Proposition: Let $T_1$ and $T_2$ be stopping times with respect to a filtration $\{\mathcal{F}_n\}_{n \ge 0}$, such that $T_1 \ge T_2$. The difference $T_1 - T_2$ is not necessarily a stopping time.
Let $(X_n)_{n \ge 0}$ be a Simple Random Walk (SRW) on $\mathbb{Z}$ starting at the origin ($X_0 = 0$). Let $\{\mathcal{F}_n\}_{n \ge 0}$ be its natural filtration, where $\mathcal{F}_n = \sigma(X_0, X_1, \dots, X_n)$.
Define two random times:
$T_2$: The first hitting time of state 1.
$$T_2 = \inf \{ t \ge 0 : X_t = 1 \}$$$T_1$: The first time returning to state 0 after state 1 has been reached.
$$T_1 = \inf \{ t > T_2 : X_t = 0 \}$$
Both $T_1$ and $T_2$ are valid stopping times, and by definition, $T_1 > T_2$.
Let us define the difference $T = T_1 - T_2$. This random variable $T$ represents the exact number of steps it takes to transition from state 1 back to state 0.
To determine if $T$ is a stopping time, we must check if the event $\{T = n\}$ is $\mathcal{F}_n$-measurable for an arbitrary integer $n > 0$.
The event $\{T_1 - T_2 = n\}$ implies that exactly $n$ steps after the random time $T_2$, the process hits 0. The absolute time at which this occurs is $T_2 + n$.
Consider the filtration at a fixed absolute time $n$, denoted as $\mathcal{F}_n$. This $\sigma$-algebra only contains the trajectory of the random walk up to time $n$. However, the variable $T_2$ is random and unbounded. Suppose a specific sample path yields $T_2 = k$, where $k > n$.
To evaluate whether $\{T_1 - T_2 = n\}$ has occurred on this path, we must observe the state of the random walk at time $T_2 + n = k + n$. Since $k > n$, it follows that $k + n > n$. The value $X_{k+n}$ is strictly in the future relative to time $n$ and is not contained within $\mathcal{F}_n$.
Since we cannot determine the occurrence of $\{T_1 - T_2 = n\}$ using only the information available up to time $n$, we conclude that:
$$\{T_1 - T_2 = n\} \notin \mathcal{F}_n$$Thus, $T_1 - T_2$ is not a stopping time.