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Does anyone know what bullseye distribution is?

It should be a special case of Weibull distribution, but I haven't found any useful information after googled it.

Just for information, I heard this term from a german.

So it could be some uncommon translation from german to english.

And moreover, it represents a result due to a combination of independent impacts.

Thanks a lot. I would say this is a proper math question though no formulae are involved.

UPDATE: Thanks Ross Millikan for the answer!

I have some following questions.

Does Rayleigh distribtuion only represent a 2D Gaussian distribution?

For completeness, this statement can be made as following:

$X \sim N(\mu, \sigma)$ and $Y \sim N(\mu, \sigma)$

then $\sqrt{X^{2}+Y^{2}} \sim Rayleigh(\sigma)$

So, is there a distribution which can be explicitly written for multidimensional Gaussian distribtuion,lets say N-dim?

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up vote 2 down vote accepted

One guess would be the Rayleigh distribution, a two dimensional Gaussian. Because there is zero area at the origin, the distribution in radius has a maximum.

Added: yes, you can do N dimensional Gaussian distributions. If you look at the Rayleigh distribution, it is just a joint Gaussian in $x$ and $y$. The distribution in $r=\sqrt{x^2+y^2}$ comes from integrating the area of an annulus from $r$ to $r+dr$. You can do the same in more dimensions. multiplying Gaussians in $x, y, z, \ldots$. The factor $r$ will become $r^{N-1}$ and the constant will change.

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I agree with this guess. – Michael Lugo Jul 18 '11 at 15:33
hmm...that's my initial guess as well. Thanks for the confirmation. – newbie Jul 18 '11 at 15:36
As I've learned, for higher dimensions (N>2), should it be $\chi^{2}$ with n degree of freedom distributed rather than Rayleigh? – newbie Jul 18 '11 at 16:11
@newbie: yes, the Rayleigh is the special case of $N=2$ – Ross Millikan Jul 18 '11 at 17:29
@RossMillikan hmm, the ways to work things out are quite different between in academy and industry. they commonly don't ask why in industry. :S I'm gonna accept your answer. Thank you. – newbie Jul 19 '11 at 7:27

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