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Given a filtered probability space $(\Omega, \mathscr F, \{\mathscr F_n\}_{n \in \mathbb N}, \mathbb P)$ where $\mathscr F_n = \mathscr F_n^Y$,

let $Y_1, Y_2, ...$ be iid random variables w/ $P(Y_n = 1) = P(Y_n = -1) = 1/2$.

$X$ is a symmetric random walk: $X_0 = 0, X_n = Y_1 + ... + Y_n$ and is a $(\{\mathscr F_n\}_{n \in \mathbb N}, \mathbb P)$-martingale.

Prove $S = \inf\{n : X_n = 7\} (< \infty)$ and $T = 10^{12} \wedge S$ are $\{\mathscr F_n\}$-stopping times.

What I tried:

$T$ is a stopping time because it is the minimum of two stopping times.

$S$ is a stopping time because:

$$\emptyset = \{S = 0\} = \{S = 1\} = ... = \{S = 6\} = \{S = 8\} = \{S = 10\} = ... = \{S = 2n\} = ... \in \mathscr F_0 \subseteq \mathscr F_i, (i = 0, 1, 2, ..., 6, 8, 10, ..., 2n, ...)$$

$$\{S = 7\} = \{Y_1 = ... = Y_7 = 1\} \in \mathscr F_7$$

$$\{S = 9\} = \{Y_1 + ... + Y_9 = 7\} \setminus \{S = 7\} \in \mathscr F_9$$

$$\{S = 11\} = \{Y_1 + ... + Y_{11} = 7\} \setminus (\{S = 7\} \cup \{S = 9\}) \in \mathscr F_{11}$$

$$\vdots$$

$$\{S = 2n+1\} = \{Y_1 + ... + Y_{2n+1} = 7\} \setminus (\{S = 7\} \cup ... \cup \{S = 2n-1\}) \in \mathscr F_{2n+1}$$

Is that right?

BCLC
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    But if $n=11$ (for example), ${S=11}$ is a proper subset of ${Y_1+\cdots+Y_{11}=7}$: a sample path with $X_7=7, X_8=8, X_9=9, X_{10}=8, X_{11}=7$ belongs to the latter event but not to the former. – John Dawkins Dec 04 '15 at 20:58
  • Thanks @JohnDawkins. How about now? – BCLC Dec 04 '15 at 21:40
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    Much better. Alternatively, ${S=11}={X_1<7,X_2<7,\ldots,X_{10}<7,X_{11}=7}$. – John Dawkins Dec 04 '15 at 22:16
  • @JohnDawkins Oh okay, hahaha. So is that all? You can post 'Your answer is correct' if you want – BCLC Dec 04 '15 at 22:32

1 Answers1

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Based on John Dawkins' comment:

I am right. Alternative is things like:

$\{S = 11\} = \{X_1 < 7, X_2 < 7, ..., X_{10} < 7, X_{11} = 7\}$

BCLC
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