Question

Let \( X \) be a random variable with probability density function \[ f(x) = \begin{cases} \alpha \lambda x^{\alpha - 1} e^{-\lambda x^\alpha} & \text{if } x > 0, \\ 0 & \text{otherwise} \end{cases} \] where \( \alpha > 0 \) and \( \lambda > 0 \). If the median of \( X \) is 1 and the third quantile is 2, then \( (\alpha, \lambda) \) equals:

• A. \( (1, \log_e 2) \)
• B. \( (1, 1) \)
• C. \( (2, \log_e 2) \)
• D. \( (1, \log_e 3) \)

Hint:

- The median corresponds to the value where the CDF is 0.5.
- The third quantile corresponds to the value where the CDF is 0.75.
- Use the CDF and solve for the desired values to find parameters \( \alpha \) and \( \lambda \).

Answer 1

The Correct Option is A

1) Understanding the function: 
The given probability density function describes a generalized form of a Weibull distribution. For the median and third quantile, we use the cumulative distribution function (CDF). The CDF \( F(x) \) is the integral of \( f(x) \). 
2) Setting up the CDF: 
\[ F(x) = \int_0^x f(t) dt = \int_0^x \alpha \lambda t^{\alpha - 1} e^{-\lambda t^\alpha} dt \] This is a standard form whose result will lead to the calculation of the median and third quantile values. 
3) Using median and third quantile values: 
Given that the median of \( X \) is 1 and the third quantile is 2, we solve the CDF equations for these values. By substituting these into the CDF equation and solving for \( \alpha \) and \( \lambda \), we find \( \alpha = 1 \) and \( \lambda = \log_e 2 \).