Talk:Derivation of the Rayleigh Distribution Equation

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rev 13:28, 3 June 2015‎ Herb

Ok, name of page needs a bit of fixing.

more verbosity in text

I goofed in accuracy section. "PDF" functions are set up wrong. Should be something like "PDF of X as x"

Been about 45 years since I took calculus in college. I'll have to look up conversion from Cartesian to Polar coordinates, but I know the conversion will swizzle to the right answer...


This is topic that I think bears further investigation...

is it \(\sigma_h = \sigma_v\) or \(\sigma_h^2 = \sigma_v^2\) ??

Of course if they are equal then both equations are true. The problem is in pooling the values if:

\(\sigma_h \approx \sigma_v\)

which equation do we use to pool the values?

  • \((\sigma_h + \sigma_v)/2\)
  • \(\sqrt{\sigma_h^2 + \sigma_v^2}\)

I think \(\sqrt{\sigma_h^2 + \sigma_v^2}\) should be corrected by factor \(\frac{1}{\sqrt{2}}\) since \((\sigma + \sigma)/2 = \sigma\) but \(\sqrt{\sigma^2 + \sigma^2} = \sigma \sqrt{2}\)

In general it would seem that the ratio \((\sigma_h / \sigma_v)\) could be useful as a guide to stay out of trouble. Obviously the ration should depend on sample size, n. but something like \( .66\sigma \leq (\sigma_h / \sigma_v) \leq 1.5\sigma \) is the idea...


I also really don't like using a \(\sigma\) as a factor in the equation. If you think about radial values then there is a \(\sigma\) which can be calculated from the \(r_i\) values. The two \(\sigma\)'s aren't equal.


The !@#$%^&* literature really messes up calculation of radial standard deviation. needs \(\frac{1}{\sqrt{2}\) correction.