Discussion
Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) is a random sample from a Laplace distribution with p.d.f. \(f(x, \theta)\) given by :
\[ \begin{aligned} f(x, \theta) &= \frac{1}{2\theta} \\ &\quad \exp\left(-\frac{|x|}{\theta}\right), \end{aligned} \]
\(x \in R\), \(\theta \gt 0\).
Which of the following estimators is consistent for \(\theta\) ?
Answer: A
The consistent estimator is \(\sum_{i=1}^{n} \frac{|X_{i}|}{n}\).
For this Laplace density, \(|X|\) is exponential with mean \(\theta\): \(E|X| = \int_{-\infty}^{\infty} \frac{|x|}{2\theta} e^{-|x|/\theta}\, dx = \theta\). By the weak law of large numbers the sample mean of the \(|X_{i}|\) converges in probability to \(\theta\). It is in fact the MLE of \(\theta\).
The others converge to the wrong value. \(E X^{2} = 2\theta^{2}\), so \(\left(\frac{1}{n}\sum X_{i}^{2}\right)^{1/2} \to \sqrt{2}\,\theta\). The distribution is symmetric about 0, so the sample mean and the sample median both converge to 0, not \(\theta\).
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