On page 180 of Robust Statistics: The Approach Based on Influence Functions one finds the following question:

  • 16: Show that for location-invariant estimators always $\varepsilon^*\leq\frac{1}{2}$. Find the corresponding upper bound on the finite-sample breakdown point $\varepsilon^*_n$, both in the case where $n$ is odd or $n$ is even.

The second part (after the period) is actually trivial (given the first) but I can't find a way to prove the first part (sentence) of the question.

In the section of the book pertaining to this question one finds (p98):

Definition 2: The finite-sample breakdown point $\varepsilon^*_n$ of an estimator $T_n$ at the sample $(x_l,\ldots, x_n)$ is given by:

where the sample $(z_1,\ldots,z_n)$ is obtained by replacing $m$ data points $x_{i_1},\ldots,x_{i_m}$ by arbitrary values $y_1,\ldots,y_m.$

The formal definition of $\varepsilon^*$ itself runs for almost a page, but can be thought of as $$\varepsilon^*=\underset{n\rightarrow\infty}{\lim}\varepsilon^*_n$$ Although not defined explicitly, one can guess that location-invariant means that $T_n$ must satisfy $$T_n(x_1,\ldots,x_n)= T_n(x_1+c,\ldots,x_n+c), \text{ for all } c\in \Bbb{R}$$

I (try to) answer whuber's question in the comment below. The book defines estimator $T_n$ is several pages, starting at p82, I try to reproduce the main parts (I think it will answer whuber's question):

Suppose we have one-dimensional observations $(X_1,\ldots,X_n)$ which are independent and identically distributed (i.i.d.). The observations belong to some sample space $\mathcal{H}$, which is a subset of the real line $\mathbb{R}$ (often $\mathcal{H}$ simply equals $\mathbb{R}$ itself, so the observations may take on any value). A parametric model consists of a family of probability distributions $F_\theta$, on the sample space, where the unknown parameter $\theta$ belongs to some parameter space $\Theta$


We identify the sample $(X_1,\ldots,X_n)$ with its empirical distribution $G_n$, ignoring the sequence of the observations (as is almost always done). Formally, $G_n$, is given by $(1/n)\sum_{i=1}^n\Delta_{x_i}$ where $\Delta_{X}$, is the point mass 1 in $X$. As estimators of $\theta$, we consider real-valued statistics $T_n=T_n(X_1,\ldots,X_n)=T_n(G_n)$. In a broader sense, an estimator can be viewed as a sequence of statistics $\{T_n,n\geq 1\}$ , one for each possible sample size $n$. Ideally, the observations are i.i.d. according to a member of the parametric model $\{F_\theta;\theta\in\Theta\}$ , but the class $\mathcal{F}(\mathcal{H})$ of all possible probability distributions on $\mathcal{H}$ is much larger.

We consider estimators which are functionals [i.e., $T_n(G_n)=T(G_n)$ for all $n$ and $G_n$] or can asymptotically be replaced by functionals. This means that we assume that there exists a functional $T:\mbox{domain}(T)\rightarrow\mathbb{R}$ [where the domain of $T$ is the set of all distributions $\mathcal{F}(\mathcal{H})$ for which $T$ is defined] such that $$T_n(X_1,\ldots,X_n)\underset{n\rightarrow\infty}{\rightarrow}T(G)$$ in probability when the observations are i.i.d. according to the true distribution $G$ in $\mbox{domain}(T)$. We say that $T(G)$ is the asymptotic value of $\{T_n;n\geq 1\}$ at $G$.


In this chapter, we always assume that the functionals under study are Fisher consistent (Kallianpur and Rao, 1955): $$T(F_\theta)=\theta\;\mbox{ for all } \theta\in\Theta$$ which means that at the model the estimator $\{T_n;n\geq 1\}$ asymptotically measures the right quantity. The notion of Fisher consistency is more suitable and elegant for functionals than the usual consistency or asymptotic unbiasedness.

