I hope you can help me formalize some things about the shift operator. So let $(\theta _{n})_{n\geq0}$ be the shift operator - that is $\theta _{n}\omega(k)=\omega(k+n)$. I'm using the Durrett setup where $X_n=\omega_n$ which makes $X_n\circ\theta_k=X_{n+k}$, obviously (I hope).
Now let $T_A=\inf\{n\gt0|X_n\in A\}$ be the hitting time of $A$, one would expect $T_A\circ\theta_n$ to be "n times" after $X_n$ hits $A$. How do I show it? And how do I show it's even a stopping time? (that might follow from the answer to the first question I guess).
Now let $\mathcal{F}_n =\sigma(\{X_k|0\le k \le n\})$ one would then expect that $\theta_k^{-1}(\mathcal{F}_n)\subseteq\mathcal{F}_{n+k}$ but how do I show that? (I know I can "exchange function and sigma", but I guess I'm not sure what the preimage of a function by the shift operator is, formally).
Let me know if I haven't been precise enough with my questions or given enough information.
Best regards,
Edit: Thanks David. I think you made a minor error (but helped me do it right) since: $ T_{A} \circ \theta_{k}\left(\omega\right)=\inf\left\{ m>0\,|\, X_{m}\circ\theta_{k}\left(\omega\right)\in A\right\} \\=\inf\left\{ m>0\,|\, X_{m+k}\left(\omega\right)\in A\right\} \\=\inf\left\{ n>k\,|\, X_{n}\left(\omega\right)\in A\right\} -k$ - one really has to be careful when substituting i guess.
Question update: Is it possible to get for a general stopping time $S$ that $S\circ\theta_{n}$ is a stopping time - actually I more specifically would like an argument that gives $\{S\circ \theta _k = n-k \}\in\mathcal{F}_n$