= Solution
Put $S=\sum_{j=1}^n|X_j|$. The <log-likelihood> for the <Laplace distribution> is
$$
\ell(\theta)=-n\log(2\theta)-S/\theta,\qquad\ell'(\theta)=-n/\theta+S/\theta^2.
$$
For $S>0$ the derivative is positive for $\theta<S/n$ and negative for $\theta>S/n$, so the unique <maximum-likelihood estimator> is
$$
\boxed{\widehat\theta=\frac1n\sum_{j=1}^n|X_j|.}
$$
The exceptional data set $S=0$ has <probability> zero; there the likelihood has its supremum at the excluded boundary $\theta=0$. For the <Fisher information>, $|X_j|$ has exponential density $\theta^{-1}e^{-y/\theta}$ for $y>0$, hence mean $\theta$. Since $\ell''(\theta)=n/\theta^2-2S/\theta^3$,
$$
\boxed{I_n(\theta)=-\mathbb E_\theta\ell''(\theta)=\frac n{\theta^2}.}
$$
This also follows from the score <variance>, because $\operatorname{Var}(S)=n\theta^2$.
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