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I don't understand how to show that some stochastic processes have the Markov property. For example, if I have the following process:

$$(\Omega, \mathcal{F}, (X_t)_{t \geq 0}, P^y)$$ where $\Omega = \mathbb{R}$, $\mathcal{F} = \mathcal{B}_{\mathbb{R}}$, $P^y = \delta_y$ (Dirac probability), $y \in \mathbb{R}$ and for every $t \geq 0$ $$ Y_t(\omega) = \omega + t, \omega \in \Omega $$

how can I show one of the equivalent forms of Markov Property? For example: $$ \mathbb{P}(X_t \in A | \mathcal{F}_t^X) = \mathbb{P} (X_t \in A | X_s) $$ where $s, t \geq 0$ and $s < t$ or, equivalently, $$ \mathbb{P} (A \cap B | X_t) = \mathbb{P} (A | X_t) \mathbb{P} (B | X_t) $$ for all $A \in \mathcal{F}_t^X$, $B \in \sigma (X_s : s \geq t)$?

Thank you!

g.pomegranate
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    In general, write $X_t=G_{s,t}(X_s,Z_{s,t})$ for some measurable function $G_{s,t}$ and $Z_{s,t}$ independent of $\mathcal F_s$. In your case, note that $Y_t=Y_s+t-s$ hence $Y$ is Markov. – Did Jan 15 '15 at 17:49
  • I don't understand why Y is Markov. Can you be a little more explicit? Thank you! – g.pomegranate Jan 19 '15 at 18:13
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    Because $Y_t=G_{s,t}(Y_s)$ for some measurable function $G_{s,t}$ hence we are in a special case of my previous, general, comment. – Did Jan 19 '15 at 20:12

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