# $(X_{z+t}-X_{z})_{t\geq 0}$ satisfies “Strong Markov Property” where $X$ is càdlag process and $z$ discrete stopping time.

I'm reading about strong Markov property. In the text there is the next proposition which I'm, trying to prove:

If $X$ is a càdlag process and $z$ is a discrete stopping time, then $(X_{z+t}-X_{z})_{t\geq 0}$ satisfies:

a) $(X_{z+t}-X_{z})$ has the same distribution as $X_{t}$ for all $t\geq 0.$

b) $(X_{z+t}-X_{z})$ is independent of $\mathcal{F}_{z}.$

I'm stuck, because I can't see how to get the relation between right-continuity and the discrete stopping time. I thought that was necessary that $X$ be a Levy process, but in this case we have a kind of strong Markov property and my feeling is this proposition can be used to prove it.

How Could be proved this proposition?

Any kind of help is thanked in advanced.

• If $X$ is just some cadlag process, then $X_{z+t}-X_t$ has, in general, not the same distribution as $X_t$. You have indeed to assume that $X$ is a Lévy process. – saz Apr 2 '18 at 9:51
• Sorry @saz. I have made a mistake. Inmediatly I fixed it. – Squird37 Apr 2 '18 at 14:04
• Yeah, sure, but that doesn't change anything. You still have to assume that $X$ is Lévy; otherwise the assertion is wrong. – saz Apr 2 '18 at 17:09
• @saz How can We prove it if $X$ is Levy Process? – Squird37 Apr 2 '18 at 19:20