Conditional Expectation With Respect To Stopped Sigma Algebras Commute

Introduction

In the realm of probability theory, conditional expectation plays a crucial role in understanding the behavior of random variables. When dealing with stopped sigma algebras, it is essential to establish the relationship between conditional expectations with respect to these algebras. In this article, we will explore the concept of conditional expectation with respect to stopped sigma algebras and demonstrate that they commute.

Background

Let $(\Omega,\mathcal{F}, \mathbb{P}, (\mathcal{F_t})_{t\ge0})$ be a filtered probability space. This means that we have a probability space $(\Omega,\mathcal{F}, \mathbb{P})$ equipped with a filtration $(\mathcal{F_t})_{t\ge0}$, which is a family of sigma algebras $\mathcal{F_t}$ that satisfy certain properties. A stopping time $S$ is a random variable that takes values in $[0,\infty]$ and satisfies the property that for each $t\ge0$, the event $\{S\le t\}$ is in $\mathcal{F_t}$.

Stopped Sigma Algebras

Given a stopping time $S$, the stopped sigma algebra $\mathcal{F_S}$ is defined as the collection of all events $A\in\mathcal{F}$ such that $A\cap\{S\le t\}\in\mathcal{F_t}$ for all $t\ge0$. In other words, $\mathcal{F_S}$ consists of all events that are "visible" at time $S$. Similarly, for a stopping time $T$, the stopped sigma algebra $\mathcal{F_T}$ is defined as the collection of all events $A\in\mathcal{F}$ such that $A\cap\{T\le t\}\in\mathcal{F_t}$ for all $t\ge0$.

Conditional Expectation

The conditional expectation of a random variable $X$ with respect to a sigma algebra $\mathcal{G}$ is a random variable $\mathbb{E}[X|\mathcal{G}]$ that satisfies certain properties. Specifically, for any $A\in\mathcal{G}$, we have:

$$\int_A \mathbb{E}[X|\mathcal{G}] d\mathbb{P} = \int_A X d\mathbb{P}$$

Conditional Expectation with Respect to Stopped Sigma Algebras

Let $S$ and $T$ be two stopping times. We want to show that for any integrable random variable $X$, the following equality holds:

$$\mathbb{E}[\mathbb{E}[X|\mathcal{F_S}]\mathbb{E}[X|\mathcal{F_T}]] = \mathbb{E}[\mathbb{E}[X|\mathcal{F_T}]\mathbb{E}[X|\mathcal{F_S}]]$$

Proof

To prove this equality, we will use the following steps:

1. Step 1: Show that for any $A\in\mathcal{F_S}$, we have:

$$\int_A \mathbb{E}[X|\mathcal{F_T}] d\mathbb{P} = \int_A X d\mathbb{P}$$

2. Step 2: Show that for any $B\in\mathcal{F_T}$, we have:

$$\int_B \mathbb{E}[X|\mathcal{F_S}] d\mathbb{P} = \int_B X d\mathbb{P}$$

3. Step 3: Use the results from Steps 1 and 2 to show that:

$$\mathbb{E}[\mathbb{E}[X|\mathcal{F_S}]\mathbb{E}[X|\mathcal{F_T}]] = \mathbb{E}[\mathbb{E}[X|\mathcal{F_T}]\mathbb{E}[X|\mathcal{F_S}]]$$

Step 1

Let $A\in\mathcal{F_S}$. Then, by definition of $\mathcal{F_S}$, we have:

$$A\cap\{S\le t\}\in\mathcal{F_t}$$

for all $t\ge0$. Now, let $B\in\mathcal{F_T}$. Then, by definition of $\mathcal{F_T}$, we have:

$$B\cap\{T\le t\}\in\mathcal{F_t}$$

for all $t\ge0$. Since $A\cap B\in\mathcal{F_S}$, we have:

$$A\cap B\cap\{S\le t\}\in\mathcal{F_t}$$

for all $t\ge0$. Now, we can write:

$$\int_A \mathbb{E}[X|\mathcal{F_T}] d\mathbb{P} = \int_{A\cap B} \mathbb{E}[X|\mathcal{F_T}] d\mathbb{P}$$

$$= \int_{A\cap B} X d\mathbb{P}$$

$$= \int_A X d\mathbb{P}$$

This shows that for any $A\in\mathcal{F_S}$, we have:

$$\int_A \mathbb{E}[X|\mathcal{F_T}] d\mathbb{P} = \int_A X d\mathbb{P}$$

Step 2

The proof of Step 2 is similar to the proof of Step 1. Let $B\in\mathcal{F_T}$. Then, by definition of $\mathcal{F_T}$, we have:

$$B\cap\{T\le t\}\in\mathcal{F_t}$$

for all $t\ge0$. Now, let $A\in\mathcal{F_S}$. Then, by definition of $\mathcal{F_S}$, we have:

$$A\cap\{S\le t\}\in\mathcal{F_t}$$

for all $t\ge0$. Since $A\cap B\in\mathcal{F_T}$, we have:

$$A\cap B\cap\{T\le t\}\in\mathcal{F_t}$$

for all $t\ge0$. Now, we can write:

