Question

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

\[ f_X(x) = \begin{cases} cx^3 + 0.25, & \text{for } 0 \leq x \leq 1, \ c \in \mathbb{R} \\ 0, & \text{elsewhere} \end{cases} \]

Then the value of \( c \) is ________________ (in integer).

Hint:

To find unknown constants in a PDF, integrate the function over its defined interval and equate the result to 1 — this ensures it satisfies the total probability rule.

Answer 1

Step 1: Use the property of a probability density function (PDF).
The total area under the PDF over its domain must equal 1: \[ \int_{-\infty}^{\infty} f_X(x) \, dx = 1 \] Since the function is non-zero only for \( 0 \leq x \leq 1 \), we evaluate: \[ \int_{0}^{1} (cx^3 + 0.25) \, dx = 1 \] Step 2: Integrate the function. \[ \int_{0}^{1} (cx^3 + 0.25) \, dx = c \int_{0}^{1} x^3 \, dx + \int_{0}^{1} 0.25 \, dx \] \[ = c \cdot \left[\frac{x^4}{4}\right]_{0}^{1} + 0.25 \cdot \left[x\right]_{0}^{1} = c \cdot \frac{1}{4} + 0.25 \cdot 1 = \frac{c}{4} + 0.25 \] Step 3: Set the integral equal to 1 and solve for \( c \). \[ \frac{c}{4} + 0.25 = 1 \quad \Rightarrow \quad \frac{c}{4} = 0.75 \quad \Rightarrow \quad c = 3 \] Hence, the value of \( c \) is 3.