Question

A probability distribution function is given as:

\[ p(x) = \begin{cases} \dfrac{1}{a}, & x \in (0, a) \\ 0, & \text{otherwise} \end{cases} \]

Where \( a \) is a positive constant. For a function \( f(x) = x^2 \), the expectation of \( f(x) \) is:

• A. \( \frac{a^2}{3} \)
• B. \( \frac{a^3}{3} \)
• C. \( \frac{2a^2}{3} \)
• D. \( \frac{2a^3}{3} \)

Hint:

For continuous random variables, the expectation of a function is calculated by integrating the product of the function and the probability density function over the relevant range.

Answer 1

The Correct Option is A

The expectation \( E[f(x)] \) for a function \( f(x) = x^2 \) under the given probability distribution is calculated as: \[ E[x^2] = \int_0^a x^2 \cdot \frac{1}{a} \, dx = \frac{1}{a} \int_0^a x^2 \, dx \] Evaluating the integral: \[ \int_0^a x^2 \, dx = \frac{x^3}{3} \bigg|_0^a = \frac{a^3}{3} \] Thus, the expected value is: \[ E[x^2] = \frac{1}{a} \cdot \frac{a^3}{3} = \frac{a^2}{3} \] This confirms that option (A) is correct, with the expectation of \( x^2 \) being \( \frac{a^2}{3} \).