Influence and Breakdown — Examples

By [F1], the mean influence function exists for every $x\in \mathbb{R}$ and equals $x-\mu$.

The sample-mean replacement breakdown fraction is $1\mathrm{/}n$ by [F3], independently of the distribution $F$.

The values $\mathrm{\mid }x-\mu \mathrm{\mid }$ are unbounded as $x$ ranges over $\mathbb{R}$; hence [F2] gives ${\gamma }^{\ast }(T,F)=+\mathrm{\infty }$.

Thus in this example the population gross-error sensitivity is infinite, while one arbitrary coordinate replacement already makes the finite-sample mean unbounded.

By [F1], the influence values have absolute value at most $1\mathrm{/}(2f(m))$, and their supremum is exactly that value, which is finite because $f(m)>0$.

By [F2], the finite-sample replacement breakdown fraction is $(r+1)\mathrm{/}(2r+1)=1\mathrm{/}2+1\mathrm{/}[2(2r+1)]$.

Since $1\mathrm{/}[2(2r+1)]\to 0$ as $r\to \mathrm{\infty }$, the odd-sample median breakdown fractions tend to $1\mathrm{/}2$.

The bounded pointwise sensitivity and the separate replacement count follow from different definitions and theorems; here the median has both properties under the stated density hypothesis.

By [F14], every Borel set is Lebesgue measurable; since [F5] makes $[-1,1]$ Borel and Lebesgue measurable with measure $2$, each $A\cap [-1,1]$ is measurable for Borel $A$. For pairwise disjoint Borel sets $A_j$, these intersections are disjoint, so countable additivity of $\lambda$ gives countable additivity of $P(A)=\lambda (A\cap [-1,1])\mathrm{/}2$. Also $P(\mathrm{\varnothing })=0$ because $\lambda (\mathrm{\varnothing })=0$, and $P(\mathbb{R})=1$ by [F5]. Thus [F1] and [F2] make $P$ a Borel probability law. For $-1\le t\le 1$, [F5] gives $F_P(t)=P((-\mathrm{\infty },t])=(t+1)\mathrm{/}2$; for $t<-1$ the intersection $(-\mathrm{\infty },t]\cap [-1,1]$ is empty. It follows that $m(P)=0$, the density near zero is $f=1\mathrm{/}2$, and $P$ has no atoms, the latter by the degenerate-interval case of [F5].

Fix any point $z\in \mathbb{R}$. By [F9], $P_{ϵ,z}$ is carried by a bounded interval containing $[-1,1]$ and $z$, and its only possible atom location is $z$, because [F5] gives $P(\{z\})=0$. The trigger in $T$ requires two distinct atom locations, so it is false and $T(P_{ϵ,z})=m(P_{ϵ,z})$ for all sufficiently small $ϵ\ge 0$. Therefore [F8] and [F12] give $\mathrm{IF}(z;T,P)=\mathrm{IF}(z;m,P)$, equal to $-1$ for $z<0$, $+1$ for $z>0$, and $0$ for $z=0$.

Fix odd $n=2r+1\ge 5$ and a clean sample $x\in [-1,1{]}^n$. Its empirical median is the middle order statistic by [F13] and lies in $[-1,1]$, so every atom location of $P_n$ is within distance $2$ of that median. The trigger is false and $T_n(x)$ is the sample median.

After at most one replacement, at least $n-1=2r$ observations remain in $[-1,1]$. At least $r+1$ resulting observations are at most $1$ and at least $r+1$ are at least $-1$, so the middle order statistic, hence the lower empirical median by [F13], remains in $[-1,1]$. All unchanged atom locations are within distance $2$ of it, and only the single replacement can add a distinct outlying atom. Thus the trigger is still false and $T_n$ remains the sample median in $[-1,1]$ for every one-coordinate replacement; its maximum displacement is bounded.

The absolute values in step 1.2 have supremum $1$ over $z\in \mathbb{R}$, so [F10] gives ${\gamma }^{\ast }(T,P)=1<\mathrm{\infty }$.

Replace two coordinates by distinct values $M$ and $2M$ with $M>3$. The remaining $2r-1$ observations lie in $[-1,1]$, so the middle order statistic (position $r+1$) is still one of them and lies in $[-1,1]$; by [F13] this is the lower empirical median. Both new atom locations are more than $2$ above that median, so the trigger is true and $T_n=2M$. Letting $M\to +\mathrm{\infty }$ makes the maximum displacement unbounded. Since step 1.4 ruled out one replacement and [F11] defines the first unbounded count, the breakdown fraction is exactly $2\mathrm{/}n$.

Along odd $n\to \mathrm{\infty }$, $2\mathrm{/}n\to 0$, while step 2.1 gives finite gross-error sensitivity at the fixed law $P$. Hence this functional has bounded influence and an empirical plug-in statistic whose replacement breakdown is not bounded away from zero, refuting the Statement.