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Let's go through each part of this problem, which involves 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 is a **linearity property** of conditional expectation. The conditional expectation operator preserves linear combinations of random variables.

**Proof:**
Let \( a, b \in \mathbb{R} \), and consider the following:
\[
\mathbb{E}[aY + bZ \mid X] = a \mathbb{E}[Y \mid X] + b \mathbb{E}[Z \mid X].
\]
Since conditional expectation is a linear operator, we can pull the constants \(a\) and \(b\) outside, and split the expectation:
\[
\mathbb{E}[aY + bZ \mid X] = a \mathbb{E}[Y \mid X] + b \mathbb{E}[Z \mid X].
\]

### (b) \( \mathbb{E}[Y \mid X] \geq 0 \text{ if } Y \geq 0 \)
This is a property of non-negative random variables. If \( Y \geq 0 \), the conditional expectation of \( Y \) given \( X \) is also non-negative.

**Proof:**
Let \( Y \geq 0 \). The conditional expectation \( \mathbb{E}[Y \mid X] \) is essentially a weighted average of \( Y \) over the possible values of \( X \). Since \( Y \) takes non-negative values, the weighted average (conditional expectation) also must be non-negative:
\[
Y \geq 0 \implies \mathbb{E}[Y \mid X] \geq 0.
\]

### (c) \( \mathbb{E}[1 \mid X] = 1 \)
This is because the conditional expectation of a constant is the constant itself, regardless of \( X \).

**Proof:**
For any random variable \( X \), the expectation of the constant function 1 is simply 1, since:
\[
\mathbb{E}[1 \mid X] = 1.
\]
This is because \( \mathbb{E} \) is the expectation operator, and conditioning on \( X \) does not change the value of constants.

### (d) If \( X \) and \( Y \) are independent, \( \mathbb{E}[X \mid Y] = \mathbb{E}[X] \)
This is a property of conditional expectation under independence.

**Proof:**
If \( X \) and \( Y \) are independent, then knowing \( Y \) provides no additional information about \( X \). Therefore, the conditional expectation of \( X \) given \( Y \) is simply the unconditional expectation of \( X \):
\[
\mathbb{E}[X \mid Y] = \mathbb{E}[X].
\]

### (e) \( \mathbb{E}[Y g(X) \mid X] = g(X) \mathbb{E}[Y \mid X] \) for any suitable \( g \)
This is the **iterated expectations property** applied when \( g(X) \) is a function of \( X \) alone.

**Proof:**
For any function \( g(X) \) that is measurable with respect to \( X \), we can factor it out of the conditional expectation:
\[
\mathbb{E}[Y g(X) \mid X] = g(X) \mathbb{E}[Y \mid X].
\]
This follows because \( g(X) \) is known given \( X \), so it behaves like a constant when taking the conditional expectation.

---

Would you like more detailed steps or explanations for any of these? 

### Further Questions:
1. Can you show a detailed step-by-step proof of the linearity of conditional expectation?
2. How does the property in (b) hold when dealing with negative values for \( Y \)?
3. What are some examples where \( \mathbb{E}[1 \mid X] \) might be useful in probability theory?
4. Can you explain the relationship between independence and conditional expectations in more detail for part (d)?
5. How would we apply part (e) when \( g(X) \) is not a simple function but something more complex?

### Tip:
For conditional expectations, always keep in mind whether you're conditioning on a variable or event. It affects how constants and random variables interact during the calculation.

Explore the key properties of conditional expectation, including linearity, independence, and the iterated expectations theorem. Detailed step-by-step explanations of each concept using mathematical proofs. Suitable for university-level students.

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