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Suppose a configuration of points is given in $X\in\mathbb{R}^{n\times 2}$. Given that the configuration has zero column means, the variance of each axis, $x$ and $y$, can be expressed as

$\mathrm{var}(x)=\frac{1}{n}x^Ty\qquad\text{and}\qquad\mathrm{var}(y)=\frac{1}{n}y^Ty.$

If $\mathrm{var}(x)>\mathrm{var}(y)$, does that means that the configuration has larger spread along $x$ axis, i.e. the configuration is stretched along $x$?

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    yes, I meant the range. So, it might not imply the above, but it only relates to dispersion (meaning that dispersion in $x$ is larger than dispersion in $y$).2012-07-26

1 Answers 1

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Discussion in the comments has revealed that the intended question is whether $\operatorname{var}(x) > \operatorname{var}(y)$ implies that $\operatorname{range}(x) > \operatorname{range}(y)$. The answer is no; for an extreme counterexample, let $\begin{align} x &= [\underbrace{-1, -1, \ldots, -1}_{\text{$n/2$ elements}}, \underbrace{1, 1, \ldots, 1}_{\text{$n/2$ elements}}],\\ y &= [-k, \underbrace{0, 0, \ldots, 0}_{\text{$n-2$ elements}}, k], \end{align}$

Then $x$ has variance $1$ and range $2$, while $y$ has variance $2k^2/n$ and range $2k$. Pick any $k$ between $1$ and $\sqrt{n/2}$, and you get $\operatorname{var}(x) > \operatorname{var}(y)$ but $\operatorname{range}(x) < \operatorname{range}(y)$. The point is that the variance averages over all the elements, while the range only looks at the two most extreme ones, so they can disagree when the two extreme points are not representative of the rest.

(In the counterexample in my comment, I had switched $x$ and $y$ by mistake.)