📂Probability Theory

Properties of Stopping Times

Theorem

Let’s assume we have a probability space $( \Omega , \mathcal{F} , P)$ and a martingale $\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\}$. For a given stopping time $\tau$, $\mathcal{F}_{\tau}:= \left\{ A \in \mathcal{F}: A \cap ( \tau = n ) \in \mathcal{F}_{n} \right\}$ is referred to as the sigma field induced by $\tau$.

• [1]: $\mathcal{F}_{\tau}$ is a sigma field.
• [2]: $\tau$ is a $\mathcal{F}_{\tau}$-measurable function.
• [3]: For a martingale $\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\}$, $X_{\tau}$ is a $\mathcal{F}_{\tau}$-measurable function.
• [4]: $\mathbb{1}_{(\sigma = n)}$ is a $\mathcal{F}_{\sigma}$-measurable function.
• [5]: If $Z_{n}$ is a $F_{n}$-measurable function, then $Z_{n} \mathbb{1}_{\sigma = n}$ is both a $\mathcal{F}_{\sigma}$-measurable function and a $\mathcal{F}_{n}$-measurable function. Moreover, $Z_{n} \mathbb{1}_{(\sigma = n)} = Z_{\sigma} \mathbb{1}_{(\sigma = n)}$ holds.
• [6]: $E(X_{\tau} | \mathcal{F}_{\sigma} ) \mathbb{1}_{(\sigma = n)} = E(X_{\tau} | \mathcal{F}_{n} ) \mathbb{1}_{(\sigma = n)} \text{ a.s.}$

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• Concerning Borel sets $B \in \mathcal{B}(\mathbb{R})$, with $(\tau \in B) = \tau^{-1} (B)$, $(\tau = n)$ is equivalent to $\tau^{-1} ( \left\{ n \right\} )$.
• That $\tau$ is a $\mathcal{F}_{n}$-measurable function means for all Borel sets $B \in \mathcal{B}(\mathbb{R})$, it implies $\tau^{-1} (B) \in \mathcal{F}_{n}$.
• Given a stochastic process $\left\{ X_{n} \right\}_{n \in \mathbb{N}_{0}}$, regarding $\omega \in \Omega$, $X_{\tau}$ signifies the following: $$X_{\tau} = X_{\tau} ( \omega )= X_{\tau (\omega)} ( \omega )$$

Explanation

A stopping time can essentially be understood as a random variable that represents a timing we are interested in.

Proof

[1]

To check if $\mathcal{F}_{\tau}$ satisfies the conditions of a sigma field, we proceed as follows:

Part (i). $\emptyset \in \mathcal{F}_{\tau}$

Since $\mathcal{F}$ and $\mathcal{F}_{n}$ are sigma fields, they include $\emptyset$. Therefore, $\emptyset \in \mathcal{F}_{\tau}$ holds.

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Part (ii). $A \in \mathcal{F}_{\tau} \implies A^{c} \in \mathcal{F}_{\tau}$

$$A^{c} \cap ( \tau = n ) = ( \tau = n ) \setminus \left[ A \cap ( \tau = n ) \right]$$ where $( \tau = n) \in \mathcal{F}_{n}$, and if $A \in \mathcal{F}_{\tau}$, then by the definition of $F_{\tau}$, $A \cap ( \tau = n) \in \mathcal{F}_{n}$ follows, thus $A^{c} \in \mathcal{F}_{\tau}$

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Part (iii). $\displaystyle \left\{ A_{n} \right\}_{n=1}^{\infty} \subset \mathcal{F}_{\tau} \implies \bigcup_{i=1}^{\infty} A_{i} \in \mathcal{F}_{\tau}$

For $i = 1 , 2, \cdots$, if $A_{i} \in \mathcal{F}_{\tau}$ then by the definition of $F_{\tau}$, $A_{i} \cap ( \tau = n) \in \mathcal{F}_{n}$ follows. Thus, $$\bigcup_{i=1}^{\infty} A_{i} \cap (\tau = n ) = \bigcup_{i=1}^{\infty} \left[ A_{i} \cap ( \tau = n) \right] \in \mathcal{F}_{n}$$ Once again, by the definition of $F_{\tau}$, $$\bigcup_{i=1}^{\infty} A_{i} \in \mathcal{F}_{\tau}$$

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[2]

According to the equivalent conditions of a measurable function, to verify if $\tau$ is a $\mathcal{F}_{\tau}$-measurable function, it’s enough to check that for all $k \in \mathbb{R}$, $( \tau \le k ) \in \mathcal{F}_{\tau}$ holds. $$( \tau \le k ) \cap ( \tau = n) = \begin{cases} \emptyset &, k < n \\ (\tau = n) &, k \ge n \end{cases}$$ Here, $\emptyset \in \mathcal{F}_{n}$ and $(\tau =n ) \in \mathcal{F}_{n}$, so regardless of what $k \in \mathbb{R}$ is, $( \tau \le k ) \cap ( \tau = n) \in \mathcal{F}_{n}$ holds. Therefore, for all $k \in \mathbb{R}$, $( \tau \le k ) \in \mathcal{F}_{\tau}$ holds, and $\tau$ is a $\mathcal{F}_{\tau}$-measurable function.

