Question

Let \( x_1 = 1.1, x_2 = 0.5, x_3 = 1.4, x_4 = 1.2 \) be the observed values of a random sample of size four from a distribution with the probability density function
\[ f(x|\theta) = \begin{cases} e^{-\theta x}, & \text{if } x \geq \theta \\ 0, & \text{otherwise}, \quad \theta \in (-\infty, \infty) \end{cases} \] Then the maximum likelihood estimate of \( \theta^2 \) is

• A. 0.5
• B. 0.25
• C. 1.21
• D. 1.44

Hint:

For maximum likelihood estimation, the likelihood function is maximized with respect to the parameter, and this involves taking the minimum or maximum of the observed values, depending on the distribution.

Answer 1

The Correct Option is B

Step 1: Understand the likelihood function.
The likelihood function for the random sample \( x_1, x_2, x_3, x_4 \) is given by: \[ L(\theta) = \prod_{i=1}^{4} f(x_i | \theta) \] For the given probability density function, the likelihood function is: \[ L(\theta) = \prod_{i=1}^{4} e^{-\theta x_i} = e^{-\theta (x_1 + x_2 + x_3 + x_4)} \] This is valid for \( \theta \leq \min(x_1, x_2, x_3, x_4) \).
Step 2: Maximizing the likelihood function.
To maximize the likelihood, we need to minimize the sum of the observed values. The maximum likelihood estimate of \( \theta \) is \( \min(x_1, x_2, x_3, x_4) \). In this case, \( \theta_{\text{MLE}} = 0.5 \).
Step 3: Estimating \( \theta^2 \).
Thus, the maximum likelihood estimate of \( \theta^2 \) is \( (0.5)^2 = 0.25 \).