Question

Let $X_{1}, X_{2}, \dots, X_{n}$ constitute a random sample of size "n" from a population having density $f_{X}(x)=\begin{cases} e^{-(x-\theta)} & x > \theta \\ 0 & \text{otherwise} \end{cases}$ then
A. $X_{(1)}$ is sufficient for $\theta$

B. $X_{(1)}$ is consistent for $\theta$

C. $X_{(1)}$ is unbiased for $\theta$

D. $MSE(X_{(1)}) = \frac{2}{n^2}$

• A. A, C only
• B. B, C only
• C. A, B, D only
• D. A, B, C, D only

Hint:

$X_{(1)}$ is the MLE for $\theta$ in a shifted exponential. Like many MLEs, it is biased but consistent. To make it unbiased, one would use $X_{(1)} - 1/n$.

Answer 1

The Correct Option is C

We evaluate the properties of the minimum order statistic $X_{(1)}$ for the shifted exponential distribution.

Step 1: \color{redCheck for Sufficiency (A)
The joint PDF is $L(\theta) = e^{-\sum (x_i - \theta)} = e^{-\sum x_i} e^{n\theta}$ for $x_i > \theta$.
This can be written as $L(\theta) = e^{-\sum x_i} e^{n\theta} I(X_{(1)} > \theta)$.
By the Factorization Theorem, $X_{(1)}$ is a sufficient statistic for $\theta$.
Result: Correct.

Step 2: \color{redCheck for Unbiasedness (C)
The distribution of $X_{(1)}$ is $f_{X_{(1)}}(x) = n e^{-n(x-\theta)}$ for $x > \theta$.
$E(X_{(1)}) = \int_{\theta}^{\infty} x \cdot n e^{-n(x-\theta)} dx$.
Let $u = n(x-\theta)$, then $dx = du/n$.
$E(X_{(1)}) = \int_{0}^{\infty} (\theta + \frac{u}{n}) e^{-u} du = \theta + \frac{1}{n} \int_{0}^{\infty} u e^{-u} du = \theta + \frac{1}{n}$.
Since $E(X_{(1)}) \ne \theta$, it is biased. Bias = $1/n$.
Result: Incorrect.

Step 3: \color{redCheck for Consistency (B)
As $n \to \infty$, $E(X_{(1)}) \to \theta$ and $Var(X_{(1)}) = 1/n^2 \to 0$.
Thus, $X_{(1)}$ converges in mean square (and therefore in probability) to $\theta$.
Result: Correct.

Step 4: \color{redCheck Mean Square Error (D)
$MSE(X_{(1)}) = Var(X_{(1)}) + [Bias(X_{(1)})]^2$.
$Var(X_{(1)}) = \frac{1}{n^2}$ (standard variance for exponential with rate $n$).
$Bias^2 = (1/n)^2 = \frac{1}{n^2}$.
$MSE = \frac{1}{n^2} + \frac{1}{n^2} = \frac{2}{n^2}$.
Result: Correct.
Conclusion: A, B, and D are correct.