Question

Let \( X = aZ + b \), where \( Z \) is a standard normal random variable, and \( a, b \) are two unknown constants. It is given that \[ E[X] = 1, \quad E[(X - E[X]) | Z] = -2, \quad E[(X - E[X])^2] = 4, \] where \( E[X] \) denotes the expectation of random variable \( X \). The values of \( a, b \) are:

• A. \( a = -2, b = 1 \)
• B. \( a = 2, b = -1 \)
• C. \( a = -2, b = -1 \)
• D. \( a = 1, b = 1 \)

Hint:

When dealing with linear transformations of normal random variables, use the properties of expectation and variance to derive the unknown constants.

Answer 1

The Correct Option is A

We are given that \( X = aZ + b \), where \( Z \sim N(0, 1) \) (a standard normal random variable), and we need to find the values of \( a \) and \( b \) based on the provided conditions.

1. Condition 1: \( E[X] = 1 \)
We know that the expectation of \( X \) is given by: \[ E[X] = E[aZ + b] = aE[Z] + b = 0 + b = b. \] Thus, from \( E[X] = 1 \), we conclude: \[ b = 1. \]

2. Condition 2: \( E[(X - E[X]) | Z] = -2 \)
Substituting \( E[X] = 1 \) and \( X = aZ + b \), we get: \[ E[(X - 1) | Z] = E[aZ + b - 1 | Z] = aE[Z | Z] + b - 1 = aZ + b - 1. \] Since \( E[Z | Z] = Z \), we simplify this to: \[ aZ + b - 1 = -2. \] Substituting \( b = 1 \), we get: \[ aZ + 1 - 1 = -2 \quad \Rightarrow \quad aZ = -2. \] Therefore, \( a = -2 \).

3. Condition 3: \( E[(X - E[X])^2] = 4 \)
Finally, we calculate: \[ E[(X - E[X])^2] = E[(aZ + b - 1)^2] = E[(aZ)^2] = a^2E[Z^2]. \] Since \( E[Z^2] = 1 \) for a standard normal variable, we have: \[ a^2 \cdot 1 = 4 \quad \Rightarrow \quad a^2 = 4. \] Thus, \( a = -2 \) (as determined previously).

Therefore, the values of \( a \) and \( b \) are \( a = -2 \) and \( b = 1 \).