X_0\right]=\kappa_t(X_0,B)\;\;\;\text{almost surely for all }B\in\mathcal B(E)\tag1$$ for all $s,t\ge0$
  • $f:E\to\mathbb R$ be bounded and continuous

Are we able to show that $$[0,\infty)\ni t\mapsto(\kappa_t f)(x):=\int\kappa_t(x,{\rm d}y)f(y)\tag1$$ is continuous for all $x\in\mathbb R$?

Let's take a look: Let $(t_n)_{n\in\mathbb N}\subseteq[0,\infty)$ and $t\ge0$ with $t_n\xrightarrow{n\to\infty}t$. For simplicity, assume $t=0$. By the dominated convergence theorem, $$\operatorname E\left[f\left(X_{t_n}\right)\mid X_0\right]\xrightarrow{n\to\infty}\operatorname E\left[f\left(X_0\right)\mid X_0\right]=f(X_0)\;\;\;\text{almost surely}\tag2$$ and hence $$\operatorname E\left[f\left(X_{t_n}\right)\mid X_0=\;\cdot\;\right]\xrightarrow{n\to\infty}\operatorname E\left[f(X_0)\mid X_0=\;\cdot\;\right]=f\;\;\;\operatorname P\circ\:X_0^{-1}\text{-almost surely}.\tag3$$ By $(1)$, $$\operatorname E\left[f\left(X_{t_n}\right)\mid X_0=\;\cdot\;\right]=\kappa_{t_n}f\;\;\;\text{for all }n\in\mathbb N\;\operatorname P\circ\:X_0^{-1}\text{-almost surely}.\tag4$$ Thus, $$\kappa_{t_n}f\xrightarrow{n\to\infty}f\;\;\;\operatorname P\circ\:X_0^{-1}\text{-almost surely}.\tag4$$

The problem with $(4)$ is the dependence of the null set on $(t_n)_{n\in\mathbb N}$.

However, $\left|\kappa_{t_n}\right|\le\left\|f\right\|_\infty$ for all $n\in\mathbb N$ and hence another application of the dominated convergence theorem yields $$\operatorname E\left[\left(\kappa_{t_n}f\right)(X_0)\right]\xrightarrow{n\to\infty}\operatorname E\left[f(X_0)\right]\tag5.$$

If we fix an $x\in E$ and assume $\operatorname P\circ\:X_0^{-1}=\delta_x$, the continuity of $(1)$ is obvious from $(5)$.

Now, Mars Plastic wrote in his answer that the assumption $\operatorname P\circ\:X_0^{-1}=\delta_x$ is no restriction.

Why?

I have some crazy idea: Assume $E$ is separable. Let $D([0,\infty),E)$ denote the space of càdlàg functions $[0,\infty)\to E$ equipped with the Skorohod topology and $$\pi_t:D([0,\infty),E)\to E\;,\;\;\;x\mapsto x(t)$$ for $t\ge0$. By separability, $$\mathcal B\left(D([0,\infty),E)\right)=\sigma(\pi_t:t\ge0)\tag6.$$ Now, assume that there is a Markov kernel $\kappa$ with source $(E,\mathcal E)$ and target $$(\tilde \Omega,\tilde{\mathcal A}):=\left(D([0,\infty),E),\mathcal B\left(D([0,\infty),E)\right)\right)$$ with $$\kappa(x,\;\cdot\;)\circ\left(\pi_{t_0},\ldots,\pi_{t_n}\right)^{-1}=\delta_x\otimes\bigotimes_{i=1}^n\kappa_{t_i-t_{i-1}}\;\;\;\text{for all }n\in\mathbb N\text{ and }0=t_0<\cdots<t_n\tag7$$ for all $x\in E$. Then, if $\mu$ is any probability measureon $(E,\mathcal E)$ and $$\tilde{\operatorname P}[\tilde A]:=(\mu\kappa)(\tilde A)=\int\mu({\rm d}x)\kappa(x,\tilde A)\;\;\;\text{for }\tilde A\in\tilde{\mathcal A},$$ then it's easy to see that $(\pi_t)_{t\ge0}$ is a càdlàg Markov process on $(\tilde\Omega,\tilde{\mathcal A},\tilde{\operatorname P})$ with transition semigroup $(\kappa_t)_{t\ge0}$ and initial distribution $\operatorname P\circ\:\pi_0^{-1}=\mu$.

Clearly, we could choose $\mu=\delta_x$ for some fixed $x\in E$ and immediately obtain the continuity of $(1)$ by the same argumentation as before.

First of all, I have no idea if we can prove the existence of $\kappa$. As you may guess, my idea was inspired by the usual existence proof of a Markov process with a given transition semigroup and initial distribution. Therein, $E$ is assumed to be a Polish space, $\pi_t$ is replaced by $E^{[0,\:\infty)}\ni x\mapsto x(t)$ and $(\tilde\Omega,\tilde{\mathcal A})$ is replaced by $\left(E^{[0,\:\infty)},\mathcal E^{\otimes[0,\:\infty)}\right)$. The definition of $\tilde{\operatorname P}$ is the same.

So, if this is the correct approach, the concrete Markov process $(X_t)_{t\ge0}$ in the question is of no use. All we need is the transition semigroup $(\kappa_t)_{t\ge0}$ satisfying the usual consistency conditions (Chapman-Kolmogorov equations) and then hope that $E$ is "nice enough" to admit the existence of $\kappa$.

So, the question is: Is this the correct approach? And if so: When is $E$ is "nice enough"?

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1 Answer 1

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Your presentation is slightly bulky, but the general argument is correct. Note that you can fix a point $x\in \mathbb R$ and that it suffices to consider $P\circ X_0^{-1}=\delta_{x}$. Then your reasoning yields that for any $(t_n)\subset\mathbb [0,\infty)$ converging to $t\in[0,\infty)$ we have \begin{equation} (\kappa_{t_n}f)(x)=E[f(X_{t_n})|X_0=x] \to E[f(X_t)|X_0=x]=(\kappa_tf)(x), \quad n\to\infty, \end{equation} and that's all you need. There are no null sets which depend on anything that bothers us.

PS: Since you are in a time-homogeneous setting, it is indeed irrelevant if you consider $t=0$ or any other $t\in [0,\infty)$.

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  • $\begingroup$ Why can we assume $\operatorname P\circ X_0^{-1}=\delta_x$? $\endgroup$
    – 0xbadf00d
    Jan 21, 2019 at 15:51
  • $\begingroup$ Assume $t=0$. In general, it's clear to me that $$\kappa_{t_n}f\xrightarrow{n\to\infty}f\;\;\;\operatorname P\circ\:X_0^{-1}\text{-almost surely}.\tag5$$ However, the null set in $(5)$ should depend on $(t_n)_{n\in\mathbb N}$. Applying the bounded convergence theorem yields $$\operatorname E\left[\kappa_{t_n}f(X_0)\right]\xrightarrow{n\to\infty}\operatorname E\left[f(X_0)\right]\tag6$$ and hence the claim is clear to me, if $\operatorname P\circ\:X_0^{-1}=\delta_x$ for a fixed $x\in E$. But how does the general case follow? $\endgroup$
    – 0xbadf00d
    Jan 22, 2019 at 13:07

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