Math Problem Statement

Show the following: (a) E[aY + bZ | X] = aE[Y | X] + bE[Z | X] for a, b ∈ R (b) E[Y | X] ≥ 0 if Y ≥ 0 (c) E[1 | X] = 1 (d) If X and Y are independent, E[X | Y] = E[X] (e) E[Y g(X) | X] = g(X) E[Y | X] for any suitable g.

Solution

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

(a) E[aY+bZX]=aE[YX]+bE[ZX]\mathbb{E}[aY + bZ \mid X] = a \mathbb{E}[Y \mid X] + b \mathbb{E}[Z \mid X] for a,bRa, 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,bRa, 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) E[YX]0\mathbb{E}[Y \mid X] \geq 0 if Y0Y \geq 0

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

  1. If Y0Y \geq 0, then for any event AA in the σ\sigma-algebra generated by XX: E[Y1A]0.\mathbb{E}[Y \mathbf{1}_A] \geq 0.
  2. By definition, E[YX]\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 E[Y1A]0\mathbb{E}[Y \mathbf{1}_A] \geq 0, it follows that E[YX]0\mathbb{E}[Y \mid X] \geq 0.

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

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

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

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

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

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

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

  1. Given that g(X)g(X) is a function of XX, 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 σ\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.

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Math Problem Analysis

Mathematical Concepts

Conditional Expectation
Linearity
Independence
Measurable Functions

Formulas

E[aY + bZ | X] = aE[Y | X] + bE[Z | X]
E[Y | X] ≥ 0 if Y ≥ 0
E[1 | X] = 1
If X and Y are independent, E[X | Y] = E[X]
E[Y g(X) | X] = g(X) E[Y | X]

Theorems

Linearity of Conditional Expectation
Independence and Conditional Expectation
Measurable Functions and Factorization

Suitable Grade Level

Undergraduate Level (Probability and Statistics)