Is this a stopping time or not?

Question

Let $(\xi_n)_{n\in\mathbb{N}_0}$ be a sequence of independent identically distributed random variables that take values in $\left\{-1,1\right\}$ with equal probabilities. Define $(X_n)_{n\in\mathbb{N}_0}$ recursively by $X_0:=0, X_{n+1}:=X_n+\xi_n$. Which of the following are stopping times?

(a) $T_0:=\inf\left\{n\geq 4\text{ such that }X_n\text{ is even}\right\}$

(b) $T_1:=T_0-1,~~T_2:=T_0-2,~~T_3:=T_0+1$

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Here is our definition of stopping time:

Let $(X_n)_{n\in\mathbb{N}_0}$ be a Markov chain with state space $E$ on the probability space $(\Omega,\mathcal{A},\mathbb{P})$. A random variable $\tau\colon\Omega\to\mathbb{N}_0\cup\left\{\infty\right\}$ is called a stopping time if $\forall~n\in\mathbb{N}$ $$ \left\{\tau\leq n\right\}\in\sigma(X_0,\ldots,X_n):=\sigma\left(\bigcup_{k=0}^n \left\{X_k^{-1}(\mathcal{P}(E))\right\}\right). $$

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Using this definition, I got that $T_0$ is no stopping time:

Edit

Here the state space is $E=\mathbb{Z}$.

First of all I think it is $$ T_0=\inf\left\{n\geq 4\text{ such that }X_n\text{ is even}\right\}=\inf\left\{n\geq 4+2\mathbb{N}_0\text{ such that }X_n\text{ is even}\right\}, $$ because as far as I see $X_n$ for $n$ not even cannot be even. So for $n\in\mathbb{N}$ it is $$ \left\{T_0\leq n\right\}=\begin{cases}\emptyset, & n<4\\\bigcup_{m\in\left\{4,6,...,n\right\}}\left\{T_0=m\right\}, & n\geq 4, n\text{ is even}\\\bigcup_{m\in\left\{4,6,...,n-1\right\}}\left\{T_0=m\right\}, & n\geq 4, n\text{ is not even}\end{cases} $$ Thus if $n<4$, then $\left\{T_0\leq n\right\}\in\sigma(X_0,\ldots,X_n)$, because the empty set is in every $\sigma$-algebra.

But if $n\geq 4$ and $n$ even, then \begin{align} \left\{T_0=m\right\}&=\left\{X_4\in 2\mathbb{Z}+1,X_6\in 2\mathbb{Z}+1,\ldots,X_{m-2}\in 2\mathbb{Z}+1,X_m\in 2\mathbb{Z}\right\}\\ &=\underbrace{X_4^{-1}(2\mathbb{Z}+1)}_{\in\sigma(X_4,...,X_m)}\cap \underbrace{X_6^{-1}(2\mathbb{Z}+1)}_{\in\sigma(X_4,...,X_m)}\cap\cdots\cap \underbrace{X_{m-2}^{-1}(2\mathbb{Z}+1)}_{\in\sigma(X_4,...,X_m)}\cap \underbrace{X_m^{-1}(2\mathbb{Z})}_{\in\sigma(X_4,...,X_m)}. \end{align} So as finite intersection of elements of $\sigma(X_4,...,X_m)$, it is $\left\{T_0=m\right\}\in\sigma(X_4,...,X_m)$. And from this it follows that $\bigcup_{m\in\left\{4,6,...,n\right\}}\left\{T_0=m\right\}\in\sigma(X_4,...,X_m)\subseteq\sigma(X_4,...,X_n)\subseteq\sigma(X_0,...,X_n)$, because it is a finite union of elements in $\sigma(X_4,...,X_m)$. So for $n\geq 4$ and $n$ even, it is $$ \left\{T_0\leq n\right\}\in\sigma(X_0,...,X_n). $$

If $n\geq 4$ and $n$ not even, then it is (same argument) $$ \left\{T_0\leq n\right\}\in\sigma(X_4,...,X_{n-1})\subseteq\sigma(X_4,...,X_n)\subseteq\sigma(X_0,...,X_n), $$ that is $\left\{T_0\leq n\right\}\in\sigma(X_0,...,X_n)$.

All together, $T_0$ is a stopping time.

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Please tell me if I am right.

Answer 1

(One has) $\left\{T_0\leq n\right\}\in\sigma(X_4,...,X_n), $ but in general not $\left\{T_0\leq n\right\}\in\sigma(X_0,...,X_n)$

This is absurd: $\sigma(X_4,...,X_n)\subseteq\sigma(X_0,...,X_n)$.

Answer 2

The intuition here is that you can know whether $T_0 \leq n$ by knowing $X_0,X_1,\dots,X_n$. You can do this by observing only up to time $n$, so $T_0$ is a stopping time.

To know $T_1 \leq n$, you need information about $X_0,\dots,X_{n+1}$. You can't do this using information up to time $n$, so $T_1$ is not a stopping time. The situation with $T_2$ is very similar.

$T_3$ is different: to know whether $T_3 \leq n$ you don't even need to know anything about $X_n$, you only need to know about the process up to $X_{n-1}$. So $T_3$ is also a stopping time.