Let $E$ be a compact complete separable metric space and $X^n,X$ be $E$-valued càdlàg strong Markov processes on a probability space $(\Omega,\mathcal A,\operatorname P)$.

Question 1: Let $\tau_n$ be a finite stopping on $(\Omega,\mathcal A)$ time wrt the filtration generated by $X^n$ and $(h_n)_{n\in\mathbb N}\subseteq(0,\infty)$ with $h_n\xrightarrow{n\to\infty}0$. Why it sufficient to show $$d\left(X^n_0,X^n_{h_n}\right)\xrightarrow{n\to\infty}0\;\;\;\text{in probability},\tag1$$ in order to conclude $$d\left(X^n_{\tau_n},X^n_{\tau_n+h_n}\right)\xrightarrow{n\to\infty}0\;\;\;\text{in probability}?\tag2$$ I guess we somehow need to utilize the strong Markov property. (If necessary, assume that $\sup_{n\in\mathbb N}\sup_{\omega\in\Omega}\tau_n(\omega)<\infty$.)

I've tried to use the strong Markov property in the following way: Let $(\kappa^n_t)_{t\ge0}$ denote the transition semigroup of $X^n$ and $\mathcal F_{\tau_n}$ the $\sigma$-algebra of the $\tau_n$-past. Then $$\operatorname P\left[d\left(X^n_{\tau_n},X^n_{\tau_n+h_n}\right)>\varepsilon\mid\mathcal F_{\tau_n}\right]=\kappa^n_{h_n}\left(X^n_{\tau_n},\left\{y\in E:d\left(X^n_{\tau_n},y\right)>\varepsilon\right\}\right)\;\;\;\text{almost surely}\tag0,$$ but that doesn't seem to help.

Question 2: By compactness of $E$, we know from Prohorov's theorem that there is a probability measure $\nu$ on $(E,\mathcal B(E))$ with $$X_0^n\xrightarrow{n\to\infty}\nu\;\;\;\text{in distribution}.\tag3$$ Suppose we know that $$\operatorname E\left[f(X^n_0)g\left(X^n_{h_n}\right)\right]\xrightarrow{n\to\infty}\operatorname E\left[(fg)(X_0)\right]\tag4$$ as long as $X_0$ is distributed according to $\nu$ from which we could conclude $$\left(X^n_0,X^n_{h_n}\right)\xrightarrow{n\to\infty}(X_0,X_0)\;\;\;\text{in distribution}\tag5$$ and hence $$\rho\left(X^n_0,X^n_{h_n}\right)\xrightarrow{n\to\infty}\rho(X_0,X_0)=0\;\;\;\text{in distribution}\tag6$$ which in turn yields $(1)$. Why is this enough, i.e. why is it no restriction to assume that $X_0$ is distributed according to $\nu$?


Your Answer

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

Browse other questions tagged or ask your own question.