Showing a Random Walk’s First Hitting Time Is a Stopping Time
Summary
The document considers a walk formed by adding independent steps of +1 or −1 and defines the first time it reaches a fixed positive level. The key proof is to describe the event that this first occurs at time n: the walk must not have reached the level at any earlier time, and must equal it at time n. Each condition depends only on observations available by n, so the event belongs to the filtration at n. This establishes the stopping-time property directly.
The question’s attempted case split by the parity of the target is unnecessary. The event-based argument works for every positive integer target and every time n. The document provides a concise measurability proof, not an analysis of the walk’s hitting probability or expected hitting time; the martingale mentioned in the question is not needed for this result.
Key ideas
- A first hitting time is a stopping time if each event that it equals n is measurable using information available by n.
- The event of first reaching the target at n requires avoiding it at all earlier times and reaching it at n.
- This argument applies without splitting into odd and even target levels.
- The martingale property of the centered walk is not needed to prove the stopping-time claim.
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Full text
# Asymmetric Random Walk / Prove that $T:= \inf\{n: X_n = b\}$ is a $\{\mathscr F_n\}_{n \in \mathbb N}$-stopping time
# Asymmetric Random Walk / Prove that $T:= \inf\{n: X_n = b\}$ is a $\{\mathscr F_n\}_{n \in \mathbb N}$-stopping time
Given random variables $Y_1, Y_2, ... \stackrel{iid}{\sim} P(Y_i = 1) = p = 1 - q = 1 - P(Y_i = -1)$ where $p > q$ in a filtered probability space $(\Omega, \mathscr F, \{\mathscr F_n\}_{n \in \mathbb N}, \mathbb P)$ where $\mathscr F_n = \mathscr F_n^Y$,
define $X = (X_n)_{n \ge 0}$ where $X_0 = 0$ and $X_n = \sum_{i=1}^{n} Y_i$
Let $b$ be a positive integer and $T:= \inf\{n: X_n = b\}$.
It can be shown that the stochastic process $M = (M_n)_{n \ge 0}$ where $M_n = X_n - n(p-q)$ is a $(\{\mathscr F_n\}_{n \in \mathbb N}, \mathbb P)$-martingale.
Prove that $T$ is a $\{\mathscr F_n\}_{n \in \mathbb N}$-stopping time.
What I tried based on my previous question:
Case 1: b is odd
$$\emptyset = \{T = 0\} = \{T = 1\} = ... = \{T = b-1\} = \{T = b+1\} = ... = \{T = 2n\} = ... \in \mathscr F_0 \subseteq \mathscr F_i \ (i = 0, 1, ..., b-1, b+1, ..., 2n, ...)$$
$$\{T = b\} = \{Y_1 = ... = Y_b = 1 \} \in \mathscr F_b$$
$$\{T = b+2\} = \{Y_1 + ... = Y_{b+2} = b \} \setminus \{T = b\} \in \mathscr F_{b+2}$$
$$\vdots$$
$$\{T = 2n+1\} = \{Y_1 + ... = Y_{2n+1} = b \} \setminus (\{T = b\} \cup \{T = b+1\} \cup \{T = 2n - 1\})\in \mathscr F_{2n+1}$$
Case 2: b is even
$$\emptyset = \{T = 0\} = \{T = 1\} = ... = \{T = b-1\} = \{T = b+1\} = ... = \{T = 2n+1\} = ... \in \mathscr F_0 \subseteq \mathscr F_i \ (i = 0, 1, ..., b-1, b+1, ..., 2n+1, ...)$$
$$\{T = b\} = \{Y_1 = ... = Y_b = 1 \} \in \mathscr F_b$$
$$\{T = b+2\} = \{Y_1 + ... = Y_{b+2} = b \} \setminus \{T = b\} \in \mathscr F_{b+2}$$
$$\vdots$$
$$\{T = 2n\} = \{Y_1 + ... = Y_{2n} = b \} \setminus (\{T = b\} \cup \{T = b+1\} \cup \{T = 2n - 2\})\in \mathscr F_{2n}$$
QED
Is that right?
## Answer by Gordon (score 2, accepted)
https://quant.stackexchange.com/a/22129
For positive integer $n$, \begin{align*} \{T=n\} &= \Big(\cap_{k=1}^{n-1} \{X_k \ne b\}\Big) \cap \{X_n = b\} \in \mathscr{F}_n. \end{align*} That is, $T$ is a stopping time.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.