Show that T is a stopping time

Question

Suppose that $Y_{i}$ are IID with $P(Y_{i}=1)=\frac{1}{2}$ and $P(Y_{i}=-1)=\frac{1}{2}$.

Let $K_{n}=\sum_{i=1}^{n}Y_{i}$. Show that $T=min\{n: K_{n}=-c$ or $K_{n}=c\}$ is a stopping time.

I would appreciate any help with this problem.

Answer 1

A random variable $T$ is a stopping time (wrt a filtration$\{\mathcal F_n\}$) if $\{T\le n\} \in \mathcal F_n$ for all $n.$ Here, we're presumably using the natural filtration $\mathcal F_n =\sigma(Y_1,\ldots, Y_n).$

Recall that, intuitively $\mathcal F_n$ is the set of events for which we know for sure whether or not they have occurred at time $n.$ In this problem, that's any event that is known from the values $Y_1,\ldots, Y_n.$

So $\{T\le n\}\in \mathcal F_n$ means that from the information we have at time $n$ we know whether or not $T$ is less than or equal to $n.$

Now let's see about this example. Here $T$ is the smallest time at which $K_n$ is either $c$ or $-c.$ At any given time $n$ can we tell whether $T\le n?$ Yes, we can. At time $n$ we know whether the sum $K_n$ has ever become $c$ or $-c.$

If this seems too obvious, here's an example of a random time that is not a stopping time: $$\max\{n : |K_n|\le c \}$$ in other words the last time at which $K_n$ is between $-c$ and $c.$ At time $n$ $K_n$ may be outside the range, but you're not sure whether it's going to come back or not, so you're not sure if the last time $K$ was in the range was $T$ or if $T$ is sometime in the future.

Okay, so that's the intuition, now for the formal stuff. We need to show $$ \{T\le n\}\in \mathcal F_n.$$ So first we need to know what $\mathcal F_n$ looks like. It is the sigma algebra generated by $Y_1,\ldots,Y_n,$ which is defined to be the sigma algebra generated by the sets $\{Y_i \le c\}$ for $i=1,\ldots, n$ and $c\in \mathbb R.$ It is easy to show that (and I will assume) that this sigma algebra also contains any set of the form $\{Y_i = c\},$ $\{Y_i\ne c\},$ $\{Y_i\ge c\},$ $\{Y_i

Now can we express $\{T\le n\}$ in terms of these generating sets? Sure. $T\le n$ whenever $K_1,\ldots K_n$ are all not equal to $\pm c$.

For $n=1$ we have $T\le 1$ if and only if $Y_1 \ne \pm c.$ We can write $$ \{Y_1\ne \pm c\}=\{Y_1\ne + c\}\cap \{Y_1\ne - c\}$$ which is in $\mathcal F_1$ because of the intersection property of sigma algebras.

Now for $n=2$ $$ \{K_2 =c\} = \bigcup_{n} \{Y_1 = n\}\cap \{Y_2 = c-n\}$$ which is in $\mathcal F_2$ by the properties of sigma algebras (since $\{Y_1 = n\}$ and $\{Y_2 = c-n\}$ are in $\mathcal F_2 = \sigma(Y_1,Y_2)$). Then of course $\{K_2 \ne c\} = \{K_2 =c\}^c$ is also in. Similarly for $\{K_2\ne -c\}.$ Then we still have $\{K_1\ne c\}\in \mathcal{F}_2$ so $\{K_1\ne c\}\cap\{K_2\ne c\} \in \mathcal F_2.$

It should be apparent that we can extend this to any $n$ and I'll leave it up to you.

The formal stuff here is one of those things that, in my opinion, you only need to do once to make sure you get how it links up to the set theory. It's much more important to be able to identify a stopping time intuitively.