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I'd like to calculate a radial profile of a 2D Gaussian.

2D-Plot

it should be a half of a Gaussian, maximum of about 3000 at $R=0$.

If I plot radial positions $\left(\sqrt{x^2+y^2} \right)$ of every point, i get second distribution:

rad-uncorrelated

Is weighting with $\tfrac{1}{R}$ correct (picture below)? Should it also be weighted with integral area of the initial Gaussian?

rad-cor

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    Migrating to math, because this question explicitly asks for "a mathematical reason" in a situation not especially statistical in nature. (It appears to be asking why the area element in polar coordinates $(r, \theta)$ equals $r dr d\theta$ and not $dr d\theta$.)2011-08-17
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    I am aware of infinitesimal polar area, I am only asking about verification/falsification - radial plot is seemingly of wrong magnitude as well2011-08-17
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    It would help to explain how you are computing these plots. For instance, the middle one clearly does not plot the radial function divided by $R$ (because that would have a positive vertical asymptote at zero). It appears instead to be *integrating* the density along the angular coordinate (and then "weighting" by $1/R$).2011-08-17
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    I'm guessing what you have is a large number of *samples* from a 2D Gaussian, and the figures you have shown are *histograms*, is that correct?2011-08-17
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    yes, histograms are derived from the 2D Gaussian2011-08-17
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    @troyaner: The first figure is also a two-dimensional histogram, yes? (P.S. It would be nice if you included "@Rahul" in your comment replies to me, so that I get notified of them. You always get notified of comments on your own question.)2011-08-17
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    @Rahul Narain: yes, it is a histogram of ~1e7 independently sampled Gaussian pairs. (thx for the @-tipp!)2011-08-17
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    $e^{- {(x^2+y^2)\over 2 \sigma^2}}dx dy = e^{-{r^2 \over 2 \sigma^2}} r dr d \theta$?2011-08-18

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