📂Probability Theory

Selective Sampling Theorem Proof

Theorem

Let’s assume there is a probability space $( \Omega , \mathcal{F} , P)$ and a supermartingale $\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\}$.

If $\tau$ and $\sigma$ are bounded stopping times with respect to $\sigma \le \tau$ and $\mathcal{F}_{n}$ then $$E \left( X_{\tau} | \mathcal{F}_{\sigma} \right) \le X_{\sigma} \text{ a.s.}$$

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• Being bounded with respect to $\mathcal{F}_{n}$ for $\tau$ means, quite literally, that there exists a $N \in \mathbb{N}$ such that for all $E \in \mathcal{F}_{n}$ it satisfies $\tau (E) \le N$.

Description

What the equation essentially says is that when there is a condition $\sigma \le \tau \le N$, the condition of the supermartingale $$E \left( X_{\sigma +1} | \mathcal{F} \right) \le X_{\sigma} \text{ a.s.}$$ even if changed to $\tau$ $$E \left( X_{\tau} | \mathcal{F} \right) \le X_{\sigma} \text{ a.s.}$$ the direction of the inequality is maintained as it is.

Proof

Part 1. $\mathbb{1}_{(\sigma = b)} \mathbb{1}_{(\tau \ge n)} = \mathbb{1}_{(\sigma = n)}$

Since $\sigma \le \tau$, it follows that $(\sigma = n ) \subset ( \tau \ge n)$, and $\mathbb{1}_{(\sigma = b)} \mathbb{1}_{(\tau \ge n)} = \mathbb{1}_{(\sigma = n)}$

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Part 2. $( \tau \ge n+1) \in \mathcal{F}_{n}$

Considering that $( \tau < n+1) = ( \tau \le n ) \in \mathcal{F}_{n}$ and given $( \tau < n+1) = ( \tau \ge n+1)^{c}$, by the definition of sigma-fields, $( \tau \ge n+1) \in \mathcal{F}_{n}$ must follow.

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Part 3. $X_{n} \mathbb{1}_{(\sigma =n)} \ge E \left( X_{\tau} | \mathcal{F}_{n} \mathbb{1}_{(\sigma = n)} \right)$

Let’s consider the scenario when $n = 1 , \cdots , N$.

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

Due to the properties of conditional expectation and indicator functions, along with Part 1 and 2, for every event $A \in \mathcal{F}_{n}$ $$\begin{align*} & \int_{A} X_{n} \mathbb{1}_{(\sigma = n)} dP - \int_{A} E \left( X_{\tau} | \mathcal{F}_{n} \right) \mathbb{1}_{(\sigma = n)} dP \\ =& \int_{A} X_{n} \mathbb{1}_{(\sigma = n)} \mathbb{1}_{(\tau \ge n)} dP - \int_{A} E \left( X_{\tau} | \mathcal{F}_{n} \right) \mathbb{1}_{(\sigma = n)} \mathbb{1}_{(\tau \ge n)} dP \\ =& \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n) } X_{n} dP - \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n) } E \left( X_{\tau} | \mathcal{F}_{n} \right) dP \\ =& \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n) } E \left( X_{n} | \mathcal{F}_{n} \right) dP - \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n) } E \left( X_{\tau} | \mathcal{F}_{n} \right) dP \\ =& \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n) } E \left( X_{n} - X_{\tau} | \mathcal{F}_{n} \right) dP \\ =& \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n) } \left( X_{n} - X_{\tau} \right) dP \end{align*}$$ Splitting the integral range from $( \tau \ge n)$ into $( \tau > n)$ and $( \tau = n)$ shows that since $(X_{\tau} - X_{n}) = 0$ from $(\tau = n)$ $$\begin{align*} & \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n) } \left( X_{n} - X_{\tau} \right) dP \\ =& \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n +1 ) } \left( X_{n} - X_{\tau} \right) dP + \int_{A \cap ( \sigma =n ) \cap ( \tau = n) } \left( X_{n} - X_{\tau} \right) dP \\ =& \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n +1 ) } \left( X_{n} - X_{\tau} \right) dP + 0 \end{align*}$$ Given that $\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\}$ is a supermartingale and $X_{\tau}$ is $\mathcal{F}_{n}$-measurable, it follows that $X_{\tau} = E \left( X_{\tau} | \mathcal{F}_{n} \right) \text{ a.s.}$ $$\begin{align*} & \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n +1 ) } \left( X_{n} - X_{\tau} \right) dP \\ =& \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n +1 ) } X_{n} dP - \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n +1 ) } X_{\tau} dP \\ \ge& \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n +1 ) } E \left( X_{n+1} | \mathcal{F}_{n} \right) dP - \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n +1 ) } E \left( X_{\tau} | \mathcal{F}_{n} \right) dP \\ =& \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n +1 ) } E \left( X_{n+1} - X_{\tau} | \mathcal{F}_{n} \right) dP \\ =& \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n +1 ) } \left( X_{n+1} - X_{\tau} \right) dP \\ \ge& \int_{A \cap ( \sigma =n ) \cap ( \tau \ge n +2 ) } \left( X_{n+2} - X_{\tau} \right) dP \\ & \vdots & \\ \ge& \int_{A \cap ( \sigma =n ) \cap ( \tau \ge N ) } \left( X_{N} - X_{\tau} \right) dP \\ =& \int_{A \cap ( \sigma =n ) \cap ( \tau = N ) } \left( X_{N} - X_{\tau} \right) dP \\ =& 0 \text{ a.s.} \end{align*}$$ Then, starting from the initial equation $$\int_{A} X_{n} \mathbb{1}_{(\sigma = n)} dP \ge \int_{A} E \left( X_{\tau} | \mathcal{F}_{n} \right) \mathbb{1}_{(\sigma = n)} dP \text{ a.s.}$$ and since $\displaystyle \forall A \in \mathcal{F}, \int_{A} f dm = 0 \iff f = 0 \text{ a.e.}$, $$X_{n} \mathbb{1}_{(\sigma =n)} \ge E \left( X_{\tau} | \mathcal{F}_{n} \right) \mathbb{1}_{(\sigma = n)} \text{ a.s.}$$

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Part 4. $E \left( X_{\tau} | \mathcal{F} \right) \le X_{\sigma} \text{ a.s.}$

Properties of Stopping Times: If $Z_{n}$ is a $F_{n}$-measurable function, then $Z_{n} \mathbb{1}_{\sigma = n}$ is a function measurable with respect to both $\mathcal{F}_{\sigma}$ and $\mathcal{F}_{n}$. Moreover, $Z_{n} \mathbb{1}_{(\sigma = n)} = Z_{\sigma} \mathbb{1}_{(\sigma = n)}$ holds true.

According to the properties of stopping times and Part 3, for $n=1,\cdots, N$ $$\begin{align*} & X_{n} \mathbb{1}_{(\sigma =n)} \ge E \left( X_{\tau} | \mathcal{F}_{n} \right) \mathbb{1}_{(\sigma = n)} \\ \iff & X_{\sigma} \mathbb{1}_{(\sigma =n)} \ge E \left( X_{\tau} | \mathcal{F}_{\sigma} \right) \mathbb{1}_{(\sigma = n)} \\ \iff & X_{\sigma} \ge E \left( X_{\tau} | \mathcal{F}_{\sigma} \right) \end{align*}$$

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