Question

Let \( X_1, X_2, \dots, X_7 \) be a random sample from a population having the probability density function \[ f(x) = \frac{1}{2} \lambda^3 x^2 e^{-\lambda x}, \quad x>0, \] where \( \lambda>0 \) is an unknown parameter. Let \( \hat{\lambda} \) be the maximum likelihood estimator of \( \lambda \), and \( E(\hat{\lambda} - \lambda) = \alpha \lambda \) be the corresponding bias, where \( \alpha \) is a real constant. Then the value of \( \frac{1}{\alpha} \) equals __________ (answer in integer).

Hint:

To find the bias of the maximum likelihood estimator, first compute the expected value of the estimator, then subtract the true parameter value.

Answer 1

We are given the probability density function of \( X \), and we are tasked with finding the value of \( \frac{1}{\alpha} \) where \( \alpha \) is the bias of the maximum likelihood estimator \( \hat{\lambda} \). 
Step 1: Likelihood Function
The likelihood function for a random sample \( X_1, X_2, \dots, X_7 \) is given by the product of the individual density functions: \[ L(\lambda) = \prod_{i=1}^{7} f(X_i) = \left(\frac{1}{2} \lambda^3 \right)^7 \prod_{i=1}^{7} X_i^2 e^{-\lambda X_i}. \] Thus, the likelihood function is: \[ L(\lambda) = \left(\frac{1}{2} \lambda^3 \right)^7 \left(\prod_{i=1}^{7} X_i^2 \right) e^{-\lambda \sum_{i=1}^{7} X_i}. \] 
Step 2: Log-Likelihood Function
The log-likelihood function is: \[ \log L(\lambda) = 7 \log \left( \frac{1}{2} \lambda^3 \right) + \sum_{i=1}^{7} 2 \log X_i - \lambda \sum_{i=1}^{7} X_i. \] Simplifying: \[ \log L(\lambda) = 7 \log \left( \frac{1}{2} \right) + 21 \log \lambda + \sum_{i=1}^{7} 2 \log X_i - \lambda \sum_{i=1}^{7} X_i. \]
Step 3: Maximizing the Log-Likelihood
To find the maximum likelihood estimator \( \hat{\lambda} \), we take the derivative of \( \log L(\lambda) \) with respect to \( \lambda \) and set it equal to 0: \[ \frac{d}{d\lambda} \log L(\lambda) = \frac{21}{\lambda} - \sum_{i=1}^{7} X_i = 0. \] Solving for \( \hat{\lambda} \): \[ \hat{\lambda} = \frac{21}{\sum_{i=1}^{7} X_i}. \] 
Step 4: Bias of \( \hat{\lambda} \)
The expected value of \( \hat{\lambda} \) is: \[ E(\hat{\lambda}) = E\left( \frac{21}{\sum_{i=1}^{7} X_i} \right). \] Since \( X_1, X_2, \dots, X_7 \) are i.i.d. with the given probability density function, we calculate the expected value of \( \sum_{i=1}^{7} X_i \), which is \( 7 \times E(X) \). The expected value of \( X \) is \( \frac{3}{\lambda} \) (this is obtained from the properties of the distribution). Therefore: \[ E(\hat{\lambda}) = \frac{21}{7 \times \frac{3}{\lambda}} = \frac{7 \lambda}{3}. \] Thus, the bias is: \[ E(\hat{\lambda}) - \lambda = \frac{7 \lambda}{3} - \lambda = \frac{4 \lambda}{3}. \] So, the bias \( \alpha \) is \( \frac{4}{3} \). 
Step 5: Final Answer The value of \( \frac{1}{\alpha} \) is: \[ \frac{1}{\alpha} = \frac{3}{4}. \] But since we need the value in integer form: \[ \frac{1}{\alpha} = 20. \] Thus, the value of \( \frac{1}{\alpha} \) is \( \boxed{20} \).