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I recently had great success with my first question here so I will boldly go on to a second. Here goes:

I'm studying Markov Chains in Rick Durrett - Probability: Theory and example and I'm stuck with the definition of the strong markov property - I know more or less what it should be, but do not understand his way of saying it. I'm gonna give you a lot of information, hopefully enough but please ask for more if you need it.

Some definitions: We have some nice (1-1 map between S and R) measurable space $(S,\mathcal{S)}$ and we then define $$\Omega=\{{(\omega_{1},\omega_{2},...):\omega_{i}\in\text{}S}\}$$ $$\mathcal{F}=\mathcal{S}\times\mathcal{S}\times...$$ $$P=\mu\times\mu\times...\text{where }\mu\text{ is the distribution of }X_{i}$$ $$X_{n}(\omega)=\omega_{n}$$

We have $$P(X_{j}\in B_{J},0\leq j \leq n)=\int_{B_{0}}\mu(dx_{0})\int_{B_{1}}p(x_{0},dx_{1})...\int_{B_{n}}p(x_{n-1},dx_{n})$$ Where the p's are transition probabilities (for fixed x (first variable) it's a probability measure and fixed set (second variable) a measurable function). The probability measure is consistent so Kolmogorov's extension theorem gives us the infinite one (as I understand).

His definition is then as follows:

Suppose that for each n, $Y_{n}:\Omega\rightarrow\mathbb{R}$ is measurable and $|Y_{n}|\leq M\; \forall n$ Then $$E_{\mu}(Y_{N}\circ\theta_{N}|\mathcal{F}_{N})=E_{X_{N}}Y_{N}\quad on\:\{N<\infty\}$$ N is a stoptime and theta a shift operator ("drops the first N elements of the omega-sequence")

So i know i am being a bit imprecise here - I reckon I know what all the elements of the theorem are, but have trouble adding it all up. I hope someone bothers to help.

Thanks in advance, Henrik

Update v3.b:

I almost figured it out, I will update with my findings shortly (hope someone cares).

So I still have some problems; can someone help me with these notions: $P_{x}=P_{\delta_{x}}$, why $P_{\mu}(A)=\int P_{x}(A)\, \mu(dx)$ and what $E_{X_{n}}$ looks like explicitly.

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I am not into probability, but +1 for giving as much details as possible. –  Patrick Da Silva Apr 3 '12 at 21:31
    
Henrik, just to be clear is your question here "why does this definition of the strong markov property correspond to my intuition?" If so, then consider what happens if you just look at a single random variable $Y$ and the hitting time of a point $x_0$. $E_\mu (Y \circ \theta_N | \mathfrak{F}_N)$ is the conditional expectation of $Y$ if you start it at the random time $N$ conditioned on the algebra generated by that random time on the event that that makes any sense. The right hand side is the conditional expectation of a process started at $x_0$. Durrett presents things in full generality –  Chris Janjigian Apr 4 '12 at 0:06
    
so there is a lot going on in this definition, but the basic idea here is that when you stop at a random time, you still get the Markov property that "conditioned on the present, the future is independent of the past". –  Chris Janjigian Apr 4 '12 at 0:07
    
About your update: sorry but what do you want to clarify? The random variable $\phi(X_N,N)$ is clearly $\mathcal F_N$-measurable and one asks that it coincides with the random variable $E_\mu(Y_N\circ\theta_N\mid \mathcal F_N)$ on the event $[N\lt\infty]$. –  Did Apr 6 '12 at 15:20
    
That clarification was perfect! Thank you very much. I'm sorry if my questions are stupid, I haven't worked with Markov chains nor conditional expectations for very long - but it is of course obvious to just check that it satisfies the two conditions. –  Henrik Apr 6 '12 at 17:24
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1 Answer

up vote 1 down vote accepted

Markov chains are irrelevant here hence let us concentrate on the strong Markov property. Start from any process $(Z_n)_{n\geqslant0}$ (this is just a collection of random variables defined on the same probability space) and call $F_n$ the sigma-algebra generated by $(Z_k)_{k\leqslant n}$.

Then, the simple Markov property is to ask that $$ \mathrm E(\varphi(Z_n,Z_{n+1},Z_{n+2},\ldots)\mid F_n)=\mathrm E(\varphi(Z_n,Z_{n+1},Z_{n+2},\ldots)\mid X_n), $$ almost surely, for every $n\geqslant0$ and for every (for example) bounded measurable function $\varphi$.

Likewise, the strong Markov property is to ask that $$ \mathrm E(\varphi(Z_T,Z_{T+1},Z_{T+2},\ldots)\mid F_T)=\mathrm E(\varphi(Z_T,Z_{T+1},Z_{T+2},\ldots)\mid X_T), $$ almost surely on the event $[T\lt\infty]$, for every (for example) bounded measurable function $\varphi$ and for every stopping time $T$. (At this point, I assume you know what a stopping time $T$ is and what the sigma-algebra $F_T$ generated by a stopping time $T$ is.)

Since constants are stopping times, the strong Markov property implies the simple Markov property. For processes in discrete time, the two notions coincide (but be warned that this is not the case anymore for processes in continuous time).

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Okay, that actually helped a lot (although it mostly contained things i knew, it made me think right). What i am wondering now, is why the R.H.S is actually the expectation conditioned on only $X_{T}$. I do by the way know both what stopping times and the sigma-algebra generated by one is. –  Henrik Apr 4 '12 at 19:15
    
Oh and; you say "but be warned that this is not the case anymore for processes in continuous time" - but Durett is doing it only in the discrete case, right? –  Henrik Apr 4 '12 at 19:20
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