# Stopping time $\max \{n: S_n \le t\}$

Let $\{X_n\}_{n \in \mathbb{N}}$ be iid and non-negative.

Let $$N(t): = \max \{n: \ X_1 + \cdots + X_n \le t \}.$$

Is $N(t)$ a stopping moment with respect to the natural filtration $\{\mathcal{F}_n\}_{n \in \mathbb{N}}$ generated by $\{X_n\}_{n \in \mathbb{N}}$ ? How about $N(t) + 1$?

I think that both are stopping moments.

Here is my solution:

$$\{N(t) \le k \} = \{N(t) =0 \} \cup \cdots \cup \{N(t) = k \}$$

Is $\{N(t) =0 \} = \{ X_0 \le t\} \cap \bigcap_{n \ge 1} \{S_n > t\} \in \mathcal{F}_0$? We know for sure that $\{ X_0 \le t\} \in \mathcal{F}_0$.

Next, $$\{N(t) =1 \} = \{ X_0 \le t\} \cap \{ X_0 + X_1 \le t\} \cap \bigcap_{n \ge 2} \{S_n > t\}$$

And again, by measurability of $X_0, X_1, \ldots$ and the fact that we have the natural filtration, we have: $\{ X_0 \le t\}\in \mathcal{F}_0 \subset \mathcal{F}_1$ and $\{ X_0 + X_1 \le t\} \in \mathcal{F}_1$.

$$\{N(t) = k \} = \bigcap_{0 \le n \le k} \{S_n \le t\} \cap \bigcap _{n \ge k+1} \{S_n > t\},$$ where $\bigcap_{0 \le n \le k} \{S_n \le t\} \in \mathcal{F}_n$ but I'm not sure about the second intersection.

When it comes to that infinite intersections, I know that each of $\{S_n > t \}$ is measurable in $\mathcal{F}_n$, but for the filtration $\{\mathcal{F}_n \} _{n \in \mathbb{N}}$ we have that $\mathcal{F}_n \subset \mathcal{F}_{n+1}$ not the other way around.

And when it comes to $N(t) + 1$ it is a general fact that $\tau$ - stopping time $\implies \tau +1$ is a stopping time.

So could you tell me what I can do with the infinite intersections?

Or maybe my reasoning is incorrect and there should be no infinite intersections here?

No, $N(t)$ is, in general, not a stopping time. The condition $\{N(t) \leq k\} \in \mathcal{F}_k$ means, roughly, that we can decide whether $N(t)(\omega) \leq k$ given the information up to time $k$, i.e. the values $X_1(\omega),\ldots,X_k(\omega)$. This is obviously not satisfied here because in order to decide whether $N(t)(\omega) \leq k$ we need to know whether $S_{k+1}(\omega) > t$ (and for this we also need $X_{k+1}(\omega)$).

As you already noted, we have

$$\{N(t)=0\} = \{X_0 \leq t\} \cap \bigcap_{n \geq 1} \{S_n>t\}.$$

Since $\{S_1>t\} \subseteq \{S_2>t\} \subseteq \dots$ (because of the non-negativity of $X_n$), this simplifies to

$$\{N(t)=0\} = \{X_0 \leq t\} \cap \{S_1>t\} = \{X_0 \leq t\} \cap \{X_0+X_1 > t\};$$

this set is, in general, not contained in $\mathcal{F}_0$. This implies that $N(t)$ is not a stopping time. However, using the formula

$$\{N(t)=k\} = \bigcap_{0 \leq n \leq k} \{S_n \leq t\} \cap \{S_{k+1}>t\}$$

it is not difficult to see that $N(t)+1$ is a stopping time.

• Thank you. Could you tell me why $\{X_0 + X_1 \}$ doesn't have to be in $\mathcal{F}_0$? Does it suffice to say that $\mathcal{F}_0$ is such that $X_0$ (and not necessarily $X_1$) is measurable with respect to $\mathcal{F}_0$ ? Or do we need to construct a counterexample? – Don Nov 11 '15 at 19:32
• @Don In my oppinion, it suffices to say that $X_1$ need not to be measurable wrt to $\mathcal{F}_0$. If you are looking for a concrete counterexample: Set e.g. $X_0 := 0$ and let $X_n \sim \frac{1}{2} \delta_0 + \frac{1}{2} \delta_1$ iid. Then $\mathcal{F}_0$ is the trivial $\sigma$-algebra; however, $X_1$ cannot be measurable wrt to $\mathcal{F}_0$ because only constant random variables are measurable wrt to the trivial $\sigma$-algebra. – saz Nov 11 '15 at 19:46
• Great, thank you a lot for your help! – Don Nov 11 '15 at 20:33
• @Don You are welcome. – saz Nov 11 '15 at 20:34