Question

If a random variable \( X \), defined such that \( E(X-1)^2 = 10 \) and \( E(X-2)^2 = 6 \) then mean \( (\mu) \) and variance \( (\sigma^2) \) are

• A. \( \frac{2}{7}, \frac{4}{15} \)
• B. \( \frac{2}{7}, \frac{15}{4} \)
• C. \( \frac{7}{2}, \frac{4}{15} \)
• D. \( \frac{7}{2}, \frac{15}{4} \)

Hint:

Always use the linearity of expectation first to clear out polynomial brackets. Treating \( E(X) \) and \( E(X^2) \) as simple standalone algebraic variables like \( u \) and \( v \) makes solving these systems quick and straightforward.

Answer 1

The Correct Option is D

Concept: The expectation operator \( E[\cdot] \) is linear, meaning it satisfies the following properties:
• \( E[X + Y] = E[X] + E[Y] \)
• \( E[cX] = cE[X] \) where \( c \) is any constant.
• \( E[c] = c \) where \( c \) is any constant. The mean \( \mu \) and variance \( \sigma^2 \) of a random variable \( X \) are defined by: \[ \mu = E(X) \] \[ \sigma^2 = E(X^2) - [E(X)]^2 = E(X^2) - \mu^2 \]

Step 1: Expanding the first expectation condition equation.
We are given our first constraint equation as: \[ E(X-1)^2 = 10 \] Expanding the squared term inside the expectation brackets algebraically: \[ E(X^2 - 2X + 1) = 10 \] Using the linearity property of expectation to separate the terms: \[ E(X^2) - 2E(X) + 1 = 10 \] Isolating the variable expectations by subtracting 1 from both sides: \[ E(X^2) - 2E(X) = 9 \quad \cdots (1) \]

Step 2: Expanding the second expectation condition equation.
We are given our second constraint equation as: \[ E(X-2)^2 = 6 \] Expanding the squared term inside the expectation brackets algebraically: \[ E(X^2 - 4X + 4) = 6 \] Using the linearity property of expectation to separate the terms: \[ E(X^2) - 4E(X) + 4 = 6 \] Isolating the variable expectations by subtracting 4 from both sides: \[ E(X^2) - 4E(X) = 2 \quad \cdots (2) \]

Step 3: Solving the system of linear equations to find \( E(X) \).
Let us subtract equation (2) directly away from equation (1) to eliminate the quadratic expectation term \( E(X^2) \): \[ [E(X^2) - 2E(X)] - [E(X^2) - 4E(X)] = 9 - 2 \] Simplifying the left-hand side by combining like terms: \[ -2E(X) + 4E(X) = 7 \quad \Rightarrow \quad 2E(X) = 7 \] Dividing by 2 gives the value for \( E(X) \), which is our mean \( \mu \): \[ \mu = E(X) = \frac{7}{2} \]

Step 4: Determining the value of \( E(X^2) \).
Substitute our calculated value \( E(X) = \frac{7}{2} \) back into equation (1): \[ E(X^2) - 2\left(\frac{7}{2}\right) = 9 \] Simplifying the multiplication: \[ E(X^2) - 7 = 9 \quad \Rightarrow \quad E(X^2) = 9 + 7 = 16 \]

Step 5: Calculating the statistical variance \( \sigma^2 \).
Now, we substitute our values for \( E(X^2) \) and \( \mu \) into the standard formula for variance: \[ \sigma^2 = E(X^2) - \mu^2 \] \[ \sigma^2 = 16 - \left(\frac{7}{2}\right)^2 = 16 - \frac{49}{4} \] Finding a common denominator to subtract the terms: \[ \sigma^2 = \frac{16 \times 4 - 49}{4} = \frac{64 - 49}{4} = \frac{15}{4} \] Thus, our calculated parameters are \( \mu = \frac{7}{2} \) and \( \sigma^2 = \frac{15}{4} \), which corresponds exactly to Option (D).