Question

Let \( X_1, X_2, \dots, X_n \) be a random sample from a distribution with the probability density function \[ f(x; \theta) = \begin{cases} c(\theta) e^{-(x - \theta)}, & x \geq 2\theta, \\ 0, & \text{otherwise}, \end{cases} \] where \( \theta \in \mathbb{R} \) is the unknown parameter. Then which of the following statement(s) is (are) true? 
 

• A. The maximum likelihood estimator of \( \theta \) is \( \dfrac{\min{\{X_1, X_2, \dots, X_n\}}}{2} \)
• B. \( c(\theta) = 1 \), for all \( \theta \in \mathbb{R} \)
• C. The maximum likelihood estimator of \( \theta \) is \( \min{\{X_1, X_2, \dots, X_n\}} \)
• D. The maximum likelihood estimator of \( \theta \) does not exist

Hint:

For many distributions, the maximum likelihood estimator can be found by maximizing the likelihood function, often resulting in the minimum or maximum of the sample values.

Answer 1

The Correct Option is A

Step 1: Find $c(\theta)$

For this to be a valid PDF: $$\int_{2\theta}^{\infty} c(\theta) e^{-(x-\theta)} dx = 1$$

Let $u = x - \theta$, then $du = dx$. When $x = 2\theta$, $u = \theta$: $$c(\theta) \int_{\theta}^{\infty} e^{-u} du = c(\theta) \left[-e^{-u}\right]_{\theta}^{\infty} = c(\theta) e^{-\theta} = 1$$

Therefore: $c(\theta) = e^{\theta}$

So: $f(x;\theta) = e^{\theta} e^{-(x-\theta)} = e^{-x+2\theta}$ for $x \geq 2\theta$

Check Option (B): $c(\theta) = e^{\theta} \neq 1$

Option (B) is FALSE 

Step 2: Find the likelihood function

For a sample $(x_1, x_2, ..., x_n)$, the likelihood is: $$L(\theta) = \prod_{i=1}^n f(x_i;\theta) = \prod_{i=1}^n e^{-x_i+2\theta} = e^{2n\theta} e^{-\sum_{i=1}^n x_i}$$

This is valid only when $x_i \geq 2\theta$ for all $i$, which means: $$\theta \leq \frac{\min{x_1, x_2, ..., x_n}}{2}$$

Step 3: Maximize the likelihood

The log-likelihood is: $$\ell(\theta) = 2n\theta - \sum_{i=1}^n x_i$$

Taking the derivative: $$\frac{d\ell}{d\theta} = 2n > 0$$

Since the derivative is always positive, $\ell(\theta)$ is strictly increasing in $\theta$.

Therefore, the likelihood is maximized at the largest possible value of $\theta$, which is: $$\hat{\theta}_{\text{MLE}} = \frac{\min{x_1, x_2, ..., x_n}}{2}$$

Step 4: Verify the options

(A) The MLE of $\theta$ is $\frac{\min{x_1,x_2,...,x_n}}{2}$

This matches our result.

Option (A) is TRUE 

(C) The MLE of $\theta$ is $\min{x_1, x_2, ..., x_n}$

This would violate the constraint $x_i \geq 2\theta$ for all $i$. If $\theta = \min{x_i}$, then for the minimum value $x_{\min}$, we'd have $x_{\min} \geq 2x_{\min}$, which is impossible for positive values.

Option (C) is FALSE 

(D) The MLE of $\theta$ does not exist

We found that the MLE exists and equals $\frac{\min{x_1,x_2,...,x_n}}{2}$.

Option (D) is FALSE 

Answer: (A)