# Exercise on stopping time

Can somebody help me with this exercise? Thanks a lot.

Let $$X_1, X_2, \cdots$$ be a sequence of non-negative independent random variables and consider $$N(t) = \max\{n : X_1 + X_2 + \cdots + X_n \le t\}$$. Define an appropriate filtration and show that $$N(t) + 1$$ is a stopping time with respect to the filtration.

• Hint: $N(t)+1=\max\{n : X_1+\cdots+X_{n-1} \leq t\}$. Thus $N(t)+1=m$ if and only if $X_1+\cdots+X_{m-1}\leq t$ but $X_1+\cdots+X_m>t$. – kccu Jan 5 at 14:15
• Thank you but I'm still unable to complete the exercise. I haven't understood at all the meaning of stopping times and how to use them. :( – Prettymath77 Jan 5 at 14:35
• $N(t)+1$ is a stopping time iff the event $\{N(t)+1=m\}$ is measurable with respect to $\mathcal{F}_m$. I just wrote down the event $\{N(t)+1=m\}$ in terms of two events involving the $X_i$'s, so now you need to define $\mathcal{F}_m$ in such a way that those events are measurable. – kccu Jan 5 at 14:40
• More intuitively, if $T$ is a stopping time, then using only the information in the sigma algebra $\mathcal{F}_k$, we have enough information to determine whether $T=k$ or not. When you have a sequence of random variables $X_1,X_2,\dots$ it is quite common to take $\mathcal{F}_k = \sigma(X_1,\dots,X_k)$. With this definition, $N(t)$ is not a stopping time. $\mathcal{F}_k$ lets us determine whether $X_1+\cdots+X_k \leq t$, but it cannot tell us whether $k$ is the largest index such that this is true. So we can't determine whether $N(t)=k$. Think about why it's different for $N(t)+1$. – kccu Jan 5 at 14:46

For each $$t\geqslant 0$$, we have $$\{N(t)+1\leqslant t\} = \{\max_n X_1+\cdots+X_{n-1}\leqslant t\}$$ which is $$\mathcal F_{n-1}$$-measurable, and is in fact predictable as well as a stopping time.