The problem asks to prove the following equality for bounded random variables $X$ and $Y$:

[ \mathbb{E}[Y \mathbb{E}[X \mid \mathcal{G}]] = \mathbb{E}[X \mathbb{E}[Y \mid \mathcal{G}]]. ]

Explanation and Proof Strategy:

This is an expectation identity involving conditional expectations. To prove this, we can apply the following properties of conditional expectations:

1. Iterated Expectation (Law of Total Expectation): [ \mathbb{E}[\mathbb{E}[X \mid \mathcal{G}]] = \mathbb{E}[X]. ]

2. Conditional Independence: If $X$ and $Y$ are both bounded, it will allow us to exchange the expectations under certain conditions.

The proof generally involves expanding both sides of the equation and showing that they are equal by applying the properties of conditional expectation. Here's a step-by-step outline:

Step 1: Write the left-hand side

Consider the left-hand side: [ \mathbb{E}[Y \mathbb{E}[X \mid \mathcal{G}]]. ] Since $\mathbb{E}[X \mid \mathcal{G}]$ is a $\mathcal{G}$-measurable function, you can take it out of the conditional expectation: [ \mathbb{E}[Y \mathbb{E}[X \mid \mathcal{G}]] = \mathbb{E}[\mathbb{E}[Y X \mid \mathcal{G}]] = \mathbb{E}[YX]. ]

Step 2: Write the right-hand side

Now, consider the right-hand side: [ \mathbb{E}[X \mathbb{E}[Y \mid \mathcal{G}]]. ] Similarly, $\mathbb{E}[Y \mid \mathcal{G}]$ is $\mathcal{G}$-measurable, so you can again apply the property of conditional expectation: [ \mathbb{E}[X \mathbb{E}[Y \mid \mathcal{G}]] = \mathbb{E}[\mathbb{E}[X Y \mid \mathcal{G}]] = \mathbb{E}[XY]. ]

Step 3: Conclude the proof

Since both sides simplify to $\mathbb{E}[XY]$, we conclude that: [ \mathbb{E}[Y \mathbb{E}[X \mid \mathcal{G}]] = \mathbb{E}[X \mathbb{E}[Y \mid \mathcal{G}]]. ]

This completes the proof.

Would you like more details on any specific part of this proof, or do you have any questions? Here are some related questions you might find helpful:

1. What is the meaning of conditional expectation $\mathbb{E}[X \mid \mathcal{G}]$?
2. How does the Law of Total Expectation work in probability theory?
3. Why does the boundedness of $X$ and $Y$ matter in this proof?
4. Can this result be generalized to unbounded random variables?
5. How do you apply the tower property of conditional expectations in this context?

Tip: Always try to simplify complex conditional expectations by leveraging properties like iterated expectation and measurability.