Question

Let \( 5, 10, 4, 15, 6 \) be an observed random sample of size 5 from a distribution with probability density function \[ f(x; \theta) = \begin{cases} e^{-(x - \theta)}, & x \ge \theta \\ 0, & \text{otherwise} \end{cases} \] where \( \theta \in (-\infty, 3] \) is unknown. Then, the maximum likelihood estimate (MLE) of \( \theta \) based on the observed sample is equal to ..............

Hint:

For exponential-type distributions with a lower bound parameter, the MLE of the location parameter \(\theta\) is the minimum of the sample.

Answer 1

Correct Answer: 3

Step 1: Write the likelihood function.
For \(x_1, x_2, ..., x_5\) independent observations: \[ L(\theta) = \prod_{i=1}^{5} e^{-(x_i - \theta)} \, I(x_i \ge \theta). \] This simplifies to: \[ L(\theta) = e^{-\sum x_i + 5\theta} \, I(\theta \le \min x_i). \]
Step 2: Determine the range for \(\theta\).
The likelihood is non-zero only when \(\theta \le \min(x_i)\).
Step 3: Maximize \(L(\theta)\).
Since \(L(\theta)\) increases with \(\theta\) (because of \(e^{5\theta}\)), the maximum occurs at the largest possible value of \(\theta\) satisfying \(\theta \le \min(x_i)\).
Step 4: Compute the MLE.
\[ \hat{\theta} = \min(x_1, x_2, x_3, x_4, x_5) = \min(5, 10, 4, 15, 6) = 4. \] Final Answer: \[ \boxed{4} \]