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Euclidean Embedding of Co-occurr ence Data under Section 2, the problem formulation is given. In the paper link, the Authors find the distribution for lower embedding for the squared distances between 2 points : $d^2_{x,y} = ||\psi(x) - \xi(y)||^2$

I am having difficulty in understanding why the probability is considered exponential. In this answer What is the PDF when X and Y are non standard distribution is different. In my case, I need to find the pdf of the square root of the distances of the error points X,Y in higher embedding and they can be correlated. How do I begin? I am new to statistics and have limited knowledge in joint distribution and having a tough time in understanding these complicated joint distribution. Please help.

Srishti M
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1 Answers1

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It seems that they choose the exponential distribution. That is, it's a modeling decision, rather than the consequence of some rule or property. However, they state in section 2 that "this reflects the intuition that closer objects should co-occur more frequently than distant objects"

Recall that the exponential distribution has very high density at low values, and the density decays exponentially. It's just a smooth and mathematically convenient way to model "small [distance] is more likely than large" which is what the authors are asserting when objects co-occur.

The rest of the section is dedicated to working around the issue that this implies: if x is unlikely, then x and y together must also be unlikely. But then x and y must also be far apart, which is not the intended model.

As far as the square root the distance goes: if distance has an exponential distribution, root distance has a Rayleigh distribution.

shadowtalker
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  • The point of section 2 is that there's more than one way to define the joint distribution. The last paragraph is a brief comparison which is a summary of the rest of the paper. Unfortunately this isn't something I know anything about -- I got my answer right out of the paper -- so I wouldn't know how to help you further. – shadowtalker Oct 26 '14 at 13:01
  • If you could go into more detail in the question about your application, maybe I could help – shadowtalker Oct 26 '14 at 13:03