# Solution to exercice 2.2a.16 of "Robust Statistics: The Approach Based on Influence Functions"

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:
$$\varepsilon^*_n(T_n;x_i,\ldots,x_n):=\frac{1}{n}\max\{m:\max_{i_1,\ldots,i_m}\sup_{y_1,\ldots,y_m}\;|T_n(z_1,\ldots,z_n)|<\infty\}$$

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.

• 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).
– whuber
Jan 9 '15 at 18:52
• 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!
– whuber
Jan 9 '15 at 20:05
• @whuber: Thanks, I understand your counter-example. I think I will contact the author and ask for more information... 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

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

and

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

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 $\varepsilon^{*} \le 1/2.$

### References

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 $x,a$.

• 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. Jan 10 '15 at 16:07
• 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. Jan 15 '15 at 23:04