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Jun 22, 2021 at 12:36 comment added Iosif Pinelis @Joe_Affine : It probably should be an exercise/remark in some book on Markov processes/chains, but I don't know such a reference.
Jun 22, 2021 at 12:34 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 22, 2021 at 12:26 comment added Joe_Affine Do you know were I can find a reference to this later fact?
Jun 22, 2021 at 12:15 comment added Iosif Pinelis That's right, here basically we use the fact that, if $(X_t)$ is Markov, then $(f(X_t))$ does not have to be Markov.
Jun 22, 2021 at 12:12 comment added Joe_Affine Amazing example, but if I left out $f$ and instead considered $(\mathbb{E}(X_t)|mathcal{G}_t])_t$ would this always be Markovian (since it seems the trucation of $\max\{0,\cdot\}$ causes the issue.
Jun 22, 2021 at 12:05 vote accept Joe_Affine
Jun 22, 2021 at 12:02 history answered Iosif Pinelis CC BY-SA 4.0