Influence and Breakdown

Each summand in $P_n(A)$ is nonnegative, $P_n(\mathrm{\varnothing })=0$, and $P_n(X)=n^{-1}{\sum }_{i=1}^{n}1=1$ by [F1].

Let $(A_j{)}_{j\ge 1}$ be pairwise disjoint. For each $j$, let $I_j:=\{i\in \{1,\dots ,n\}:x_i\in A_j\}$. The sets $I_j$ are pairwise disjoint, and at most $n$ of them are nonempty, because each index $i$ belongs to at most one $I_j$. Their union is $\{i:x_i\in {\bigcup }_j{A}_j\}$. Hence the number of sample indices in ${\bigcup }_j{A}_j$ is ${\sum }_j\mathrm{\#}{I}_j$, and dividing by $n$ gives $P_n({\bigcup }_j{A}_j)={\sum }_j{P}_n(A_j)$.

Steps 1.1–1.2 show that $P_n$ is a nonnegative countably additive set function with total mass one; by [F2] it is a probability measure.

If $t<x_{(r+1)}$, at most $r$ sample coordinates are at most $t$. Therefore $P_n((-\mathrm{\infty },t])\le r\mathrm{/}(2r+1)<1\mathrm{/}2$, so no such $t$ belongs to the defining set in [F2].

At $t=x_{(r+1)}$, at least $r+1$ sample coordinates are at most $t$. Hence $P_n((-\mathrm{\infty },t])\ge (r+1)\mathrm{/}(2r+1)>1\mathrm{/}2$, so $x_{(r+1)}$ belongs to that defining set.

Steps 1.1–1.2 show that the defining set has no point below $x_{(r+1)}$ and contains $x_{(r+1)}$ itself. Its infimum is therefore $x_{(r+1)}$, as claimed.

With no replacements the mean displacement is zero. Replacing coordinate $i$ by $M$ changes the sample mean to $\bar{x}+(M-x_i)\mathrm{/}n$, whose displacement is unbounded as $M\to +\mathrm{\infty }$. Thus the smallest breaking count for the mean is one, and [F1] gives $1\mathrm{/}n$.

Let $n=2r+1$. If at most $r$ coordinates are replaced, at least $r+1$ original observations remain. Writing $x_{\min }={\min }_i{x}_i$ and $x_{\max }={\max }_i{x}_i$, at least $r+1$ resulting observations are at most $x_{\max }$ and at least $r+1$ are at least $x_{\min }$. Therefore the $(r+1)$-st order statistic lies in $[x_{\min },x_{\max }]$, so its displacement from the original median is bounded uniformly over all such replacements.

Replace $r+1$ coordinates by the same value $M>x_{\max }$. The $r$ unchanged observations are the first $r$ order statistics and the $r+1$ copies of $M$ occupy positions $r+1$ through $2r+1$. The median is therefore $M$, which is unbounded as $M\to +\mathrm{\infty }$.

Steps 1.2–1.3 show that the smallest breaking count for the odd-sample median is $r+1$. By [F1] its breakdown fraction is $(r+1)\mathrm{/}(2r+1)=\lfloor (n+1)\mathrm{/}2\rfloor \mathrm{/}n$.

For $t\in I$, the mean value theorem [F3] gives $\mathrm{\mid }\psi (y-t)-\psi (y-\theta )\mathrm{\mid }\le \mathrm{\mid }t-\theta \mathrm{\mid }g(y)$ for $F$-almost every $y$, so $J(t)$ is finite. For any $t\in I$ and nonzero $h$ with $t+h\in I$, the difference quotient of $\psi (y-t)$ in $t$ converges pointwise to $-{\psi }^{\mathrm{\prime }}(y-t)$ and is bounded in absolute value by $g(y)$, again by [F3]. Dominated convergence [F2] therefore gives $J^{\mathrm{\prime }}(t)=-\int {\psi }^{\mathrm{\prime }}(y-t)dF(y)$. If $t_j\to t$ in $I$, continuity of ${\psi }^{\mathrm{\prime }}$ gives pointwise convergence of ${\psi }^{\mathrm{\prime }}(y-t_j)$ to ${\psi }^{\mathrm{\prime }}(y-t)$, and the same majorant with [F2] shows $J^{\mathrm{\prime }}(t_j)\to J^{\mathrm{\prime }}(t)$. Thus $J$ is $C^1$ on $I$.