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[3]

For any Borel set $B \in \mathcal{B}(\mathbb{R})$, the following holds: $$(X_{\tau} \in B) \cap (\tau = n) = (X_{n} \in B) \cap (\tau = n)$$ This is because the reason $(X_{\tau} \in B)$ and $(X_{n} \in B)$ are considered only when taking the intersection with $(\tau = n)$ results in only $\tau = n$. It’s similar to how we only consider fixed cases when computing conditional expectations $E(Y|X)$ as in $X=x$. Meanwhile, according to the definition of a martingale, $(X_{n} \in B) \in \mathcal{F}_{n}$ must hold, and by the definition of stopping time, $(\tau = n ) \in \mathcal{F}_{n}$ holds. Therefore, $\left[ (X_{\tau} \in B) \cap (\tau = n) \right] \in \mathcal{F}_{\tau}$, and $X_{\tau}$ is a $\mathcal{F}_{\tau}$-measurable function.

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[4]

For event $A \in \mathcal{F}$, $\mathbb{1}_{A}$ is represented as one of three possibilities: $$( \mathbb{1}_{A} \le a ) = \begin{cases} \Omega &, a \ge 1 \\ A^{c} &, a \in [0,1) \\ \emptyset &, a < 0 \end{cases}$$ Thus, given event $A = (\sigma = n)$, for all Borel sets $B \in \mathcal{B}(\mathbb{R})$, $\emptyset, (\sigma \ne n), \Omega \in \mathcal{F}_{\sigma}$ holds, so $( \mathbb{1}_{(\sigma = n)} \le a ) \in \mathcal{F}_{\sigma}$. In other words, $\mathbb{1}_{(\sigma = n)}$ is a $\mathcal{F}_{\sigma}$-measurable function.

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[5]

$$Z_{n} \mathbb{1}_{(\sigma = n)} = \begin{cases} Z_{\sigma} \cdot 1 &, \sigma = n \\ 0 &, \sigma \ne 0 \end{cases}$$ thus $Z_{n} \mathbb{1}_{(\sigma = n)} = Z_{\sigma} \mathbb{1}_{(\sigma = n)}$ holds. [3] implicates that $Z_{\sigma}$ is a $\mathcal{F}_{\sigma}$-measurable function, and [4] implies that $\mathbb{1}_{(\sigma = n)}$ is also a $\mathcal{F}_{\sigma}$-measurable function, so their product, $Z_{n} \mathbb{1}_{(\sigma = n)}$, is also a $\mathcal{F}_{\sigma}$-measurable function.

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[6]

Since $E \left( X_{\tau} | \mathcal{F}_{n} \right)$ is $\mathcal{F}_{n}$-measurable, $E \left( X_{\tau} | \mathcal{F}_{n} \right) \mathbb{1}_{ \sigma = n}$ is $\mathcal{F}_{\sigma}$-measurable by [4]. Meanwhile, $\mathbb{1}_{(\sigma=n)}$, due to the smoothing property, is a $\mathcal{F}_{\sigma}$-measurable function, thus moving in and out of $E ( \cdot | \mathcal{F}_{\sigma} )$ is permissible.

Smoothing property: If $X$ is $\mathcal{G}$-measurable, then $E(XY | \mathcal{G}) = X E (Y | \mathcal{G}) \text{ a.s.}$

Property of conditional expectation: If $X$ is $\mathcal{F}$-measurable, then $E(X|\mathcal{F}) =X \text{ a.s.}$

Furthermore, according to the properties above, for all $A \in \mathcal{F}_{\sigma}$, $$\begin{align*} \int_{A} E \left( X_{\tau} | \mathcal{F}_{\sigma} \right) \mathbb{1}_{(\sigma = n)} dP =& \int_{A} E \left( X_{\tau} \mathbb{1}_{(\sigma = n)} | \mathcal{F}_{\sigma} \right) dP \\ =& \int_{A} X_{\tau} \mathbb{1}_{(\sigma = n)} dP \\ =& \int_{A \cap (\sigma = n)} X_{\tau} dP \\ =& \int_{A} X_{\tau} \mathbb{1}_{(\sigma = n)} dP \\ =& \int_{A \cap (\sigma = n)} E \left( X_{\tau} | \mathcal{F}_{n} \right) dP \\ =& \int_{A } E \left( X_{\tau} | \mathcal{F}_{n} \right) \mathbb{1}_{(\sigma = n)} dP \end{align*}$$ $\displaystyle \forall A \in \mathcal{F}, \int_{A} f dm = 0 \iff f = 0 \text{ a.e.}$ hence, $$E(X_{\tau} | \mathcal{F}_{\sigma} ) \mathbb{1}_{(\sigma = n)} = E(X_{\tau} | \mathcal{F}_{n} ) \mathbb{1}_{(\sigma = n)} \text{ a.s.}$$

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