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Let's go through each part of the problem, showing the expected properties of conditional expectations:

### (a) \(\mathbb{E}[aY + bZ \mid X] = a \mathbb{E}[Y \mid X] + b \mathbb{E}[Z \mid X]\) for \(a, b \in \mathbb{R}\)

This property is known as the *linearity* of conditional expectation. To prove this, we can use the linearity of the expectation itself:

1. Given \(a, b \in \mathbb{R}\):
   \[
   \mathbb{E}[aY + bZ \mid X] = \mathbb{E}[aY \mid X] + \mathbb{E}[bZ \mid X].
   \]
2. Using the property of conditional expectation with constants:
   \[
   \mathbb{E}[aY \mid X] = a \mathbb{E}[Y \mid X] \quad \text{and} \quad \mathbb{E}[bZ \mid X] = b \mathbb{E}[Z \mid X].
   \]
3. Therefore:
   \[
   \mathbb{E}[aY + bZ \mid X] = a \mathbb{E}[Y \mid X] + b \mathbb{E}[Z \mid X].
   \]

### (b) \(\mathbb{E}[Y \mid X] \geq 0\) if \(Y \geq 0\)

This property uses the fact that conditional expectation preserves the non-negativity of a random variable:

1. If \(Y \geq 0\), then for any event \(A\) in the \(\sigma\)-algebra generated by \(X\):
   \[
   \mathbb{E}[Y \mathbf{1}_A] \geq 0.
   \]
2. By definition, \(\mathbb{E}[Y \mid X]\) is the function that satisfies:
   \[
   \mathbb{E}[\mathbb{E}[Y \mid X] \mathbf{1}_A] = \mathbb{E}[Y \mathbf{1}_A] \quad \text{for all events } A.
   \]
3. Since \(\mathbb{E}[Y \mathbf{1}_A] \geq 0\), it follows that \(\mathbb{E}[Y \mid X] \geq 0\).

### (c) \(\mathbb{E}[1 \mid X] = 1\)

This is a property of the indicator function and the definition of conditional expectation:

1. Consider \(Y = 1\):
   \[
   \mathbb{E}[1 \mid X] = \mathbb{E}[\mathbb{P}(1 \mid X)].
   \]
2. Since \(\mathbb{P}(1 \mid X)\) is trivially 1 (as it refers to the entire sample space having probability 1):
   \[
   \mathbb{E}[1 \mid X] = 1.
   \]

### (d) If \(X\) and \(Y\) are independent, \(\mathbb{E}[X \mid Y] = \mathbb{E}[X]\)

1. If \(X\) and \(Y\) are independent, the distribution of \(X\) does not depend on \(Y\).
2. Thus, the conditional expectation \(\mathbb{E}[X \mid Y]\) should be the same as the unconditional expectation \(\mathbb{E}[X]\):
   \[
   \mathbb{E}[X \mid Y] = \mathbb{E}[X].
   \]

### (e) \(\mathbb{E}[Y g(X) \mid X] = g(X) \mathbb{E}[Y \mid X]\) for any suitable \(g\)

This property uses the fact that \(g(X)\) is measurable with respect to \(X\):

1. Given that \(g(X)\) is a function of \(X\), it can be factored out of the conditional expectation:
   \[
   \mathbb{E}[Y g(X) \mid X] = g(X) \mathbb{E}[Y \mid X].
   \]

Would you like a more detailed explanation of any part? 

Here are some related questions that might help deepen your understanding:

1. How does linearity of conditional expectations extend to more than two random variables?
2. What are some examples where conditional expectations are not linear?
3. How is conditional expectation related to the concept of sufficiency in statistics?
4. Can the properties of conditional expectation be applied when dealing with continuous random variables?
5. What role do \(\sigma\)-algebras play in defining conditional expectations?

**Tip:** Conditional expectation is crucial in probability theory, especially in scenarios involving random processes. Understanding its properties helps in solving problems related to Markov processes, martingales, and more.

Explore the fundamental properties of conditional expectation, including linearity, non-negativity, and factorization with measurable functions. Learn how independence influences conditional expectations and why E[1 | X] always equals 1. Suitable for undergraduate students studying probability theory.

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