Question

Let \( X \) be a random variable with the moment generating function \[ M(t) = \frac{(1 + 3e^t)^2}{16}, \quad -\infty < t < \infty. \]
Let \( \alpha = E(X) - \text{Var}(X) \). Then the value of \( 8\alpha \) is equal to __________ (answer in integer).

Answer 1

Correct Answer: 9

1. Moment Generating Function: - Expand \( M(t) \): \[ M(t) = \frac{(1 + 3e^t)^2}{16} = \frac{1 + 6e^t + 9e^{2t}}{16}. \] 2. First Moment (\( E(X) \)): - Differentiate \( M(t) \) and evaluate at \( t = 0 \): \[ M'(t) = \frac{6e^t + 18e^{2t}}{16}, \quad M'(0) = \frac{6 + 18}{16} = \frac{24}{16} = \frac{3}{2}. \] Thus, \( E(X) = \frac{3}{2} \). 3. Second Moment (\( E(X^2) \)): - Differentiate \( M'(t) \) to find \( M''(t) \): \[ M''(t) = \frac{6e^t + 36e^{2t}}{16}, \quad M''(0) = \frac{6 + 36}{16} = \frac{42}{16} = \frac{21}{8}. \] Thus, \( E(X^2) = \frac{21}{8} \). 4. Variance (\( \text{Var}(X) \)): - Compute \( \text{Var}(X) \): \[ \text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{21}{8} - \left(\frac{3}{2}\right)^2 = \frac{21}{8} - \frac{9}{4} = \frac{3}{8}. \] 5. Compute \( \alpha \): - \( \alpha = E(X) - \text{Var}(X) \): \[ \alpha = \frac{3}{2} - \frac{3}{8} = \frac{12}{8} - \frac{3}{8} = \frac{9}{8}. \] 6. Compute \( 8\alpha \): - Multiply \( \alpha \) by 8: \[ 8\alpha = 8 \cdot \frac{9}{8} = 9. \]