Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 223 (Robust Statistics)

Let $m=F^{-1}(1/2)$. Since $F$ has an everywhere positive density, it is continuous and strictly increasing, so $F(m)=1/2$. Membership in the Kolmogorov neighborhood of a distribution givesBy the symmetry of the standard normal distribution,The stated asymptotic-bias formula for the sample median therefore yields

WriteIf $X_{+}\sim F_{+}$, then $X_{+}-2b_1\sim F_{-}$ because $F_{-}(t)=F_{+}(t+2b_1)$. The translation-invariant estimator property givesFor every real $u$,Taking $u={\mu }_{+}$ shows that every admissible estimator has maximum asymptotic bias at least $b_1$ on the pair $\{F_{+},F_{-}\}$. ThereforeTogether with part a, this proves the minimax asymptotic bias optimality of the median.

DefineThus $F_{+}$ is the equal mixture of $N(0,1)$ and $N(2b_1,1)$, while $F_{-}$ is the equal mixture of $N(0,1)$ and $N(-2b_1,1)$. They have finite variance and everywhere positive densities. Normal symmetry shows that $F_{+}$ is symmetric about $b_1$ and $F_{-}$ about $-b_1$.

For every $t$,where the maximum occurs at $t=b_1$. ConsequentlyThe same argument, shifted and reflected, applies to $F_{-}$. Hence both distributions belong to ${\mathcal{P}}_{\epsilon }^K(\Phi )\cap \mathcal{M}$ and satisfy the required translation relation.

Put $q_s=F^{-1}(s)$ and $k=-F^{-1}(\alpha )>0$. Symmetry gives $F^{-1}(1-\alpha )=k$. Differentiating the trimmed mean functional along $F_t=(1-t)F+t{\Delta }_x$ givesUnder the substitution $y=q_s$, symmetry implieswhileEvaluating the last integral in the three regions $x<-k$, $|x|\le k$, and $x>k$ yieldsThis is the influence function of a trimmed mean, namely the Huber score with clipping parameter $k$, divided by $1-2\alpha$.

Let $m=\lfloor \alpha n\rfloor$. If at most $m$ observations are replaced arbitrarily, every order statistic retained between ranks $m+1$ and $n-m$ remains between the minimum and maximum of the unreplaced observations. The trimmed mean is therefore bounded as the replacement values diverge.

If more than $m$ observations are replaced by a common value tending to $+\infty$, at least one replacement remains after the largest $m$ observations are trimmed, and the trimmed mean tends to $+\infty$. The analogous construction tends to $-\infty$. Thus the largest fraction of arbitrary replacements for which boundedness is guaranteed isThis is the convention for the finite-sample replacement breakdown point used in the question; the alternative convention based on the smallest breaking fraction reports $(m+1)/n$.

Write $y_i=|x_i|$. If $n$ is odd, the sample median $t=T_n$ equals one observation, with equally many $y_i$ below and above it; the corresponding terms $\mathrm{sign}(y_i/t-1)$ cancel and the median term is zero. If $n$ is even, the conventional median lies strictly between the two middle observations when the $y_i$ are distinct, so exactly half the terms are $-1$ and half are $+1$. In either caseHence $T_n$ is a Scale M-estimator.

If $X\sim N(0,{\theta }^2)$, then $Y=|X|$ has a scaled half-normal distribution withPut $c={\Phi }^{-1}(3/4)$. The population median is $t_0=\theta c$, and the asymptotic distribution of a sample median gives

Under $H_0:{\theta }^2=1$, positivity of the scale means $\theta =1$. If $z_{1-\alpha }$ is the $(1-\alpha )$-quantile of the standard normal distribution, an asymptotically level-$\alpha$ test rejects forLarger scale makes the population median $\theta c$ larger, so this is the appropriate one-sided rejection region.

The upper envelope for $\psi$ impliesWriting ${\mu }_i=\mathbb{E}[X_i]$, taking expectations, and using $1+u\le e^u$ givesIndependent random variables then yieldApplying the lower envelope to $-\psi (\theta x)$ similarly gives

The Markov inequality applied to the first exponential-moment bound givesThe second bound controls the lower tail. The union bound therefore givesSinceputting $L=\log (2/\delta )$,makes the exponent equal to $-L$. It follows that the Catoni mean estimator satisfies