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I want to prove the Markov property of the process $X$ defined by $$X_t:=(t-\tau)^+$$ where $\tau\sim \exp(1)$ with respect to the filtration $(\mathcal{F}_t)_{\geq 0}$ defined by $\mathcal{F}_t=\sigma\{X_k:0\leq k\leq t\}\vee N_0 $ with $N_0$ as the $P$-null sets (this process is an example of a markov-process, which does not have the strong markov property).

In order to do that, one has to show that for $t>s$ the equality $P(X_t\in A\mid\mathcal{F}_s)=P(X_t\in A\mid X_s)$ holds for $A\in\mathcal{B}(\mathbb{R})$. It should then be enough to prove $$E[P(X_t\leq x|\mathcal{F}_s);\{X_k\leq y\}]=E[P(X_t\leq x|X_s);\{X_k\leq y\}]$$ for $k\leq s$.

If I start on the left side, I get $E[P(X_t\leq x|\mathcal{F}_s);\{X_k\leq y\}]=P[\{X_t\leq x\}\cap\{X_k\leq y\}]=P[\{(t-\tau)^+\leq x\}\cap\{(k-\tau)^+\leq y\}]$. I am not sure how to go on?

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1 Answer 1

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Define the process $X_t(\omega):=(t-\tau(\omega))_+.$ Then for $r\leq s$, we have $X_r(\omega)=(X_s(\omega)-(s-r))_+.$ This shows that ${\cal F}_s:=\sigma(X_r: r\leq s)=\sigma(X_s),$ and so $\mathbb{E}(f(X_t)\mid{\cal F}_s)=\mathbb{E}(f(X_t)\mid X_s)$ trivially. Therefore $(X_t)$ is a Markov process.

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    $\begingroup$ Thus the distribution of $\tau$ is irrelevant. $\endgroup$
    – Did
    Jan 23, 2017 at 17:04

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