Question

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?

• A. \(\hat{\mu}\) will always exist and \(\hat{\mu} = \bar{X}\)
• B. \(\hat{\mu}\) may not always exist
• C. \(\hat{\mu}\) always exists but it may not be unique
• D. \(\hat{\mu}\) is a consistent estimator of \(\mu\)

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.

Answer 1

The Correct Option is C, D

Step 1: Concept.
Laplace distribution symmetric about \(\mu\), median=mean=\(\mu\).
Step 2: Log-likelihood.
\[ \ell(\mu) = -n\ln 2 - \sum_{i=1}^{n} |X_i - \mu|. \] Maximizing = minimizing \(\sum|X_i-\mu|\), solved by sample median.
Step 3: Median minimizing property.
Odd n: unique. Even n: any value in \([X_{(n/2)},X_{(n/2+1)}]\) works.
Step 4-7: Check options.
(A) FALSE, MLE is median not mean. (B) FALSE, always exists. (C) TRUE, exists but non-unique for even n. (D) TRUE, sample median consistent for population median = \(\mu\).
Final Answer: \[ \boxed{\text{(C) and (D)}} \]