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 ratio should depend on sample size, n. Think this is sort of the idea, but it doesn't right (two limits should converge) ...

    \( .33\frac{(\sigma_h + \sigma_v)}{\sqrt{n}} \leq (\sigma_h / \sigma_v) \leq 0.75\frac{(\sigma_h + \sigma_v)}{\sqrt{n}} \)

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.