Define $H_x(e,t):=(1-e)J(t)+e\psi (x-t)$ for real $e$ near $0$ and $t\in I$. By step 1.1 and $\psi \in C^1$, $H_x$ is $C^1$; also $H_x(0,\theta )=0$ and ${\mathrm{\partial }}_t{H}_x(0,\theta )=J^{\mathrm{\prime }}(\theta )=-A\mathrm{\ne }0$. Applying [F4] gives a unique local $C^1$ root $t={\theta }_x(e)$ through $(0,\theta )$. For $e\ge 0$ small, [F1] identifies its equation with the stated M-location equation under $F_{e,x}$.

At $(0,\theta )$, ${\mathrm{\partial }}_e{H}_x=\psi (x-\theta )-J(\theta )=\psi (x-\theta )$ and ${\mathrm{\partial }}_t{H}_x=-A$. The derivative formula in [F4] therefore yields ${\theta }_x^{\mathrm{\prime }}(0)=\psi (x-\theta )\mathrm{/}A$. Since a two-sided derivative implies the same right derivative, [F5] identifies this value with $\mathrm{IF}(x;T,F)$.

If $\psi (u)=u$, then $J(t)=\int ydF(y)-t=\mu -t$, so its unique root is $\theta =\mu$ and $A=\int 1dF=1$. Substitution in step 3.1 gives $\mathrm{IF}(x;T,F)=x-\mu$.

Since $I$ is open, $f$ is continuous at $m$, and $f(m)>0$, choose $\delta >0$ and $c>0$ so that $[m-\delta ,m+\delta ]\subset I$ and $f(u)\ge c$ throughout this closed interval. For $s<t$ in it, the density identity gives $F(t)-F(s)={\int }_s^{t}f(u)du\ge c(t-s)>0$. Thus $F$ is continuous and strictly increasing there, with $F(m)=1\mathrm{/}2$.

Suppose first that $x<m$, and shrink $\delta$ so that $m-\delta >x$. For sufficiently small $ϵ>0$, the number $r_{ϵ}=(1\mathrm{/}2-ϵ)\mathrm{/}(1-ϵ)$ lies strictly between $F(m-\delta )$ and $1\mathrm{/}2$. Continuity and strict increase give a unique $q_{ϵ}\in (m-\delta ,m)$ with $F(q_{ϵ})=r_{ϵ}$. Since $r_{ϵ}\uparrow 1\mathrm{/}2$ as $ϵ\downarrow 0$, for every $0<\eta <\delta$ eventually $r_{ϵ}>F(m-\eta )$, so $m-\eta <q_{ϵ}<m$ and $q_{ϵ}\to m$. Since $q_{ϵ}>x$, the mixture cdf at $q_{ϵ}$ is $(1-ϵ)F(q_{ϵ})+ϵ=1\mathrm{/}2$. For $t<x$, the contamination term is zero and, by monotonicity and $F(m)=1\mathrm{/}2$, the mixture cdf is $(1-ϵ)F(t)\le (1-ϵ)\mathrm{/}2<1\mathrm{/}2$. At $t=x$, monotonicity gives $F(x)\le F(m-\delta )<r_{ϵ}$, so $(1-ϵ)F(x)+ϵ<(1-ϵ)r_{ϵ}+ϵ=1\mathrm{/}2$. For $x<t<q_{ϵ}$, the contamination term is $ϵ$ and $F(t)<F(q_{ϵ})=r_{ϵ}$: below the local interval this follows from monotonicity and $F(t)\le F(m-\delta )<F(q_{ϵ})$, and inside it from strict increase. Therefore the mixture cdf at such $t$ is $(1-ϵ)F(t)+ϵ<(1-ϵ)r_{ϵ}+ϵ=1\mathrm{/}2$. Together with the cases $t<x$ and $t=x$, this puts the mixture cdf below $1\mathrm{/}2$ for every $t<q_{ϵ}$, while it equals $1\mathrm{/}2$ at $q_{ϵ}$, so [F2] gives $m(F_{ϵ,x})=q_{ϵ}$.

