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In my study of strong Markov property of an RCLL canonical Markov process I encounter the following definition:

Suppose $Y_t:\omega\rightarrow \omega(t)$ is canonical Markov process with respect to its raw filtration $\mathbb{F}^0$, with transition probability kernel $(K_t)_t$, where $\omega$ is RCLL function of $t$ taking values in polish space S, then $Y$ has strong Markov property wrt to the RC filtration $\mathbb{F}$ if for all $\mathbb{F}$-stopping time $\tau$,

$$E[f(Y_{\cdot+\tau})\mid\mathcal{F}_\tau]=E_{Y_\tau}[f(Y)]$$ on $\{\tau<\infty\}$, where $E_\mu$ refer to taking expectation conditional on initial value $\mu$ and the same transition probability, and $f(\omega)$ bounded measurable real positive RV of the sample path.

Then there is this claim that the above condition is equivalent to for all $\mathbb{F}$-stopping time $\tau$,

$$E[f(Y_{\cdot+\tau}){\bf 1}_{\tau<\infty}]=E[E_{Y_\tau}[f(Y)]{\bf 1}_{\tau<\infty}],$$

i.e., the first equation only need to hold in integrated form. I tried to prove this claim but was not able to make any progresses. Any suggestions? Thank you.

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  • $\begingroup$ Is it not just taking the expectation of both sides and applying iterated conditioning? Oh wait, that holds in one direction. Are you sure about the equivalency part? $\endgroup$
    – Calculon
    Aug 5, 2015 at 12:18
  • $\begingroup$ I think so. In the next step this equivalence was used to prove RCLL Feller processes have strong MP, using countable approximations to stopping times and the regular MP. $\endgroup$
    – user138668
    Aug 5, 2015 at 13:15
  • $\begingroup$ @user138668 By finite stopping time, do you mean it only takes finitely many values? Or do you mean a bounded stopping time? There is a big difference and I think bounded is what you want. Could you tell us the name of your textbook? $\endgroup$
    – user940
    Aug 5, 2015 at 13:51
  • $\begingroup$ @ByronSchmuland, Yes I refer to bounded stopping times. This is from our lecture notes rather than a textbook, many lemmas are left as exercises. $\endgroup$
    – user138668
    Aug 5, 2015 at 14:08

1 Answer 1

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Let $\tau$ be a bounded $\mathbb{F}$-stopping time and $F \in \mathcal{F}_{\tau}$. We define a new stopping time by setting

$$\varrho(\omega) := \begin{cases} \tau(\omega), & \omega \in F, \\ \infty, &\text{otherwise} \end{cases}.$$

Then, by assumption,

$$\mathbb{E}(f(Y_{\cdot+\varrho}) 1_{\{\varrho<\infty\}}) = \mathbb{E}(\mathbb{E}_{Y_{\varrho}}(f(Y)) 1_{\{\varrho<\infty\}}),$$

i.e.

$$\mathbb{E}(f(Y_{\cdot+\tau}) 1_F) = \mathbb{E}(\mathbb{E}_{Y_{\tau}}(f(Y)) 1_F).$$

Since this holds for any $F \in \mathcal{F}_{\tau}$, we conclude

$$\mathbb{E}(f(Y_{\cdot+\tau}) \mid \mathcal{F}_{\tau}) = \mathbb{E}_{Y_{\tau}}(f(Y)).$$

Remark: A very similar result holds true for martingales. In fact, an adapted integrable process $(M_t,\mathcal{F}_t)_{t \geq 0}$ is a martingale if and only if

$$\mathbb{E}(M_{\varrho}) = \mathbb{E}(M_{\tau})$$

for all bounded ($\mathcal{F}_t)$-stopping times $\varrho$ and $\tau$.

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    $\begingroup$ Your $\varrho$ is not a stopping time. Usually we set $\varrho$ to be equal to $\tau$ on $F$, and $\infty$ otherwise, not zero otherwise. With $\infty$, this new stopping time is not bounded, or finite (whichever of these the OP wants) $\endgroup$
    – user940
    Aug 5, 2015 at 14:04
  • $\begingroup$ @ByronSchmuland, sorry, I made a bad mistake in the statement of the problem, causing your confusion. Now it has been fixed. $\endgroup$
    – user138668
    Aug 5, 2015 at 14:58
  • $\begingroup$ @saz, following your approach with Byron Schmuland's comments in mind, I was able to fix the problem statement and make the proof. Thank you guys very much! $\endgroup$
    – user138668
    Aug 5, 2015 at 15:29
  • $\begingroup$ @ByronSchmuland Argh, you are right of course; thanks for pointing this out. $\endgroup$
    – saz
    Aug 5, 2015 at 16:40
  • $\begingroup$ @saz +1 Glad we got it all straightened out! $\endgroup$
    – user940
    Aug 5, 2015 at 17:04

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