Proving a non-stopping time

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. –  KingOliver Sep 9 '12 at 16:15