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

Randomly partition the observations into $k$ groups of size $m=n/k$, form the group means ${\overline{X}}_1,\dots ,{\overline{X}}_k$, and define the median-of-means estimator byChebyshev givesA binomial tail bound shows that at least half the groups are good with probability at least $1-e^{-k/8}\ge 1-\delta$. Thereforewith probability at least $1-\delta$.

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The analogous statement is false under only a finite-variance assumption. Let $X=0$ with probability $3/4$ and $X=4$ with probability $1/4$. Then $\mu =1$, while the unique population median is zero. With fixed $k$ and group size $n/k\to \infty$, every group median converges in probability to zero, so their mean also converges to zero. Its error from $\mu$ tends to one rather than having order $n^{-1/2}$. Medians inside groups estimate the population median, while means inside groups preserve the population mean.

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Write $k=2r+1$ and $m=n/k$. The central limit theorem gives jointlywhere the $Z_j$ are independent standard normal variables. HenceIts limiting cumulative distribution function is

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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 varianceThis suggests the conjectureprovided a sufficiently uniform central and local limit approximation controls the triangular array.

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