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This is a follow up discussion of https://mathoverflow.net/q/43833

The examples there are interesting, but it seems that they all rely on an "ambiguous" transition function, i.e., there is some state $x$, $p(X_t\in A|X_0=x)$ may have different values, depending on how we arrived at $x$ in the first place.

My question: can someone provide an example of markov process which has well defined transition function but without strong Markov property?

Essentially, I am asking why continuity assumption in Feller process is necessary in proving the strong markov property.

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  • $\begingroup$ What do you mean by "continuity assumption"? $\endgroup$ – saz Nov 18 '15 at 17:56
  • $\begingroup$ @saz I mean the requirement that $\endgroup$ – Jay.H Nov 18 '15 at 17:58
  • $\begingroup$ @saz I mean the requirement that $P_tf(x)\to f(x)$ as $t \to 0$ $\endgroup$ – Jay.H Nov 18 '15 at 18:00
  • $\begingroup$ @Jay.H I don't understand. None of the examples in the MO answer have ambiguous transition functions. In no case does $p(X_t\in A\mid X_0=x)$ depend on how we arrived at $x$. $\endgroup$ – user940 Nov 18 '15 at 18:23
  • $\begingroup$ @Jay.H I've posted an answer to what I think is your question. Let me know if it doesn't address your concerns. $\endgroup$ – user940 Nov 18 '15 at 18:58
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To prove that a Feller process is strong Markov, it is not the continuity of $P_tf(x)$ in $t$ that is important, rather the continuity in $x$. That is, we use that $P_t$ maps bounded continuous functions back into bounded continuous functions.

Here is a sketch of the usual proof: Let $T$ be $({\cal G}_{t})$-optional, and let $T_n$ be the standard dyadic approximations to $T$, so that, each $T_n$ is $({\cal G}_t)$-optional and $T_n\downarrow T$ on $\{T<\infty\}$. Fix $u\geq 0$ and $f\in C_b(E)$. Then, by the Markov property over $({\cal G}_t)$ we have \begin{eqnarray*} \mathbb{E}(f(X_{u+T_n})1_{\{T<\infty\}}) &=&\sum_{k=1}^\infty \mathbb{E}(f(X_{u+k/2^n})1_{\{T_n=k/2^n\}})\\ &=&\sum_{k=1}^\infty \mathbb{E}(P_uf(X_{k/2^n})1_{\{T_n=k/2^n\}})\\[5pt] &=& \mathbb{E}(P_uf(X_{T_n})1_{\{T<\infty\}}). \end{eqnarray*}

Since $P_uf$ is a bounded continuous function, and $(X_t)$ has right continuous paths, letting $n\to\infty$ gives $$\mathbb{E}(f(X_{u+T})1_{\{T<\infty\}})= \mathbb{E}(P_uf(X_{T})1_{\{T<\infty\}}).$$

This shows that $(X_t)$ is a strong Markov process.


By way of contrast, consider the Markov process $(X_t)$ on $E=[0,\infty)$ that (starting at the origin) spends an exponential amount of time at the origin, and then moves to the right deterministically at unit speed. It has transition kernel $$p_t(x,A)=\cases{e^{-t}\delta_0(A)+\int_0^t e^{-z}\delta_{t-z}(A)\,dz&if $x=0$\cr \delta_{t+x}(A)&if $x>0$.}$$

For a bounded, measurable $f$ on $E$ we have $$p_tf(x)=\cases{e^{-t}f(0)+\int_0^t e^{-z} f(t-z)\,dz&if $x=0$\cr f(t+x)&if $x>0$.}$$

Notice that even if $f$ is continuous, $p_tf(x)$ will typically have a discontinuity at $x=0$ for $t>0$. The right hand limit is $\lim_{x\downarrow 0}p_tf(x)=f(t)$, while its value $p_tf(0)$ is a weighted average of $f$ over the interval $[0,t]$.

This process $(X_t)$ has continuous sample paths, is not Feller, nor is it strong Markov.

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  • $\begingroup$ Using your formula for $p_t(x)$, it seems to satisfy all conditions of a Feller process. $\endgroup$ – Jay.H Nov 19 '15 at 0:33
  • $\begingroup$ So I am now totally confused. Also, there's this result for Feller process: it can leave a holding point only by a jump(see example, Revuz and Yor, page 110). $\endgroup$ – Jay.H Nov 19 '15 at 0:39
  • $\begingroup$ So, one question I have, the example you provide, are we sure it has continous path? If not, then $X_\tau$ might not be 0, and this will not really an example of Markovian but not strong Markovian. Haven't thought every carefully yet, apologize for any obvious errors that I do not realize yet. $\endgroup$ – Jay.H Nov 19 '15 at 0:44
  • $\begingroup$ @Jay.H For a Feller process, if $f$ is continuous then so is $p_tf$. That is not the case here, so $(X_t)$ is not, I repeat, not Feller. $\endgroup$ – user940 Nov 19 '15 at 1:13
  • $\begingroup$ Got it! Thanks! $\endgroup$ – Jay.H Nov 19 '15 at 1:46

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