  • 1
    $\begingroup$ How exactly does this book define "estimator"? It seems to me that any bounded estimator $T_n$ must have a breakdown point of $1$, so surely it is placing some kind of special restrictions on $T_n$; and there always exist bounded location-invariant estimators (they will include the constants). $\endgroup$
    – whuber
    Jan 9 '15 at 18:52
  • 1
    $\begingroup$ Thank you for the expanded material. It still seems there are plenty of counterexamples. A simple one is the constant estimator $T_n(X_1,\ldots,X_n)=1$ for the one-parameter family of normal distributions of variance $1$. This is a location-invariant estimator of the variance. Its breakdown point is $1$. It is Fisher consistent (trivially), but I need to interpret the definition carefully: "$\theta$" cannot refer necessarily to all the parameters, for then no location-invariant estimator could be consistent! $\endgroup$
    – whuber
    Jan 9 '15 at 20:05
  • $\begingroup$ @whuber: Thanks, I understand your counter-example. I think I will contact the author and ask for more information... $\endgroup$
    – user603
    Jan 9 '15 at 20:19

Older statistics books used "invariant" in a slightly different way than one might expect; the ambiguous terminology persists. A more modern equivalent is "equivariant" (see the references at the end of this post). In the present context it means

$$T_n(X_1+c,X_2+c,\ldots,X_n+c) = T_n(X_1,X_2,\ldots,X_n) + c$$

for all real $c$.

To address the question, then, suppose that $T_n$ has the property that for sufficiently large $n$, all real $c$, and all $m \le \varepsilon^{*}n$,

$$|T_n(\mathbf{X + Y}) - T_n(\mathbf{X})| = o(|c|)$$

whenever $\mathbf Y$ differs from $\mathbf{X}$ by at most $c$ in at most $m$ coordinates.

(This is a weaker condition than assumed in the definition of breakdown bound. In fact, all we really need to assume is that when $n$ is sufficiently large, the expression "$o(|c|)$" is some value guaranteed to be less than $|c|/2$ in size.)

The proof is by contradiction. Assume, accordingly, that this $T_n$ is also equivariant and suppose $\varepsilon^{*} \gt 1/2$. Then for sufficiently large $n$, $m(n) = \lfloor \varepsilon^{*}n\rfloor$ is an integer for which both $m(n)/n \le \varepsilon^{*}$ and $(n-m(n))/n \le \varepsilon^{*}$. For any real numbers $a,b$ define

$$t_n(a, b) = T_n(a, a, \ldots, a,\ b, b, \ldots, b)$$

where there are $m(n)$ $a$'s and $n-m(n)$ $b$'s. By changing $m(n)$ or fewer of the coordinates we conclude both

$$|t(a,b) - t(0,b)| = o(|a|)$$


$$|t(a,b) - t(a,0)| = o(|b|).$$

For $c\gt 0$ the triangle inequality asserts

$$\eqalign{ c = |t_n(c, c) - t_n(0, 0)| &\le |t_n(c, c) - t_n(c, 0)| + |t_n(c, 0) - t_n(0,0)| \\&= o(c) + o(c) \\&\lt c/2 + c/2 \\ &= c}$$

The strict inequality on the penultimate line is assured for sufficiently large $n$. The contradiction it implies, $c \lt c$, proves $\varepsilon^{*} \le 1/2.$


E. L. Lehmann, Theory of Point Estimation. John Wiley 1983.

In the text (chapter 3, section 1) and an accompanying footnote Lehmann writes

An estimator satisfying $\delta(X_1+a, \ldots, X_n+a) = \delta(X_1,\ldots,X_n)+a$ for all $a$ will be called equivariant ...

Some authors call such estimators "invariant." Since this suggests that the estimator remains unchanged under $X_i^\prime = X_i+a$, it seems preferable to reserve that term for functions satisfying $u(x+a)=u(x)$ for all $x,a$.

  • 1
    $\begingroup$ yes I have contacted the main author of the book yesterday with the same question about the actual definition of invariance used (I looked in the index and I could not find it explicit in the book). I upvoted because I think your answer is the correct one, but will give the author a couple of days to be sure before accepting it. $\endgroup$
    – user603
    Jan 10 '15 at 16:07
  • 1
    $\begingroup$ I didn't receive an answer from the author but the arguments presented above (in the answer and the comment) convinced me that this must indeed be the correct interpretation of the problem. $\endgroup$
    – user603
    Jan 15 '15 at 23:04

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