Question

Let $X_1, X_2, \ldots, X_n$ be a random sample from $U(\theta - 0.5, \theta + 0.5)$ distribution, where $\theta \in \mathbb{R}$. If $X_{(1)} = \min(X_1, X_2, \ldots, X_n)$ and $X_{(n)} = \max(X_1, X_2, \ldots, X_n)$, then which one of the following estimators is NOT a maximum likelihood estimator (MLE) of $\theta$?
 

• A. $\dfrac{1}{2}(X_{(1)} + X_{(n)})$
• B. $\dfrac{1}{4}(3X_{(1)} + X_{(n)} + 1)$
• C. $\dfrac{1}{4}(X_{(1)} + 3X_{(n)} - 1)$
• D. $\dfrac{1}{2}(3X_{(n)} - X_{(1)} - 2)$

Hint:

For uniform distributions, the MLE lies midway between sample minimum and maximum because likelihood is constant only within this range.

Answer 1

The Correct Option is D

Step 1: Recall MLE property for uniform distributions.
For $U(\theta - 0.5, \theta + 0.5)$, the likelihood is nonzero only if \[ X_{(1)} \ge \theta - 0.5, X_{(n)} \le \theta + 0.5. \] Thus, $\theta$ must satisfy \[ X_{(n)} - 0.5 \le \theta \le X_{(1)} + 0.5. \] The MLE of $\theta$ is the midpoint of these bounds: \[ \hat{\theta} = \frac{X_{(1)} + X_{(n)}}{2}. \]

Step 2: Compare given estimators.
Option (A) matches the MLE exactly. Options (B) and (C) are affine transformations preserving the same range and location (still MLEs under reparameterization). Option (D) does not lie symmetrically between bounds and thus violates the MLE condition.

Step 3: Conclusion.
Hence, the estimator in (D) is NOT an MLE.