Question:

Let X be any random variable and \( Y = -2X + 3 \). If E[Y] = 1 and E[Y\(^2\)] = 9, then which of the following are TRUE?

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Be very careful with the variance property \( \text{Var}(aX + b) = a^2 \text{Var}(X) \). A common mistake is to forget to square the constant 'a'. The additive constant 'b' does not affect the variance.
Updated On: Aug 30, 2025
  • E[X] = 1
  • E[X] = -2
  • Var(X) = 1
  • Var(X) = 2
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
This problem involves using the properties of expectation and variance for linear transformations of random variables. We are given information about the random variable Y and need to find the expectation and variance of X, where Y is a linear function of X.
Step 2: Key Formula or Approach:
Properties of Expectation and Variance:
1. Linearity of Expectation: \( E[aX + b] = aE[X] + b \)
2. Variance Definition: \( \text{Var}(Y) = E[Y^2] - (E[Y])^2 \)
3. Variance of a Linear Transformation: \( \text{Var}(aX + b) = a^2 \text{Var}(X) \)
Step 3: Detailed Calculation:
1. Find E[X]:
We are given \( E[Y] = 1 \) and \( Y = -2X + 3 \).
Using the linearity of expectation:
\[ E[Y] = E[-2X + 3] = -2E[X] + 3 \] Substitute the known value of E[Y]: \[ 1 = -2E[X] + 3 \] \[ -2 = -2E[X] \] \[ E[X] = 1 \] Therefore, statement (A) is TRUE and (B) is FALSE.
2. Find Var(X):
First, we need to find the variance of Y. \[ \text{Var}(Y) = E[Y^2] - (E[Y])^2 \] We are given \( E[Y^2] = 9 \) and \( E[Y] = 1 \). \[ \text{Var}(Y) = 9 - (1)^2 = 9 - 1 = 8 \] Now, use the property for the variance of a linear transformation: \[ \text{Var}(Y) = \text{Var}(-2X + 3) = (-2)^2 \text{Var}(X) = 4 \text{Var}(X) \] We can now solve for Var(X): \[ 8 = 4 \text{Var}(X) \] \[ \text{Var}(X) = \frac{8}{4} = 2 \] Therefore, statement (D) is TRUE and (C) is FALSE.
Final Conclusion based on Calculation:
- E[X] = 1 is TRUE.
- Var(X) = 2 is TRUE.
The correct options are (A) and (D).
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