Possibly should have posted this to the Stats board, but the question is to identify which of the following are stopping times;
\begin{align} \tau_1 &= \mathbb{inf } \{ t \in \mathbb{N}_0 ; Y_t = 10 \} \\ \tau_2 &= \mathbb{inf } \{ t \in \mathbb{N}_0 ; Y_t \in \{2,5,7\} \}\\ \tau_3 &= \mathbb{inf } \{ t \in \mathbb{N}_0 ; Y_{t+1} \in \{2,5,7\} \}\\ \tau_4 &= \mathbb{sup } \{ t \leq\mathbb{N}_0 ; Y_t = 3\}\\ \end{align}
This is in a probability space $(\Omega, \mathbb{P})$ with $(Y_t)^\infty_{t=0}$ as a sequence of iid random variables. The space is endowed with the filtration $(\mathcal{F}_t)^T_{t=0}$ with $\mathcal{F}_t = \sigma(Y_0,...,Y_t)$.
I think, though my understanding is very small, that perhaps $\tau_1$ and $\tau_2$ are, whereas $\tau_3$ and $\tau_4$ aren't. Could anyone shed some light as to whether these are correct?
My solutions would be:
$\tau_1$ is a stopping time since $\{\tau_1=t\} = \{Y_0 = ... = Y_{t-1} = 0, Y_t = 10\} = \left(\bigcap^{t-1}_{s=1}\{Y_s = 0\} \right) \cap \{Y_t = 10\}$ and both $\{Y_s = 0\} \in \mathcal{F}_t$ for $s \leq t$ and $\{Y_t = 10\} \in \mathcal{F}_t$.
$\tau_2$. For this I've been thinking if we are stopping the first time $Y_t \in \{2,5,7\}$, then surely we are just looking for the first time $Y_t \gt 2$, in which case the argument is similar to above?
$\tau_3$ is not a stopping time since $Y_{t+1} \not\in \mathcal{F}_t$.
$\tau_4$ is not a stopping time since $t$ only runs to $10$, we may not see $Y_t = 3$.
Many thanks for any help! The more I think about each one the less I am sure about the answers...