Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 223 (Robust Statistics)

The gross-error sensitivity is ${\gamma }^{\ast }(T,F)={\sup }_x|\mathrm{IF}(x;T,F)|$. At $N(\theta ,1)$, the influence function of the sample median has magnitude $1/(2\phi (0))=\sqrt{\pi /2}$, soFor the Huber location estimator, ${\psi }_k(u)=\max (-k,\min (u,k))$ and $\mathbb{E}{\psi }_k^{\prime }(Z)=\mathbb{P}(|Z|\le k)=2\Phi (k)-1$, givingFor the symmetric normal law, the trimmed population mean is $\theta$. Writing $q_{\gamma }={\Phi }^{-1}(1-\gamma )$ in the supplied influence function gives

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Let replacement tolerance mean the greatest number of observations that can be replaced while the estimator remains bounded. For $n=2m$, a sample with $m$ zeros and $m$ copies of $a$ is within $m$ replacements of both the all-zero sample and its translate by $a$. If an equivariant estimator tolerated $m$ replacements, it would remain within bounded distance of both $T(0^n)$ and $T(0^n)+a$, which is impossible as $a\to \infty$. Thus at most $m-1$ replacements are tolerable.

For $n=2m+1$, a sample with $m$ zeros and $m+1$ copies of $a$ is obtained from the all-zero sample by $m+1$ replacements and from the all-$a$ sample by $m$ replacements. Translation equivariance again forces breakdown by $m+1$ replacements. In both cases the finite-sample replacement breakdown point is at most

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The sample median attains the equivariant upper bound:For the Huber estimator, fewer than half the observations cannot overpower the bounded scores of the uncontaminated majority. Evaluating the estimating equation below $x_{(1)}-k$ or above $x_{(n)}+k$ makes every uncontaminated score have the same sign. A contaminating majority can balance these scores arbitrarily far away, soA $\gamma$-trimmed mean remains bounded while at most $\lfloor \gamma n\rfloor$ arbitrary observations are removed by each tail trim; one more arbitrarily large replacement survives. Hence

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If $v_T=\mathbb{E}[operatornameIF(X;T,N({\theta }_0,1){)}^2]$, asymptotic normality gives an asymptotically level-$\alpha$ test that rejects whenFor the median, $v_T=\pi /2$. For Huber,For the trimmed mean, with $q={\Phi }^{-1}(1-\gamma )$,These tests have bounded influence functions, so a small contamination proportion has bounded first-order effect on their statistics, asymptotic levels, and powers. Their finite-contamination protection is quantified by the breakdown points in part (c).

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For residual $r=y_i-x_i^{\top }\theta$, minimizeover $\gamma$. Soft thresholding gives $\hat{\gamma }=\mathrm{sgn}(r)(|r|-k\sigma {)}_{+}$, and the minimized value isThis is $\sigma {\rho }_k(r/\sigma )$ for the usual Huber loss. Multiplication by the positive constant $\sigma$ does not change the minimizing $\theta$, proving equivalence.

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For fixed residual $r=y_i-x_i^{\top }\theta$, choosing ${\gamma }_i=0$ costs $r^2$, while choosing ${\gamma }_i=r$ costs $\ell$. No other nonzero choice improves on ${\gamma }_i=r$. Thuswith $L=\sqrt{\ell }$. Summing over observations proves the equivalence.

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Away from the two nondifferentiable cutoffs, the skipped-mean score is ${\psi }_L(u)=2u{1}_{\{|u|\le L\}}$. The estimating equation is therefore

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Choose $v\in {\mathbb{R}}^p$ outside the finitely many hyperplanes $\{v:x_i^{\top }v=0\}$. Then $x_i^{\top }v\ne 0$ for every $i$. For ${\theta }_t=tv$, every residual $y_i-t{x}_i^{\top }v$ eventually has absolute value greater than $L$. Every indicator in the estimating equation is then zero, so the equation is satisfied for every sufficiently large $t$.

Thus arbitrarily large solutions already exist without contamination. Under the definition in the question, the skipped-mean regression estimator has breakdown point zero. This is a standard pathology of an exactly redescending score when every root is admitted as an estimator.

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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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