📂Probability Theory

Stopping Times in Stochastic Processes

Definitions

Let’s assume a probability space $( \Omega , \mathcal{F} , P)$ is given. A random variable $\tau$ with an integer value greater than or equal to $0$ for all $n \in \mathbb{N}_{0}$ that satisfies $(\tau = n) \in \mathcal{F}_{n}$ with respect to the filtration $\left\{ \mathcal{F}_{n} \right\}$ is called a Stopping Time.

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• For a Borel set $B \in \mathcal{B}(\mathbb{R})$, $(\tau \in B) = \tau^{-1} (B)$ is, therefore, the same as $\tau^{-1} ( \left\{ n \right\} )$.

Examples

The intuitive concept of stopping time refers to the moment when an event of interest occurs—being observed. For example, $\tau = 8$ means knowing the information $\mathcal{F}_{8}$ while the event of interest occurs at $n=8$. At first glance, the condition for stopping time might seem too easy. However, the challenge lies in satisfying it for all $n \in \mathbb{N}_{0}$.

Let’s say $Y_{1}, Y_{2} , \cdots \overset{iid}{\sim} B(1,p)$. In other words, each $Y_{n}$ follows the Bernoulli distribution with probability $p$, and the results up to $Y_{5}$ are as follows: $$\begin{matrix} Y_{1} & Y_{2} & Y_{3} & Y_{4} & Y_{5} \\ 0 & 0 & 1 & 0 & 1 \end{matrix}$$

(1) When it’s not a stopping time: If we set $\tau$ as $\tau:= \max \left\{ k: Y_{k} = 0 \right\}$, in the above case, $\tau$ is calculated as follows: $$\begin{matrix} Y_{1} & Y_{2} & Y_{3} & Y_{4} & Y_{5} \\ 0 & 0 & 1 & 0 & 1 \\ \tau = 1 & \tau = 2 & \tau = 2 & \tau = 4 & \tau = 5 \end{matrix}$$ Here, $\tau$ must satisfy the following to be a stopping time: $$(\tau = n ) = \left( Y_{n} = 0 , Y_{n+1} = 1 , \cdots \right)$$ This precisely means $Y_{n} = 0$, and afterwards, it must always be $1$. Regardless of what $n \in \mathbb{N}$ is, it’s impossible to know the outcome without conducting the trial. Therefore, $\tau$ cannot be a stopping time.

(2) When it becomes a stopping time: If we set $\tau$ as $\tau:= \min \left\{ k: Y_{k} = 1 \right\}$, in the above case, $\tau$ is calculated as follows: $$\begin{matrix} Y_{1} & Y_{2} & Y_{3} & Y_{4} & Y_{5} \\ 0 & 0 & 1 & 0 & 1 \\ \tau = 0 & \tau = 0 & \tau = 3 & \tau = 3 & \tau = 3 \end{matrix}$$ $\tau$ is already not concerned with what comes in the future since the event of interest has occurred at $n=3$, becoming a stopping time.

Explanation

Note in the above examples that while $\max$ was not suitable as a stopping time, $\min$ became a stopping time. In this sense, stopping time can intuitively be considered as the ’timing when something happens for the first time’. Meanwhile, one must not forget that, in a strict mathematical definition, $\tau$ is still a random variable. When a stochastic process $\left\{ X_{n} \right\}_{n \in \mathbb{N}_{0}}$ is given, the condition for $X_{\tau}$ to $\omega \in \Omega$ means the following: $$X_{\tau} = X_{\tau} ( \omega )= X_{\tau (\omega)} ( \omega )$$ For instance, if $\tau (\omega_{1}) = 5$, then it becomes an equation where $X_{\tau} (\omega_{1}) = X_{5} ( \omega_{1})$ . $\tau$ represents ‘sometime when an event may occur’, thus it’s a ‘function’ that maps all respective $\omega \in \Omega$ to some $n \in \mathbb{N}_{0}$ even before being called a ‘stopping time’. Clinging to intuitive understanding and forgetting this point will make the deployment of all formulas involving stopping time painful. Remember well.