Let \(X_1, X_2, \ldots, X_n\) \((n \geq 2)\) be a random sample from the following probability density function \[ f(x) = \frac{1}{2} e^{-|x-\mu|}, \quad -\infty < x < \infty, \] where \(\mu \in (-\infty, \infty)\) is an unknown parameter. Let \(\bar{X} = \frac{1}{n}\sum_{i=1}^{n} X_i\) and \(\hat{\mu}\) denote the maximum likelihood estimator of \(\mu\), whenever it exists. Then which of the following statements is/are correct?
Show Hint
Maximizing the Laplace likelihood is the same as minimizing the sum of absolute deviations, and that is solved by the sample median, not the sample mean.
\(\hat{\mu}\) will always exist and \(\hat{\mu} = \bar{X}\)
\(\hat{\mu}\) may not always exist
\(\hat{\mu}\) always exists but it may not be unique
\(\hat{\mu}\) is a consistent estimator of \(\mu\)
Show Solution
The Correct Option isC, D
Solution and Explanation
Step 1: Set up minimization. \(g(\mu)=\sum|X_i-\mu|\); MLE minimizes this. Step 2: Piecewise derivative. \(g'(\mu)=\sum\text{sign}(\mu-X_i)\), zero at median. Step 3: Odd/even n. Odd: unique minimizer. Even: interval of minimizers. Step 4-7: Test options. (A) FALSE median not mean. (B) FALSE always exists. (C) TRUE exists, non-unique for even n. (D) TRUE median converges in probability to population median = \(\mu\) by symmetry. \[ \boxed{\text{(C) and (D)}} \]
