= Solution
For the <normal distribution>, the density at its centre is $f(\theta)=1/(\sigma\sqrt{2\pi})$. The <sample mean> has exact <variance> $\sigma^2/n$, whereas part (b) gives the <sample median> the leading <variance> $\pi\sigma^2/(2n)$. Thus the requested <asymptotic relative efficiency> is
$$
\boxed{\operatorname{ARE}(M_X,\overline X)=\frac2\pi\approx0.637
\quad\text{for the normal model}.}
$$
At a fixed confidence level, a <confidence interval> based on the <sample mean> has a leading half-width smaller by the factor $\sqrt{2/\pi}$. With known $\sigma$, $\overline X\pm z_{1-\alpha/2}\sigma/\sqrt n$ also has exact normal coverage. \b[Prefer the sample mean under the normal model.]
For the <Laplace distribution>, symmetry puts both the <mean> and <median> at $\theta$. Its density there is $1/(2\sigma)$, and
$$
\operatorname{Var}(X_i)=\frac1\sigma\int_0^\infty y^2e^{-y/\sigma}\,dy
=2\sigma^2.
$$
Consequently the <sample mean> has <variance> $2\sigma^2/n$, while the <sample median> has leading <variance> $\sigma^2/n$. Hence
$$
\boxed{\operatorname{ARE}(M_X,\overline X)=2
\quad\text{for the Laplace model}.}
$$
A large-sample <confidence interval> based on the <sample median> has leading half-width smaller by $1/\sqrt2$; with known scale it is $M_X\pm z_{1-\alpha/2}\sigma/\sqrt n$. \b[Prefer the sample median under the Laplace model.] Consistent estimation of an unknown scale preserves these leading comparisons. The reason for the reversal is quantitative: the <median>'s precision depends on the central density $f(\theta)$, while the <sample mean>'s precision depends on the whole population <variance>.
\Image[/past-exam-of-the-statistics-course-of-the-university-of-oxford/2015-mean-median-efficiency.png]
{title=Asymptotic variance of mean and median in normal and Laplace location models}
{description=The vertical scale is $n$ times estimator variance divided by $\sigma^2$. In the normal model $\sigma$ is the population standard deviation; in the Laplace model it is the scale parameter, with population variance $2\sigma^2$. Lower bars give shorter leading confidence intervals at the same confidence level.}
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