0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $X_1+\cdots+X_m>t$. $\endgroup$ – kccu Jan 5 at 14:15
  • $\begingroup$ 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. :( $\endgroup$ – Prettymath77 Jan 5 at 14:35
  • $\begingroup$ $N(t)+1$ 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. $\endgroup$ – kccu Jan 5 at 14:40
  • $\begingroup$ 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 $N(t)+1$. $\endgroup$ – kccu Jan 5 at 14:46
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.