Question

Suppose that $X_{1}, X_{2}, \dots, X_{n}$ are independent random variables each drawn from a population having density function $f_{X}(x)=\begin{cases} \frac{1}{\theta} e^{-(x-\mu)/\theta} & \text{if } x \ge \mu \\ 0 & x < \mu \end{cases}$ where $\theta > 0$ and $\mu \in R^{+}$, then maximum likelihood estimate of $(\theta, \mu)$, when both $\theta, \mu$ are unknown is

• A. $(\overline{X}-X_{(1)}, X_{(1)})$
• B. $(\frac{1}{\overline{X}}-X_{(1)}, X_{(1)})$
• C. $(\overline{X}, X_{(1)})$
• D. $(\frac{1}{\overline{X}}, X_{(1)})$

Hint:

For distributions where the support depends on a parameter (like the lower bound $\mu$ here), the MLE of that bound is usually the extreme order statistic ($X_{(1)}$ or $X_{(n)}$).

Answer 1

The Correct Option is A

This is a shifted (two-parameter) exponential distribution. We find the MLEs for $\mu$ and $\theta$ sequentially.

Step 1: \color{redWrite the Likelihood Function
$L(\theta, \mu) = \prod_{i=1}^n \frac{1}{\theta} e^{-(x_i-\mu)/\theta}$ provided all $x_i \ge \mu$.
$L(\theta, \mu) = \frac{1}{\theta^n} \exp\left( -\frac{\sum (x_i - \mu)}{\theta} \right)$ for $\mu \le \min(x_i)$.

Step 2: \color{redMaximize with respect to $\mu$
Observe that $L(\theta, \mu)$ increases as $\mu$ increases (since the term in the exponent $-\frac{\sum x_i - n\mu}{\theta}$ becomes less negative as $\mu$ grows).
However, $\mu$ is constrained by $\mu \le x_i$ for all $i$, meaning $\mu \le X_{(1)}$, where $X_{(1)}$ is the first order statistic (minimum).
To maximize the likelihood, we choose the largest possible value for $\mu$:
$\hat{\mu} = X_{(1)}$.

Step 3: \color{redMaximize with respect to $\theta$
Now, substitute $\hat{\mu} = X_{(1)}$ and take the log of the profile likelihood:
$\ln L(\theta, \hat{\mu}) = -n \ln \theta - \frac{\sum (x_i - X_{(1)})}{\theta}$.
Differentiating with respect to $\theta$:
$\frac{\partial}{\partial \theta} \ln L = -\frac{n}{\theta} + \frac{\sum (x_i - X_{(1)})}{\theta^2} = 0$.
$\frac{n}{\theta} = \frac{\sum x_i - nX_{(1)}}{\theta^2} \implies n\theta = \sum x_i - nX_{(1)}$.
$\hat{\theta} = \frac{\sum x_i}{n} - X_{(1)} = \overline{X} - X_{(1)}$.

Step 4: \color{redCombine the Estimates
The MLE pair is $(\hat{\theta}, \hat{\mu}) = (\overline{X} - X_{(1)}, X_{(1)})$.
This corresponds to Option (1).