Now both the number of groups and their size equal $\sqrt n$. A group mean is approximately $N(\mu,\sigma^2/\sqrt n)$, whose density at $\mu$ is approximately $n^{1/4}/(\sigma\sqrt{2\pi})$. The <asymptotic distribution of a sample median> based on $\sqrt n$ such values therefore has variance
$$
\frac1{4\sqrt n,f(\mu)^2}
\sim\frac{\pi\sigma^2}{2n}.
$$
This suggests the conjecture
$$
\sqrt n(\widehat\mu_{\mathrm{MOM}}-\mu)
\Longrightarrow N\left(0,\frac{\pi\sigma^2}{2}\right),
$$
provided a sufficiently uniform central and local limit approximation controls the triangular array.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 223 / 3 / d / Solution

For an odd sample size $n$ from a distribution with positive continuous density $f$ at its median $m$,
$$
\sqrt n(\widehat m-m)\Longrightarrow N\left(0,\frac1{4f(m)^2}\right).
$$


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 Incoming links: Asymptotic distribution of a sample median