3
$\begingroup$

$X(t) := 0 \;\; (t \leq \tau),\;\; t - \tau\;\;(t \geq \tau)$ 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?

  • 0
    Hint: $X_{\tau+1}-X_\tau$ and $X_1$ have different distributions.2012-11-30

1 Answers 1

4

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.

  • 0
    @peer Not the correct argument, even intuitively.2016-12-09