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$X(t) := 0 \;\; (t \leq \tau),\;\; t - \tau\;\;(t \geq \tau)$ with $\tau$ exponentially distributed.

Then X has the Markov property but not Strong Markov Property. But why ???? Can someone kindly explain in words and maths why the strong markov property does not hold?

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Hint: $X_{\tau+1}-X_\tau$ and $X_1$ have different distributions. –  Ben Derrett Nov 30 '12 at 19:10
    
What happens with this question? Are you satisfied by the answer? –  Did Dec 19 '12 at 21:25

1 Answer 1

For every $t\geqslant0$, the distribution of $(X_{t+s})_{s\geqslant0}$ conditionally on $\sigma(X_u;u\leqslant t)$ is the Dirac distribution at $(X_t+s)_{s\geqslant0}$ on $[X_t\ne0]$ and is $\mu$ on $[X_t=0]$, where $\mu$ denotes the (unconditional) distribution of $(X_s)_{s\geqslant0}$. Thus the distribution of $(X_{t+s})_{s\geqslant0}$ conditionally on $\sigma(X_u;u\leqslant t)$ depends on $X_t$ only and $(X_t)_{t\geqslant0}$ is a Markov process.

On the other hand, $\tau$ is finite almost surely and the distribution of $(X_{\tau+s})_{s\geqslant0}$ conditionally on the past of $\tau$ is the Dirac distribution at $\xi$, where $\xi:s\mapsto s$. This is not $\mu$ hence $(X_t)_{t\geqslant0}$ is not a strong Markov process.

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