Question

Let \( X \) be a continuous random variable with the probability density function

\[ f(x) = \begin{cases} ax^2, & 0 < x < 1 \\ bx^{-4}, & x \geq 1 \\ 0, & \text{otherwise} \end{cases} \]

where \( a \) and \( b \) are positive real numbers. If \( E(X) = 1 \), then \( E(X^2) \) equals ................

Hint:

For continuous probability distributions, always ensure that the total probability is 1 by normalizing the distribution, then compute expected values using the appropriate integrals.

Answer 1

Correct Answer: 1.4

Step 1: Find the normalization constants.
For \( f(x) \) to be a valid probability density function, we must have: \[ \int_0^1 ax^2 \, dx + \int_1^\infty bx^{-4} \, dx = 1. \] First, solve for \( a \): \[ \int_0^1 ax^2 \, dx = \frac{a}{3}, \quad \int_1^\infty bx^{-4} \, dx = \frac{b}{3}. \] Thus, \[ \frac{a}{3} + \frac{b}{3} = 1 \quad \Rightarrow \quad a + b = 3. \] Step 2: Calculate \( E(X^2) \).
The expected value \( E(X^2) \) is given by: \[ E(X^2) = \int_0^1 ax^4 \, dx + \int_1^\infty bx^{-2} \, dx. \] First, solve for each integral: \[ \int_0^1 ax^4 \, dx = \frac{a}{5}, \quad \int_1^\infty bx^{-2} \, dx = \frac{b}{1}. \] Thus, \[ E(X^2) = \frac{a}{5} + b. \] Substitute \( b = 3 - a \) into the equation: \[ E(X^2) = \frac{a}{5} + (3 - a). \] Step 3: Use the condition \( E(X) = 1 \).
We know that \( E(X) = 1 \), so we can use the equation for \( E(X) \) to find \( a \) and \( b \). The calculation yields \( a = 2 \) and \( b = 1 \), so: \[ E(X^2) = \frac{2}{5} + 1 = \frac{7}{5}. \] Final Answer: \[ \boxed{\frac{7}{5}}. \]