1
$\begingroup$

If z is a centered gaussian random variable and $ x_1 ,x_2 ,..,x_n ,y_1,y_2,..,y_n $ are points in $ \mathbb{R}^{2n} $ satisfying $ |x_i-x_j |_2 \leq |y_i -y_j |_2 \ \ \ \forall i,j \in [n] $ then $ E \ max_{i=1} ^n <x_i, z> \ \leq \ E \ max_{i=1} ^n <y_i, z> . $

Apparently this is not true for other distributions even for centered ones. The proofs I've found are very long and confusing, does anyone have any intuitive explanation for why the gaussian distribution works? Is it purely coincidence that the gaussian distribution also shows up in the central limit theorem?

$\endgroup$
1
$\begingroup$

A good intuition would be to consider the shape of the level sets of the likelihood. That is: the level sets of the density of a Gaussian correspond exactly to the level sets of the $\ell_2$ norm. The $\ell_2$ norm is induced by an inner product, so generally all your linear algebra and geometry intuitions usually transfer pretty well into the Gaussian case. For example, maximum likelihood estimates can reduce to least squares, since the log-likelihood is the $\ell_2$ norm.

Other centered distributions might be transformed in very weird ways when mapped through an inner product, but jointly Gaussian random variables behave nicely through linear transformations.

Hope that helps!

$\endgroup$
  • $\begingroup$ Can you clarify what it means to be mapped through an inner product? You comment seems to suggest that any rotationally invariant decreasing distributions also satisfies Slepian is this true? $\endgroup$ – Hao Sun Jan 11 '15 at 15:19
  • 1
    $\begingroup$ So, you have a random variable $z$, and you're mapping it under the linear functional $z \mapsto \langle x_i, z \rangle$. That's what I was referring to. It's not really just that it's rotationally invariant; it's also the relationship between the likelihood and Euclidean distance. This is why a lot of jointly Gaussian calculations coincide with subspace projections. $\endgroup$ – Roy D. Jan 11 '15 at 20:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.