Question

Let \( X \) be a random variable with the probability density function \[ f(x) = \begin{cases} \dfrac{\alpha^{p}}{\Gamma(p)} e^{-\alpha x} x^{p-1}, & x \geq 0, \, \alpha > 0, \, p > 0, \\ 0, & \text{otherwise}. \end{cases} \] If \( E(X) = 20 \) and \( \text{Var}(X) = 10 \), then \( (\alpha, p) \) is 
 

• A. (2, 20)
• B. (2, 40)
• C. (4, 20)
• D. (4, 40)

Hint:

For Gamma distributions, use the relationships between the shape and rate parameters to find the mean and variance.

Answer 1

The Correct Option is B

Step 1: Understand the given conditions. 
The probability density function is of the Gamma distribution type, where the expectation and variance for a Gamma distribution with shape parameter \( p \) and rate parameter \( \alpha \) are: \[ E(X) = \frac{p}{\alpha}, \text{Var}(X) = \frac{p}{\alpha^2}. \]

Step 2: Solve for \( \alpha \) and \( p \). 
We are given \( E(X) = 20 \) and \( \text{Var}(X) = 10 \). Using the formulas: \[ \frac{p}{\alpha} = 20 \,\,\text{and}\,\, \frac{p}{\alpha^2} = 10. \] From the first equation, solve for \( p \): \[ p = 20\alpha. \] Substitute this into the second equation: \[ \frac{20\alpha}{\alpha^2} = 10, \] which simplifies to: \[ \frac{20}{\alpha} = 10 \Rightarrow \alpha = 2. \] Now, substitute \( \alpha = 2 \) into \( p = 20\alpha \): \[ p = 20(2) = 40. \] Thus, \( \alpha = 2 \) and \( p = 40 \).

Step 3: Conclusion. 
The correct answer is (B) (2, 40).