Question

Let \( X \) be a random variable with the distribution function \[ F(x) = \begin{cases} 0 & \text{if } x < 0, \\ \alpha(1 + 2x^2) & \text{if } 0 \leq x < 1, \\ 1 & \text{if } x \geq 1, \end{cases} \] where \( \alpha \) is a real constant. If the median of \( X \) is \( \frac{1}{\sqrt{2}} \), then the value of \( \alpha \) equals:

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{3} \)
• C. \( \frac{1}{4} \)
• D. \( \frac{1}{6} \)

Hint:

The median of a random variable is the value \( m \) where the cumulative distribution function \( F(m) = 0.5 \). Use this to solve for unknown parameters in the distribution function.

Answer 1

The Correct Option is C

Step 1: Understanding the Median 
The median of a random variable \( X \) is the value \( m \) such that \( F(m) = 0.5 \). We are given that the median is \( \frac{1}{\sqrt{2}} \), so we need to solve for \( \alpha \) when \( F\left( \frac{1}{\sqrt{2}} \right) = 0.5 \).

Step 2: Using the Distribution Function  
From the given distribution function, for \( 0 \leq x < 1 \), the CDF is: \[ F(x) = \alpha(1 + 2x^2). \] We substitute the median value \( x = \frac{1}{\sqrt{2}} \) into this expression: \[ F\left( \frac{1}{\sqrt{2}} \right) = \alpha \left( 1 + 2 \left( \frac{1}{\sqrt{2}} \right)^2 \right). \] Simplifying the expression: \[ F\left( \frac{1}{\sqrt{2}} \right) = \alpha \left( 1 + 2 \times \frac{1}{2} \right) = \alpha(2). \] We are told that the median is \( \frac{1}{\sqrt{2}} \), so: \[ F\left( \frac{1}{\sqrt{2}} \right) = 0.5. \] Thus, we have: \[ \alpha \times 2 = 0.5 \quad \Rightarrow \quad \alpha = \frac{0.5}{2} = \frac{1}{4}. \] Thus, the value of \( \alpha \) is \( \boxed{\frac{1}{4}} \).