$$\int_B \mathbb{E}[X|\mathcal{F_S}] d\mathbb{P} = \int_{A\cap B} \mathbb{E}[X|\mathcal{F_S}] d\mathbb{P}$$

$$= \int_{A\cap B} X d\mathbb{P}$$

$$= \int_B X d\mathbb{P}$$

This shows that for any $B\in\mathcal{F_T}$, we have:

$$\int_B \mathbb{E}[X|\mathcal{F_S}] d\mathbb{P} = \int_B X d\mathbb{P}$$

Step 3

Now, we can use the results from Steps 1 and 2 to show that:

$$\mathbb{E}[\mathbb{E}[X|\mathcal{F_S}]\mathbb{E}[X|\mathcal{F_T}]] = \mathbb{E}[\mathbb{E}[X|\mathcal{F_T}]\mathbb{E}[X|\mathcal{F_S}]]$$

Let $A\in\mathcal{F_S}$ and $B\in\mathcal{F_T}$. Then, by Steps 1 and 2, we have:

$$\int_{A\cap B} \mathbb{E}[X|\mathcal{F_S}]\mathbb{E}[X|\mathcal{F_T}] d\mathbb{P} = \int_{A\cap B} X d\mathbb{P}$$

$$\int_{A\cap B} \mathbb{E}[X|\mathcal{F_T}]\mathbb{E}[X|\mathcal{F_S}] d\mathbb{P} = \int_{A\cap B} X d\mathbb{P}$$

This shows that:

$$\mathbb{E}[\mathbb{E}[X|\mathcal{F_S}]\mathbb{E}[X|\mathcal{F_T}]] = \mathbb{E}[\mathbb{E}[X|\mathcal{F_T}]\mathbb{E}[X|\mathcal{F_S}]]$$

Conclusion

In this article, we have shown that for any integrable random variable $X$ and any two stopping times $S$ and $T$, the following equality holds:

$$\mathbb{E}[\mathbb{E}[X|\mathcal{F_S}]\mathbb{E}[X|\mathcal{F_T}]] = \mathbb{E}[\mathbb{E}[X|\mathcal{F_T}]\mathbb{E}[X|\mathcal{F_S}]]$$

Q: What is the main result of this article?

A: The main result of this article is that for any integrable random variable $X$ and any two stopping times $S$ and $T$, the following equality holds:

$$\mathbb{E}[\mathbb{E}[X|\mathcal{F_S}]\mathbb{E}[X|\mathcal{F_T}]] = \mathbb{E}[\mathbb{E}[X|\mathcal{F_T}]\mathbb{E}[X|\mathcal{F_S}]]$$

Q: What is the significance of this result?

A: This result demonstrates that conditional expectations with respect to stopped sigma algebras commute. In other words, the order in which we take the conditional expectations does not matter.

Q: What are the assumptions of this result?

A: The assumptions of this result are that we have a filtered probability space $(\Omega,\mathcal{F}, \mathbb{P}, (\mathcal{F_t})_{t\ge0})$ and two stopping times $S$ and $T$. We also assume that the random variable $X$ is integrable.

Q: How does this result relate to other results in probability theory?

A: This result is related to other results in probability theory, such as the tower property of conditional expectation. The tower property states that for any random variable $X$ and any sigma algebra $\mathcal{G}$, we have:

$$\mathbb{E}[\mathbb{E}[X|\mathcal{G}]] = \mathbb{E}[X]$$

This result can be seen as a generalization of the tower property to the case where we have two sigma algebras, $\mathcal{F_S}$ and $\mathcal{F_T}$.

Q: Can this result be extended to more general settings?

A: Yes, this result can be extended to more general settings. For example, we can consider the case where we have multiple stopping times, or where we have a more general filtration. However, the proof of the result would require additional assumptions and techniques.

Q: What are some potential applications of this result?

A: Some potential applications of this result include:

• Risk management: This result can be used to model and analyze the behavior of financial instruments that are subject to multiple stopping times.
• Insurance: This result can be used to model and analyze the behavior of insurance policies that are subject to multiple stopping times.
• Biology: This result can be used to model and analyze the behavior of biological systems that are subject to multiple stopping times.

Q: How can this result be used in practice?

A: This result can be used in practice by:

• Modeling and analyzing: Using this result to model and analyze the behavior of systems that are subject to multiple stopping times.
• Risk assessment: Using this result to assess the risk of financial instruments or insurance policies that are subject to multiple stopping times.
• Decision-making: Using this result to make informed decisions about systems that are subject to multiple stopping times.

Q: What are some potential limitations of this result?

A: Some potential limitations of this result include:

• Assumptions: The assumptions of this result may not hold in all cases, which could limit its applicability.
• Complexity: The proof of this result may be complex and difficult to understand, which could limit its accessibility.
• Generalizability: This result may not be generalizable to all settings, which could limit its applicability.

Q: What are some potential future directions for research?

A: Some potential future directions for research include:

• Extending the result: Extending the result to more general settings, such as multiple stopping times or more general filtrations.
• Applying the result: Applying the result to real-world problems, such as risk management or insurance.
• Developing new techniques: Developing new techniques for proving and applying this result.