Contamination and Robust Risk — Examples

By [F1] and [F2], the path law places weight $1-ϵ$ at $0$ and weight $ϵ$ at $a$; this also covers $a=0$, when both atoms coincide.

The identity function is measurable for the Borel sigma-algebra because each Borel set is its own preimage under the identity. Let $S=\{0,a\}$ with its full power-set sigma-algebra and weights $w(b)=F_{ϵ,H}(\{b\})$. Since the complement of $S$ has path-law measure zero, the integral of the identity is the integral of its restriction to this finite probability space: both are the same finite sum over the distinct atoms (apply the simple-function integral separately to the positive and negative parts). Thus [F3] gives $T(F_{ϵ,H})=(1-ϵ)0+ϵa=ϵa$, while $T(F)=0$.

For $ϵ>0$, the difference quotient is $(T(F_{ϵ,H})-T(F))\mathrm{/}ϵ=a$, so the right derivative at zero is $a$. Since $a$ ranges over all real numbers, the supremum of its absolute value is $+\mathrm{\infty }$.

Let $I=\{i:x_i\mathrm{\ne }{y}_i\}$, with $\mathrm{\mid }I\mathrm{\mid }\le k$. Since $x_i,y_i\in [0,1]$, each changed coordinate satisfies $\mathrm{\mid }{y}_i-x_i\mathrm{\mid }\le 1$, and $\mid N^{-1}{\sum }_i(y_i-x_i)\mid =\mid N^{-1}{\sum }_{i\in I}(y_i-x_i)\mid \le \mathrm{\mid }I\mathrm{\mid }\mathrm{/}N\le k\mathrm{/}N$.

Use a one-point probability space for each coupling. By [F3], at parameter $0$ the clean iid sample is $(0,0)$; set the observed vector to $(0,1)$. At parameter $1$ the clean sample is $(1,1)$; set the observed vector to $(0,1)$. Each coupling changes exactly one of two coordinates, which is the allowed $\lfloor (1\mathrm{/}2)2\rfloor =1$ replacement, so both observed laws are the same point mass.

A rule that sees only the observed vector receives the same value $(0,1)$ under both parameter laws, so it cannot identify both parameters. The observed vector is not iid: its first coordinate is $0$ almost surely and its second is $1$ almost surely, whereas [F2] requires identical coordinate laws. Thus this replacement mechanism is distinct from iid Huber contamination.

The weights in $P_{\theta }=(1-\theta ){\delta }_0+\theta {\delta }_1$ are nonnegative and sum to one. By [F4] and [F5], each $P_{\theta }$ is a probability law. On the full power set of $\{0,1\}$ the four events are $\mathrm{\varnothing },\{0\},\{1\}$, and $\{0,1\}$.

For the four events in step 1.1, the probability differences are either zero or $\pm ({\theta }_0-{\theta }_1)$. Thus [F3] gives $TV(P_{{\theta }_0},P_{{\theta }_1})=\mathrm{\mid }{\theta }_0-{\theta }_1\mathrm{\mid }$. Under absolute loss, the allowed parameter separation is at most $\min \{ϵ\mathrm{/}(1-ϵ),1\}$ when $ϵ<1$. If the fraction is at most $1$, the pair $0,ϵ\mathrm{/}(1-ϵ)$ attains it; if it is at least $1$, the endpoints attain separation $1$. The endpoint definition gives $\omega (1,\mathrm{\Theta })=1$.

For $0<ϵ\le 1\mathrm{/}2$, both chosen parameters lie in $[0,1]$. Under $({\theta }_0,Q_0)$ the success probability is $ϵ$; under $({\theta }_1,Q_1)$ it is $(1-ϵ)ϵ\mathrm{/}(1-ϵ)=ϵ$. Hence both mixtures are Bernoulli$(ϵ)$.

For $1\mathrm{/}2\le ϵ\le 1$, take ${\theta }_0=0,{\theta }_1=1$, set $q_0=1\mathrm{/}(2ϵ)$ and $q_1=1-1\mathrm{/}(2ϵ)$, and define $Q_j=(1-q_j){\delta }_0+q_j{\delta }_1$. The $q_j$ lie in $[0,1]$, so [F4] and [F5] make these probability laws. The mixture success probabilities are $ϵq_0=1\mathrm{/}2$ and $(1-ϵ)+ϵq_1=1\mathrm{/}2$. At $ϵ=1$ the clean part vanishes and both mixtures still equal Bernoulli$(1\mathrm{/}2)$.