Question

If \( X \) is a continuous random variable whose probability density function is given by

\[ f_X(x) = \begin{cases} \frac{1}{x^2}, & \text{for } 1 < x < \infty \\ 0, & \text{elsewhere} \end{cases} \]

Then the median of \( X \) is ________________ (in integer).

Hint:

For continuous random variables, the median is the value \( m \) such that the area under the PDF from the lower bound to \( m \) equals 0.5. Always integrate the PDF and solve for this equality.

Answer 1

Step 1: Understand the definition of the median of a continuous random variable.
The median \( m \) of a continuous random variable \( X \) satisfies: \[ P(X \leq m) = \int_{1}^{m} f_X(x) \, dx = 0.5 \] Step 2: Plug in the given probability density function. \[ \int_{1}^{m} \frac{1}{x^2} \, dx = 0.5 \] Step 3: Evaluate the integral. \[ \int_{1}^{m} x^{-2} \, dx = \left[ \frac{-1}{x} \right]_1^m = \left( -\frac{1}{m} + 1 \right) = 0.5 \] Step 4: Solve the equation. \[ 1 - \frac{1}{m} = 0.5 \quad \Rightarrow \quad \frac{1}{m} = 0.5 \quad \Rightarrow \quad m = 2 \] Hence, the median of \( X \) is 2.