Understanding how to formally use the definition of stopping time to prove something is a stopping time

342 Views Asked by At

In my notes I have the following definition A random variable $T=0,1,2,...,\infty$ is a stopping time , if $\forall n \le \infty$, the event $\{T \le n\}\in \mathscr F_n$

There are several examples (about martingales) where they first say something is a stopping time, so I just report here that initial part

1 The time $T$ of the first Head in a sequence of coinflips is a stopping time, while the time one before the first Head is not

2 A monkey repeatedly types any of the 26 letters of the English alphabet independently with equal chance. Let $T$ be the number of letters that have been typed when the entire word “ABRACADABRA” first appears. $T$ is a stopping time

3 Let $ X_k$ be iid ±1-valued with equal chance. The sum $S_n = \sum_{ k=1}^n X_k$ is called simple symmetric random walk (SSRW). Empty sums are zero, hence $S_0 = 0$. Then $S_n$ is a martingale, and $T : = \inf\{n : S_n = 1\}$ is a stopping time in the natural filtration of the walk

In all of these examples no clear explanation is given on how to prove that T is a stopping time. All they argue is that "using T you don't need to look into the future for stopping the process, so T is a stopping time". I would like to learn to prove this in a more formal way, by using the definition above. Therefore, how can I explicitly see that the event $\{T \le n\}\in \mathscr F_n = \sigma(X_0,X_1, ...,X_n)$ in each case?

These are the notes we are following:

people.maths.bris.ac.uk/~mb13434/mart_thy_notes.pdf

Edit 1 : Following the suggestion for the first case:

Considering $n=1, $ $\mathscr F_1 = \sigma(X_0,X_1)$, $\{T\le 1\}=\{T=0\}\cup \{T=1\}$. $ \{T=0\}$ is composed of the events where I get my first head in the first toss, so $\{T=0\}= \{H\}$ , $ \{T=1\}$ is composed of the events where I get my first head in the second toss so $ \{T=1\}=\{TH\}$ and therefore $\{T\le 1\}=\{H,TH\}$. How to continue from here? What is explicitly $\mathscr F_1$ and what are the values that $X_0$ and $X_1$ tan take?

Edit 2.

Actually I think the correct answer for n=1 of the first question is the following (I just hope someone can confirm the previous one was wrong) Since we are with n=1, $\mathscr F_1 = \sigma(X_0,X_1)$, $\Omega$ should be composed of two-toss events. Therefore: $\Omega=\{HH,HT,TT,TH\}$

$\{T=0\}= \{HH,HT\}$

$ \{T=1\}=\{TH\}$

$ \{T\le 1\}=\{TH,HH,HT\}$ To explicitly write $\mathscr F_1$, Let $X_i(H)=1$, $X_i(T)=0$, $i=1,2 $, and I look at all possible preimages of singletons:

$X_0^{-1}(0)=\{TH,TT\}$

$X_0^{-1}(1)=\{HH,HT\}$

$X_1^{-1}(0)=\{HT,TT\}$

$X_1^{-1}(1)=\{HH,TH\}$

Since the sigma algebra generated by $X_0$ and $X_1$ is the smallest one that make them measurable, I put all these preimages, their complements and unions in the sigma algebra, so:

$ \mathscr F_1= \sigma(X_0,X_1)=\{X_0^{-1}(0),X_0^{-1}(1),X_1^{-1}(0),X_1^{-1}(1),\Omega,\phi,\{HT,TH,TT\}, \{HH,TH,TT\}, \{HH,HT,TT\}, \{HH,HT,TH\}\}$ where I can explicitly see that $ \{T\le 1\}=\{TH,HH,HT\}\in \mathscr F_1$

I see that I still needed to take complements of the three-element sets and maybe even more sets are missing in the sigma-algebra. This rises the question of if there is an algorithmic way for listing them all without missing any? Maybe at the end I always get the power set? Now how to generalize this to arbitrary n?

1

There are 1 best solutions below

8
On

These examples don't want you to prove anything. They want you to develop intuition.

  • ${\cal F}_n$ is the amount of information available to you at time $n\,.$ Say, ${\cal F}_0=\{\emptyset,\Omega\}\,.$ That means at time $0$ you don't really know if an event happened or not. Later this event $A$ and its complement is in some ${\cal F}_n\,.$ That's when you know if it happened or not.

  • The time before the first head is clearly not a stopping time because you know if it was before the first head only later and not at that "time".

  • Typically, every random time that specifies when something happens for the first time is a stopping time.

More formally. Example 3 is the most common: By definition of $$ T=\inf\{n:S_n=1\} $$ we have $$ \{T\le k\}=\bigcup_{n\le k}\{S_n=1\}\,. $$ This shows that $\{T\le k\}\in{\cal F}_k$ and therefore $T$ is a stopping time.

About Example 1.

Let $S_n$ take value in $\{H,T\}$ and define $$ T=\inf\{n:S_n=H\} $$ the time of the first head. We know from the previous considerations that $$ \{T\le k\}=\bigcup_{n\le k}\{S_n=H\} $$ and that $T$ is a stopping time. The time before the first head is $T-1\,.$ We know $$ \{T-1\le k\}=\{T\le k+1\} $$ which is ${\cal F}_{k+1}-$ but not ${\cal F}_k$-measurable.