Stopping time max $\{n: \ X_1 + \cdots + X_n \le t \}$

Question

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 $N(t) + 1$?

It's clear to me that N(t) is not a stopping time. Since we want the max, independently when the time that "n" happens, we have to look into the future. What's not clear to me is why N(t) + 1 is a stopping time.

EDIT: What would be my solution is: $$ N(t) + 1 = t \rightarrow N(t) = t - 1 $$ Therefore:

$$ N(t) = k-1 \begin{aligned}[t] & = \bigcup \{N(t) = k-1\} \\ & = \bigcup \{N(t) \leq k-1\} \cap \{N(t) \leq k-2\}^c \end{aligned} $$ but everything is in $\mathcal{F}_k$ so ok.

EDIT 2: What I think is that the max never make this a stopping time. For example:

$$ N(4) \leq k = [N(4) = K] \cup [N(4) < k] $$ If for example K=2 we would have [N(4) =2] so the event where max{n: $S_n \leq 2$} = 4.

But this happens iff $S_4 \neq 2$ but we would have this info only at time 4, so it's not a stopping time because we need "2 more t" in order to get the info.

So why adding 1 would lead us to have a stopping time?

Answer 1

Given the question of the author in the comment I have written this answer in order to try to answer some questions

Adding $1$ to $N(t)$ is equivalent to looking at if $\{N(t) = k-1\}\in\mathcal{F}_k$ for $k\in\mathbb{N}$ to know if $N(t) + 1$ is a stopping time. But we notice that the sequence $(X_i)$ that appears in $N(t)$ is non negative. The event above means that for $n\leq k-1$ the condition imposed by $N(t)$ is respected and for $n\geq k$ the condition imposed by $N(t)$ is not respected. ($n$ here is the order of the sum of the $X_i$'s in the event $N(t)$)

But, as pointed out by geethta290krm, if we have that the condition is not respected for $k$, that is $\sum_{i=1}^{k}X_i > t$, clearly it will not be the case for $n\geq k$ by the non negativity of the $X_i$'s.

So, by considering $N(t) + 1$ you have the opportunity to consider a supplementary measurable event in the filtration (namely one concerning $X_k$) that will be enough to conclude on what happens in the future (i.e. the events concerning $X_i$'s for $i> k$) without looking in it, but just at looking at en event concerning $X_k$ which will be in $\mathcal{F}_k$.

This gives us the following (already shown by geethta290krm) :

$\{N(t) = k-1\} = \{\omega\in\Omega : max\{ n : \sum_{i=1}^nX_i(\omega)\leq t\} = k-1\}$ but this event means that the sum over the $X_i$'s cannot be of an order bigger than $k-1$, this means that we should have $\sum_{i=1}^{n}X_i>t$ for $n\geq k$ since the sequence is non negative. So the event $\{N(t) = k-1\}$ can be written

$\cup_{n=1}^{k-1}\{\omega\in\Omega : \sum_{i=1}^{n}X_i(\omega)\leq t\}\cap\{\omega\in\Omega : \sum_{i=1}^{k}X_i(\omega)>t\}\in\mathcal{F}_k$

where the event $\{\omega\in\Omega : \sum_{i=1}^{k}X_i(\omega)>t\}\in\mathcal{F}_k$ is the one that allows us to say appropriately that no $n\geq k$ would allow to verify $N(t)$.

Answer 2

This is somewhat technical, and it is false without non-negativity of $X_i$'s. I don't think it is possible to give an intuitive argument for this.

$N(t)+1=k$ iff $N(t)=k-1$ iff $X_1+X_2+\cdots+X_{k-1} \leq t$ and $X_1+X_2+\cdots+X_n >t$ for all $n \geq k$ iff $X_1+X_2+\cdots+X_{k-1} \leq t$ and $X_1+X_2+\cdots+X_k >t$ (by non-negtivity) and the event $(X_1+X_2+\cdots+X_{k-1} \leq t,X_1+X_2+\cdots+X_k >t)$ belongs to $\mathcal F_k$.

Answer 3

We claim that $$ \begin{align} \{N(t) \ge n\} \Leftrightarrow \{\sum_{i=1}^nX_i \le t\} \end{align} $$ Hence, $$ \begin{align} P\{N(t) = n\} &= P\{N(t) \ge n\} - P\{N(t) \ge n+1\} \\ &= P\{\sum^n_i X_i \le t\} - P\{\sum_i^{n+1} X_i \le t\} \\ \end{align} $$ and $$ \begin{align} P\{N(t)+1 = n\} &= P\{N(t) = n-1\}\\ &=P\{N(t) \ge n-1\} - P\{N(t) \ge n\} \\ &= P\{\sum^{n-1}_i X_i \le t\} - P\{\sum_i^{n} X_i \le t\} \\ \end{align} $$ which implies $\{N(t) = n\} \in \mathcal{F}_{n+1}, \{N(t) +1 = n\} \in \mathcal{F}_{n}$.

Answer 4

The event $N(t)+1 =k$ equals the event $$\{X_1+\cdots+X_k>t\} \cap \{X_1+\cdots+X_{k-1}\le t\}\,.$$ This depends only on $X_1,\ldots,X_k$.