Question

Let \( X_1, X_2, \dots, X_{18} \) be a random sample from the distribution \[ f(x; \theta) = \begin{cases} \frac{2x}{\theta} e^{-x^2/\theta}, & x>0,\\ 0, & x \leq 0. \end{cases} \] What is the maximum likelihood estimator (MLE) of \( \theta \)?

• A. \( \hat{\theta} = \frac{1}{18} \sum_{i=1}^{18} X_i^2 \)
• B. \( \hat{\theta} = \frac{1}{9} \sum_{i=1}^{18} X_i^2 \)
• C. \( \hat{\theta} = \frac{1}{36} \sum_{i=1}^{18} X_i^2 \)
• D. \( \hat{\theta} = \frac{1}{2} \sum_{i=1}^{18} X_i^2 \)

Hint:

The maximum likelihood estimator (MLE) for a parameter is found by differentiating the log-likelihood function and setting it equal to zero. In cases where the likelihood involves sums of squares, the MLE is typically the sample mean or a scaled version of it.

Answer 1

The Correct Option is A

Step 1: Write the likelihood function.
The likelihood function \( L(\theta) \) for the sample is given by: \[ L(\theta) = \prod_{i=1}^{18} f(X_i; \theta) = \prod_{i=1}^{18} \frac{2X_i}{\theta} e^{-X_i^2/\theta}. \] Thus, the log-likelihood function is: \[ \ell(\theta) = \sum_{i=1}^{18} \left[ \ln \left( \frac{2X_i}{\theta} \right) - \frac{X_i^2}{\theta} \right] = \sum_{i=1}^{18} \left( \ln(2X_i) - \ln(\theta) - \frac{X_i^2}{\theta} \right). \]
Step 2: Differentiate and find the MLE.
The derivative of the log-likelihood function with respect to \( \theta \) is: \[ \frac{d}{d\theta} \ell(\theta) = -\frac{18}{\theta} + \frac{1}{\theta^2} \sum_{i=1}^{18} X_i^2. \] Setting this equal to 0 and solving for \( \theta \), we get: \[ \hat{\theta} = \frac{1}{18} \sum_{i=1}^{18} X_i^2. \]
Step 3: Conclusion.
Thus, the MLE of \( \theta \) is \( \hat{\theta} = \frac{1}{18} \sum_{i=1}^{18} X_i^2 \), and the correct answer is (A).