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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?

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1 Answer 1

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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.

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  • $\begingroup$ Thanks Gordon. I admire the succinctness, but how exactly do you have the $\in \mathscr F_n$? $\endgroup$
    – BCLC
    Commented Dec 6, 2015 at 18:47
  • $\begingroup$ Wait, I think I got it. Is it really as simple as $\{X_n = b\} \in \sigma(X_n) \subseteq \mathscr F_n$? $\endgroup$
    – BCLC
    Commented Dec 6, 2015 at 18:50
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    $\begingroup$ Yes. that is correct. $\endgroup$
    – Gordon
    Commented Dec 6, 2015 at 19:09
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    $\begingroup$ Gordon, is it really that $\sigma(X_n) \subseteq \mathscr F_n$? $\endgroup$
    – BCLC
    Commented Dec 7, 2015 at 6:59
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    $\begingroup$ It is based on measure theory. Since $Y_1, Y_2, \ldots, Y_n$ are $\mathscr{F}_n$ measurable, then the sum $X_n$ is also $\mathscr{F}_n$ measurable. $\endgroup$
    – Gordon
    Commented Dec 7, 2015 at 13:44

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