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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.2012-11-30

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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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    In my lecture notes, i have the equality $E[f(X_{t+s})|\mathcal{F}_t]=E[f(s-t-\tau)^+]$ on $X_t=0$, but i don't see why this is true.2016-12-07
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    @peer The correct formula is $E(f(X_{t+s})\mid\mathcal F_t)=E(f((s-\tau)^+))$ on $[X_t=0]$. What is it that you do not understand in it?2016-12-07
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    If $X_t(\omega)=0$, then $t\leq T(\omega)$. Thus $E[f(X_{t+s})|\mathcal{F}_t]1_{X_t=0}=E[f(X_{t+s})1_{X_t=0}|\mathcal{F}_t]=E[f((t+s-T)^+)1_{X_t=0}|\mathcal{F}_t]$. What's the next step?2016-12-08
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    @peer That, conditionally on $[X_t=0]=[\tau>t]$, $t+s-\tau$ is distributed like $s-\tau$ unconditionally.2016-12-08
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    So, $P(s-\tau\leq x)$ is $exp(s-x)$ and $P(t+s-\tau\leq x\mid X_t=0)=P(\{t+s-\tau\leq x\}\cap\{\tau>t\})exp(t)$. That means $P(\{t+s-\tau\leq x\}\cap\{\tau>t\})$ has to be $exp(-t+s-x)$?2016-12-08
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    If $\tau$ is standard exponential, then $\{t+s-\tau\leqslant x\}\cap\{\tau>t\}=\{\tau>t+(s-x)^+\}$ hence the probability of this event is $\exp(-t-(s-x)^+)$ (which is not what you wrote).2016-12-08
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    Ok, looks like applying the disintegration theorem would yield the result. But in order to apply this, i have to compute $P(X_{t+s}\in\dot\mid\mathcal{F}_t)$. Is this the same as $P(X_{t+s}\in\dot\mid X_t)$?2016-12-08
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    Quote: `Then X has the Markov property`.2016-12-08
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    Ok, that's intuitivley clear since the knowledge of the state of the function $X(\omega,.)$ at time t yields the whole function before t.2016-12-09
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    @peer Not the correct argument, even intuitively.2016-12-09