If $x>m$, shrink $\delta$ so that $m+\delta <x$. For small $ϵ>0$, $s_{ϵ}=1\mathrm{/}(2(1-ϵ))$ lies strictly between $1\mathrm{/}2$ and $F(m+\delta )$. There is a unique $q_{ϵ}\in (m,m+\delta )$ with $F(q_{ϵ})=s_{ϵ}$. Since $s_{ϵ}\downarrow 1\mathrm{/}2$ as $ϵ\downarrow 0$, for every $0<\eta <\delta$ eventually $s_{ϵ}<F(m+\eta )$, so $m<q_{ϵ}<m+\eta$ and $q_{ϵ}\to m$. Now $q_{ϵ}<x$, so the mixture cdf at $q_{ϵ}$ is $(1-ϵ)F(q_{ϵ})=1\mathrm{/}2$. For every $t<q_{ϵ}$, the contamination term is zero; if $m\le t<q_{ϵ}$, strict increase on the local interval gives $F(t)<F(q_{ϵ})$, and if $t<m$ then monotonicity gives $F(t)\le F(m)<F(q_{ϵ})$. Thus the mixture cdf is below $1\mathrm{/}2$ for every $t<q_{ϵ}$, while it equals $1\mathrm{/}2$ at $q_{ϵ}$, so [F2] gives $m(F_{ϵ,x})=q_{ϵ}$.

In the case $x<m$, step 2.1 gives $F(q_{ϵ})-F(m)=-ϵ\mathrm{/}[2(1-ϵ)]$ and $q_{ϵ}\to m$. The density identity and continuity of $f$ imply $[F(q_{ϵ})-F(m)]\mathrm{/}(q_{ϵ}-m)\to f(m)$. Dividing these two relations yields $(q_{ϵ}-m)\mathrm{/}ϵ\to -1\mathrm{/}(2f(m))$.

In the case $x>m$, step 2.2 gives $F(q_{ϵ})-F(m)=ϵ\mathrm{/}[2(1-ϵ)]$ and $q_{ϵ}\to m$. The same density quotient tends to $f(m)$, hence $(q_{ϵ}-m)\mathrm{/}ϵ\to +1\mathrm{/}(2f(m))$.

If $x=m$, continuity of $F$ at $m$ gives $F_{ϵ,m}(m-)=(1-ϵ)\mathrm{/}2<1\mathrm{/}2$ and $F_{ϵ,m}(m)=(1+ϵ)\mathrm{/}2>1\mathrm{/}2$. Monotonicity then puts every $t<m$ below the median threshold, while $m$ itself is above it, so the lower median remains exactly $m$ and its right derivative is $0$. By [F3] and steps 3.1–3.2, these three cases give the displayed pointwise influence function.

The absolute influence value is $1\mathrm{/}(2f(m))$ for every $x\mathrm{\ne }m$ and zero at $m$. Both sides of $m$ are nonempty in $\mathbb{R}$, so the supremum over all directions is $1\mathrm{/}(2f(m))$ by [F4]. Since $F$ has a density near $m$, $F\{m\}=0$; therefore the conventional expression $(1\mathrm{/}2-\mathbf{1}\{x\le m\})\mathrm{/}f(m)$ agrees with this pointwise derivative $F$-almost everywhere, though its value at $x=m$ is not the literal right derivative.