Question

Let \( X \) be a random variable such that \( X \) and \( -X \) have the same distribution. Let \( Y = X^2 \) be a continuous random variable with the probability density function
\[g(y) = \begin{cases} \frac{e^{-y/2}}{\sqrt{2\pi y}}, & \text{if } y > 0 \\0, & \text{if } y \leq 0 \end{cases}.\]
Then \( E((X - 1)^4) \) is equal to:

• A. 9
• B. 10
• C. 11
• D. 12

Answer 1

The Correct Option is B

Step 1: Understand the distribution of \( X \).

Since \( X \) and \( -X \) have the same distribution, \( X \) is symmetric about 0.

The random variable \( Y = X^2 \) represents the square of \( X \), so it is always non-negative.

The given probability density function \( g(y) \) is for \( Y \), and \( g(y) = \frac{e^{-y/2}}{\sqrt{2\pi y}} \) for \( y > 0 \).

Step 2: Compute \( \mathbb{E}((X - 1)^4) \).

Expanding \( (X - 1)^4 \), we have: \[ (X - 1)^4 = X^4 - 4X^3 + 6X^2 - 4X + 1. \]

Since \( X \) is symmetric about 0, \( \mathbb{E}(X^3) = 0 \) and \( \mathbb{E}(X) = 0 \).

The expression simplifies to: \[ \mathbb{E}((X - 1)^4) = \mathbb{E}(X^4) + 6\mathbb{E}(X^2) + 1. \]

Step 3: Compute \( \mathbb{E}(X^2) \) and \( \mathbb{E}(X^4) \) using \( g(y) \).

Since \( Y = X^2 \), \( \mathbb{E}(X^2) = \mathbb{E}(Y) \) and \( \mathbb{E}(X^4) = \mathbb{E}(Y^2) \).

The expectation \( \mathbb{E}(Y) \) is: \[ \mathbb{E}(Y) = \int_{0}^{\infty} y \cdot g(y) \, dy = \int_{0}^{\infty} y \cdot \frac{e^{-y/2}}{\sqrt{2\pi y}} \, dy = \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} \sqrt{y} \cdot e^{-y/2} \, dy. \] Substituting \( u = y/2 \), \( du = dy/2 \), we get: \[ \mathbb{E}(Y) = \frac{2}{\sqrt{2\pi}} \int_{0}^{\infty} \sqrt{2u} \cdot e^{-u} \, du = \frac{2\sqrt{2}}{\sqrt{2\pi}} \int_{0}^{\infty} u^{1/2} e^{-u} \, du. \] Using the Gamma function \( \Gamma(3/2) = \frac{\sqrt{\pi}}{2} \), we get: \[ \mathbb{E}(Y) = \frac{2\sqrt{2}}{\sqrt{2\pi}} \cdot \frac{\sqrt{\pi}}{2} = 1. \] Thus, \( \mathbb{E}(X^2) = 1 \).

The expectation \( \mathbb{E}(Y^2) \) is: \[ \mathbb{E}(Y^2) = \int_{0}^{\infty} y^2 \cdot g(y) \, dy = \int_{0}^{\infty} y^2 \cdot \frac{e^{-y/2}}{\sqrt{2\pi y}} \, dy = \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} y^{3/2} e^{-y/2} \, dy. \] Substituting \( u = y/2 \), \( du = dy/2 \), we get: \[ \mathbb{E}(Y^2) = \frac{2^{5/2}}{\sqrt{2\pi}} \int_{0}^{\infty} u^{3/2} e^{-u} \, du. \] Using \( \Gamma(5/2) = \frac{3\sqrt{\pi}}{4} \), we get: \[ \mathbb{E}(Y^2) = \frac{2^{5/2}}{\sqrt{2\pi}} \cdot \frac{3\sqrt{\pi}}{4} = 3. \] Thus, \( \mathbb{E}(X^4) = 3 \).

Step 4: Substitute the values.

Substituting \( \mathbb{E}(X^4) = 3 \) and \( \mathbb{E}(X^2) = 1 \) into the simplified expression: \[ \mathbb{E}((X - 1)^4) = 3 + 6(1) + 1 = 10. \]

Conclusion: The correct answer is (B) 10