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Denote

$$\pmb x_i\sim\mbox{Ell}(\pmb 0,\pmb\varSigma)|i=1,\ldots,n$$

with $\pmb\varSigma$ symmetric positive definite and $\mbox{Ell}(\pmb 0,\pmb\varSigma)$ denotes a $p$ variate elliptical distribution. Denote $\pmb X_n=\{\pmb x_1,\ldots,\pmb x_n\}$ and $\pmb A(\pmb X_n)$ an estimator of scatter computed on $\pmb X_n$.

In page 217 of Robust Statistics: Theory and Methods the authors prove that affine equivariance of $\pmb A(\pmb X_n)\implies$ consistency of $\pmb A(\pmb X_n)$ for $\pmb\varSigma$ in the sense that:

$$\pmb A(\pmb X_\infty)=c\pmb\varSigma$$

Now, I wonder if this $\implies$ is not an $\iff$: naively suppose that I have an estimator $\pmb B(\pmb X_n)$ that is not affine equivariant so that $\pmb B(\pmb C\pmb X_n)\neq \pmb C\pmb B(\pmb X_n)\pmb C'$, doesn't this imply that either $\pmb B(\pmb X_n)$ must be inconsistent for $\pmb\varSigma$ or that $\pmb B(\pmb C\pmb X_n)$ is inconsistent for $\pmb C\pmb \varSigma\pmb C'$?

user603
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2 Answers2

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For a counterexample consider a one-dimensional sequence, $x_i\sim \mathcal N(0, \sigma^2)$, and let $S$ be the sample variance

$$S(x_1, \ldots, x_n) = \frac{1}{n}\sum_{i=1}x_i^2 - \left(\frac{1}{n}\sum_{i=1}^n x_i\right)^2.$$

Set $B=S$ except let $B=0$ whenever $|x_1+x_2+\cdots+x_n| \lt 1$. Although $S$ is equivariant, $B$ obviously is not, because rescaling the $x_i$ to be sufficiently small will turn an almost surely positive value of $S$ into a zero value. Nevertheless, almost surely the limit of $B(x_1, \ldots, x_n)$ will equal $\sigma^2$ because the chance that $B=0$ is bounded above by $\frac{1}{n}\sqrt{2/\pi}\to 0$.

whuber
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    I believe $B$ is consistent for *any* $\sigma^2$. The basis for that is (1) $S$ is consistent (because it is the MLE for $\sigma^2$) and (2) in the limit $B=S$ almost surely. – whuber Oct 14 '14 at 19:52
  • Can you please explain what an affine equivariant is? – Avv Mar 12 '21 at 21:32
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    @Avra It is explained at https://stats.stackexchange.com/questions/243133. – whuber Mar 12 '21 at 22:14
  • Thank you. I asked similar question here, so I am not sure if it's okay for you to read it:https://math.stackexchange.com/questions/4059549/covariance-matrix-in-pca – Avv Mar 12 '21 at 22:31
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Let $B(X_n)$ be an affine equivariant estimator. Let $A(X_n) = B(X_n) k(B(X_n))$ where $k$ is a scalar function such as trace. Then $A$ is consistent but not affine equivariant. This is called a weak covariance functional.

Tom Minka
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