= Solution
Here $s=\log(0.4\times0.7\times0.8\times0.9)=-1.601469743$. The <maximum-likelihood estimate> and estimated <standard error> from the <observed Fisher information> are
$$
\widehat\theta=-\frac4s-1=1.497705634,\qquad
\widehat{\operatorname{se}}(\widehat\theta)=J(\widehat\theta)^{-1/2}
=\frac{\widehat\theta+1}{\sqrt4}=1.248852817.
$$
The 95% <Wald confidence interval> is therefore
$$
\boxed{\widehat\theta\pm1.960\,\widehat{\operatorname{se}}(\widehat\theta)
=(-0.95005,\ 3.94546)}.
$$
The whole reported <confidence interval> lies in $\theta>-1$, though its lower endpoint is close to the boundary. With only four observations, the <normal approximation> underlying this <Wald confidence interval> need not provide accurate 95% coverage.
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