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Let $X_n$ be a Markov chain on the state space $\mathcal S$ and for $ y \in \mathcal S$ let $T_y = \min\{ n \ge 1 : X_n =y\}$ be the first return time to $y$. Let $W_y = T_y - 1$ be the time just before the first return to $y$

  • Explain why $W_y$ is not a stopping time

  • Show that the Strong Markov Property does not apply to $X_n$ at random time $W_y$.

My Work

When showing that $W_y$ is not a stopping time, is it sufficient to write $$W_y = \bigcap_{i = 1}^{n-1} \{X_i \ne y\} \cap X_n = y$$ and claim that since $X_n$ does not belong to the set $\{X_0, X_1, \dots, X_{n-1}\}$, we have that $W_y$ is not a stopping time?


Then, for showing that the Strong Markov Property does not apply, can I write $$\mathbf{P}(X_n = y \mid W_y = n-1, X_{n-1} = i, X_{n-2} = x_{n-2}, \dots, X_0 = y) = 1 \ne p(i, y)$$ where $p(i,y)$ is the one step transition probability from $i$ to $y$?

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What is your definition of a stopping time? –  Nate Eldredge Sep 9 '12 at 13:56
    
For $\mathbf{X} = \{X_n : n \ge 0\}$ a stochastic process, a stopping time $T$ is a random time such that for each $n \ge 0$, the event $\{T = n\}$ is completely determined by (at most) the total information known up to time n, $\{X_0, \dots, X_n\}$. I know how to informally state that $W_y$ is not a stopping time (because it depends on a time $X_n \notin \{X_0, \dots, X_{n-1}\}$ but I'm not sure how to formally prove/state this. –  Zvpunry Sep 9 '12 at 16:15
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1 Answer 1

up vote 0 down vote accepted

I'm posting an answer here to move this question off of the unanswered section as it is no longer requiring an answer, which I received on stats.SE.

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Please don't crosspost, though. It is considered impolite, is strongly discouraged, and in this case has created unnecessary fragmentation. Please edit your answer to provide a link to the one you received on stats.SE. –  cardinal Sep 15 '12 at 1:23
    
@cardinal I did not mean to offend. I will suitably edit my answer. –  Zvpunry Sep 15 '12 at 3:47
    
Hi, jmi4. No offense taken. I just wanted you to be aware. If you search on meta.SO you'll find some more explanation regarding the reasoning for this. Thanks for the edit! This way the questions are now linked so that others can find the answer more easily in the future. Cheers. –  cardinal Sep 15 '12 at 4:09
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