= Solution
Because each increment is $\pm1$, the event at time $k$ defining $T$ is exactly $\{X_{k-1}=X_k=1\}$. Therefore, for $n\geq2$,
$$
\{T\leq n\}=\bigcup_{k=2}^n\{X_{k-1}=X_k=1\}\in\mathcal F_n,
$$
and for $n<2$ this event is empty. Thus \b[$T$ is a <stopping time>] for the <natural filtration>.
On the other hand, $\{U\leq0\}=\{T=2\}=\{X_1=X_2=1\}$. This event has <probability> $1/4$, whereas the initial <sigma-algebra> $\mathcal F_0$ is trivial, even after completion up to null sets. It therefore cannot belong to $\mathcal F_0$. Hence \b[$U=T-2$ is not a <stopping time>]. Subtracting a deterministic delay from a <stopping time> can require information from the future.
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